##### Class 12^{th} Mathematics Part I CBSE Solution

**Exercise 4.1**- | cc 2&4 -5&-1 | Evaluate the determinants:
- | cc costheta &-sintegrate heta sintegrate heta | Evaluate the determinants:…
- | cc x^2 - x+1 x+1+1 | Evaluate the determinants:
- If a = [ll 1&2 4&2] then show that |2A| = 4|A|.
- If a = [lll 1&0&1 0&1&2 0&0&4] then show that |3A| = 27|A|
- Evaluate the determinants | ccc 3&-1&-2 0&1&-1 3&-5&0 |
- | ccc 3&-4&5 1&1&-2 2&3&1 | Evaluate the determinants
- | ccc 0&1&2 -1&0&-3 -2&3&0 | Evaluate the determinants
- | ccc 2&-1&-2 0&2&-1 3&-5&0 | Evaluate the determinants
- If A = [lll 1&1&-2 2&1&-3 5&4&-9] , find |A|.
- Find values of x, if | ll 2&4 5&1 | = | cc 2x&4 6 | | ll 2&4 5&1 | = | cc 2x&4…
- | ll 2&3 4&5 | = | ll x&3 2x&5 | Evaluate the determinants
- If , then x is equal toA. 6 B. 6 C. 6 D. 0

**Exercise 4.2**- | lll x+a y+b x+c | = 0 Using the property of determinants and without expanding…
- | lll a-b b-c c-a | = 0 Using the property of determinants and without expanding…
- | lll 2&7&65 3&8&75 5&9&86 | = 0 Using the property of determinants and without…
- Using the property of determinants and without expanding in prove that:…
- | ccc b+c+r+z c+a+p+x a+b+q+y | = 2 | ccc a b c | Using the property of…
- | ccc 0&-b -a&0&-c b&0 | = 0 Using the property of determinants and without…
- | ccc - a^2 ba& - b^2 ca& - c^2 | = 4a^2b^2c^2 Using the property of…
- | lll 1& a^2 1& b^2 1& c^2 | = (a-b) (b-c) (c-a) By using properties of…
- | lll 1&1&1 a a^3 & b^3 & c^3 | = (a-b) (b-c) (c-a) (a+b+c) By using properties…
- | lll x& x^2 y& y^2 z& z^2 | = (x-y) (y-z) (z-x) (xy+yz+zx) By using properties…
- | ccc x+4&2x&2x 2x+4&2x 2x&2x+4 | = (5x+4) (4-x)^2 By using properties of…
- | ccc y+k y+k y+k | = k^2 (3y+k) By using properties of determinants, show…
- | ccc a-b-c&2a&2a 2b&2b 2c&2c | = (a+b+c)^3 By using properties of…
- | ccc x+y+2z z+z+2x z+x+2y | = 2 (x+y+z)^3 By using properties of…
- | ccc 1& x^2 x^2 &1 x& x^2 &1 | = (1-x^3)^2 By using properties of…
- | ccc 1+a^2 - b^2 &2ab&-2b 2ab& 1-a^2 + b^2 &2a 2b&-2a& 1-a^2 - b^2 | = (1+a^2…
- | ccc a^2 + 1 ab& b^2 + 1 ca& c^2 + 1 | = 1+a^2 + b^2 + c^2 By using properties…
- Let A be a square matrix of order 3 × 3, then | kA| is equal toA. k|A| B. k^2…
- Which of the following is correctA. Determinant is a square matrix. B.…

**Exercise 4.3**- Find area of the triangle with vertices at the point given in each of the…
- (2, 7), (1, 1), (10, 8) Find area of the triangle with vertices at the point…
- (-2, -3), (3, 2), (-1, -8) Find area of the triangle with vertices at the point…
- Show that points A (a, b + c), B (b, c + a), C (c, a + b) are collinear.…
- (k, 0), (4, 0), (0, 2) Find values of k if area of triangle is 4 sq. units and…
- (-2, 0), (0, 4), (0, k) Find values of k if area of triangle is 4 sq. units and…
- Find equation of line joining (1, 2) and (3, 6) using determinants.…
- Find equation of line joining (3, 1) and (9, 3) using determinants.…
- If area of triangle is 35 sq units with vertices (2, -6), (5, 4) and (k, 4).…

**Exercise 4.4**- | cc 2&-4 0&3 | Write Minors and Cofactors of the elements of following…
- | ll a b | Write Minors and Cofactors of the elements of following…
- | lll 1&0&0 0&1&0 0&0&1 | Write Minors and Cofactors of the elements of…
- | ccc 1&0&4 3&5&-1 0&1&2 | Write Minors and Cofactors of the elements of…
- Using Cofactors of elements of second row, evaluate delta = | lll 5&3&8 2&0&1…
- Using Cofactors of elements of third column, evaluate delta = | lll 1 1 1 | .…
- If delta = | lll a_11_12_13 a_21_22_23 a_31_32_33 | and Aij is Cofactors of aij,…

**Exercise 4.5**- Find adjoint of each of the matrices. [ll 1&2 3&4]
- Find adjoint of each of the matrices. [ccc 1&-1&2 2&3&5 -2&0&1]
- [cc 2&3 -4&-6] Verify A (adj A) = (adj A) A = |A|
- [ccc 1&-1&2 3&0&-2 1&0&3] Verify A (adj A) = (adj A) A = |A|
- [cc 2&-2 4&3] Find the inverse of each of the matrices (if it exists)…
- [ll -1&5 -3&2] Find the inverse of each of the matrices (if it exists)…
- [lll 1&2&3 0&2&4 0&0&5] Find the inverse of each of the matrices (if it exists)…
- [ccc 1&0&0 3&3&0 5&2&-1] Find the inverse of each of the matrices (if it exists)…
- [ccc 2&1&3 4&-1&0 -7&2&1] Find the inverse of each of the matrices (if it…
- [ccc 1&-1&2 0&2&-3 3&-2&4] Find the inverse of each of the matrices (if it…
- [ccc 1&0&0 0 0 &-cosalpha] Find the inverse of each of the matrices (if it…
- Let a = [ll 3&7 2&5] b = [ll 6&8 7&9] . Verify that (AB)-1 = B-1 A-1.…
- If a = [cc 3&1 -1&2] , show that A^2 - 5A + 7I = O. Hence find A-1.…
- For the matrix a = [ll 3&1 1&2] , find the numbers a and b such that A^2 + aA +…
- For the matrix a = [ccc 1&1&1 1&2&-3 2&-1&3] Show that A^3 - 6A^2 + 5A + 11 I =…
- If a = [ccc 2&-1&1 -1&2&-1 1&-1&2] Verify that A^3 - 6A^2 + 9A - 4I = O and…
- Let A be a non-singular square matrix of order 3 × 3. Then |adj A| is equal…
- If A is an invertible matrix of order 2, then det (A-1) is equal toA. det (A)…

**Exercise 4.6**- x + 2y = 2 2x + 3y = 3 Examine the consistency of the system of equations.…
- 2x - y = 5 x + y = 4 Examine the consistency of the system of equations.…
- x + 3y = 5 2x + 6y = 8 Examine the consistency of the system of equations.…
- x + y + z = 1 2x + 3y + 2z = 2 ax + ay + 2az = 4 Examine the consistency of the…
- 3x-y - 2z = 2 2y - z = -1 3x - 5y = 3 Examine the consistency of the system of…
- 5x - y + 4z = 5 2x + 3y + 5z = 2 5x - 2y + 6z = -1 Examine the consistency of…
- 5x + 2y = 4 7x + 3y = 5 Solve system of linear equations, using matrix method.…
- 2x - y = -2 3x + 4y = 3 Solve system of linear equations, using matrix method.…
- 4x - 3y = 3 3x - 5y = 7 Solve system of linear equations, using matrix method.…
- 5x + 2y = 3 3x + 2y = 5 Solve system of linear equations, using matrix method.…
- 2x+y+z = 1 x-2y-z = 3/2 3y-5z = 9 Solve system of linear equations, using…
- x - y + z = 4 2x + y - 3z = 0 x + y + z = 2 Solve system of linear equations,…
- 2x + 3y +3 z = 5 x - 2y + z = -4 3x - y - 2z = 3 Solve system of linear…
- x - y + 2z = 7 3x + 4y - 5z = -5 2x - y + 3z = 12 Solve system of linear…
- If a = [ccc 2&-3&5 3&2&-4 1&1&-2] , find A-1. Using A-1 solve the system of…
- The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs. 60. The cost of 2 kg…

**Miscellaneous Exercise**- Prove that the determinant | ccc x heta -sintegrate heta &-x&1 costheta &1 | is…
- Without expanding the determinant, prove that | ccc a& a^2 b& b^2 c& c^2 | = |…
- Evaluate | ccc cosalpha cosbeta sinbeta &-sinalpha -sinbeta &0 sinalpha cosbeta…
- If a, b and c are real numbers, and delta = | lll b+c+a+b c+a+b+c a+b+c+a | = 0…
- Solve the equation | ccc x+a x+a x+a | = 0 , a not equal 0
- Prove that | ccc a^2 & ac+c^2 a^2 + ab & b^2 ab& b^2 + bc & c^2 | = 4a^2b^2c^2…
- If a^-1 = [ccc 3&-1&1 -15&6&-5 5&-2&2] b = [ccc 1&2&-2 -1&3&0 0&-2&1] , find…
- Let a = [ccc 1&-2&1 -2&3&1 1&1&5] . Verify that [adj A]-1 = adj (A-1)…
- Let a = [ccc 1&-2&1 -2&3&1 1&1&5] . Verify that (A-1)-1 = A
- Evaluate | ccc x+y y+y x+y |
- Evaluate | ccc 1 1+y 1+y |
- Prove that | lll alpha & alpha^2 & beta + gamma beta & beta^2 & gamma + alpha…
- Prove that | lll x& x^2 & 1+px^3 y& y^2 & 1+py^3 z& z^2 & 1+pz^3 | = (1 = pxyz)…
- Prove that | ccc 3a&-a+b&-a+c -b+a&3b&-b+c -c+a&-c+b&3c | = 3 (a+b+c)…
- Prove that | ccc 1&1+p&1+p+q 2&3+2p&4+3p+2q 3&6+3p&10+6p+3q | = 1…
- Prove that | ccc sinalpha & cos (alpha + delta) sinbeta & cos (beta + delta)…
- Solve the system of equations 2/x + 3/y + 10/z = 4 4/x - 6/y + 5/z = 1 6/x +…
- If a, b, c, are in A.P, then the determinant | lll x+2+3+2a x+3+4+20 x+4+5+2c |…
- If x, y, z are nonzero real numbers, then the inverse of matrix a = [lll x&0&0…
- Let a = [ccc 1 heta &1 -sintegrate heta &1 heta -1&-sintegrate heta &1] , where…

**Exercise 4.1**

- | cc 2&4 -5&-1 | Evaluate the determinants:
- | cc costheta &-sintegrate heta sintegrate heta | Evaluate the determinants:…
- | cc x^2 - x+1 x+1+1 | Evaluate the determinants:
- If a = [ll 1&2 4&2] then show that |2A| = 4|A|.
- If a = [lll 1&0&1 0&1&2 0&0&4] then show that |3A| = 27|A|
- Evaluate the determinants | ccc 3&-1&-2 0&1&-1 3&-5&0 |
- | ccc 3&-4&5 1&1&-2 2&3&1 | Evaluate the determinants
- | ccc 0&1&2 -1&0&-3 -2&3&0 | Evaluate the determinants
- | ccc 2&-1&-2 0&2&-1 3&-5&0 | Evaluate the determinants
- If A = [lll 1&1&-2 2&1&-3 5&4&-9] , find |A|.
- Find values of x, if | ll 2&4 5&1 | = | cc 2x&4 6 | | ll 2&4 5&1 | = | cc 2x&4…
- | ll 2&3 4&5 | = | ll x&3 2x&5 | Evaluate the determinants
- If , then x is equal toA. 6 B. 6 C. 6 D. 0

**Exercise 4.2**

- | lll x+a y+b x+c | = 0 Using the property of determinants and without expanding…
- | lll a-b b-c c-a | = 0 Using the property of determinants and without expanding…
- | lll 2&7&65 3&8&75 5&9&86 | = 0 Using the property of determinants and without…
- Using the property of determinants and without expanding in prove that:…
- | ccc b+c+r+z c+a+p+x a+b+q+y | = 2 | ccc a b c | Using the property of…
- | ccc 0&-b -a&0&-c b&0 | = 0 Using the property of determinants and without…
- | ccc - a^2 ba& - b^2 ca& - c^2 | = 4a^2b^2c^2 Using the property of…
- | lll 1& a^2 1& b^2 1& c^2 | = (a-b) (b-c) (c-a) By using properties of…
- | lll 1&1&1 a a^3 & b^3 & c^3 | = (a-b) (b-c) (c-a) (a+b+c) By using properties…
- | lll x& x^2 y& y^2 z& z^2 | = (x-y) (y-z) (z-x) (xy+yz+zx) By using properties…
- | ccc x+4&2x&2x 2x+4&2x 2x&2x+4 | = (5x+4) (4-x)^2 By using properties of…
- | ccc y+k y+k y+k | = k^2 (3y+k) By using properties of determinants, show…
- | ccc a-b-c&2a&2a 2b&2b 2c&2c | = (a+b+c)^3 By using properties of…
- | ccc x+y+2z z+z+2x z+x+2y | = 2 (x+y+z)^3 By using properties of…
- | ccc 1& x^2 x^2 &1 x& x^2 &1 | = (1-x^3)^2 By using properties of…
- | ccc 1+a^2 - b^2 &2ab&-2b 2ab& 1-a^2 + b^2 &2a 2b&-2a& 1-a^2 - b^2 | = (1+a^2…
- | ccc a^2 + 1 ab& b^2 + 1 ca& c^2 + 1 | = 1+a^2 + b^2 + c^2 By using properties…
- Let A be a square matrix of order 3 × 3, then | kA| is equal toA. k|A| B. k^2…
- Which of the following is correctA. Determinant is a square matrix. B.…

**Exercise 4.3**

- Find area of the triangle with vertices at the point given in each of the…
- (2, 7), (1, 1), (10, 8) Find area of the triangle with vertices at the point…
- (-2, -3), (3, 2), (-1, -8) Find area of the triangle with vertices at the point…
- Show that points A (a, b + c), B (b, c + a), C (c, a + b) are collinear.…
- (k, 0), (4, 0), (0, 2) Find values of k if area of triangle is 4 sq. units and…
- (-2, 0), (0, 4), (0, k) Find values of k if area of triangle is 4 sq. units and…
- Find equation of line joining (1, 2) and (3, 6) using determinants.…
- Find equation of line joining (3, 1) and (9, 3) using determinants.…
- If area of triangle is 35 sq units with vertices (2, -6), (5, 4) and (k, 4).…

**Exercise 4.4**

- | cc 2&-4 0&3 | Write Minors and Cofactors of the elements of following…
- | ll a b | Write Minors and Cofactors of the elements of following…
- | lll 1&0&0 0&1&0 0&0&1 | Write Minors and Cofactors of the elements of…
- | ccc 1&0&4 3&5&-1 0&1&2 | Write Minors and Cofactors of the elements of…
- Using Cofactors of elements of second row, evaluate delta = | lll 5&3&8 2&0&1…
- Using Cofactors of elements of third column, evaluate delta = | lll 1 1 1 | .…
- If delta = | lll a_11_12_13 a_21_22_23 a_31_32_33 | and Aij is Cofactors of aij,…

**Exercise 4.5**

- Find adjoint of each of the matrices. [ll 1&2 3&4]
- Find adjoint of each of the matrices. [ccc 1&-1&2 2&3&5 -2&0&1]
- [cc 2&3 -4&-6] Verify A (adj A) = (adj A) A = |A|
- [ccc 1&-1&2 3&0&-2 1&0&3] Verify A (adj A) = (adj A) A = |A|
- [cc 2&-2 4&3] Find the inverse of each of the matrices (if it exists)…
- [ll -1&5 -3&2] Find the inverse of each of the matrices (if it exists)…
- [lll 1&2&3 0&2&4 0&0&5] Find the inverse of each of the matrices (if it exists)…
- [ccc 1&0&0 3&3&0 5&2&-1] Find the inverse of each of the matrices (if it exists)…
- [ccc 2&1&3 4&-1&0 -7&2&1] Find the inverse of each of the matrices (if it…
- [ccc 1&-1&2 0&2&-3 3&-2&4] Find the inverse of each of the matrices (if it…
- [ccc 1&0&0 0 0 &-cosalpha] Find the inverse of each of the matrices (if it…
- Let a = [ll 3&7 2&5] b = [ll 6&8 7&9] . Verify that (AB)-1 = B-1 A-1.…
- If a = [cc 3&1 -1&2] , show that A^2 - 5A + 7I = O. Hence find A-1.…
- For the matrix a = [ll 3&1 1&2] , find the numbers a and b such that A^2 + aA +…
- For the matrix a = [ccc 1&1&1 1&2&-3 2&-1&3] Show that A^3 - 6A^2 + 5A + 11 I =…
- If a = [ccc 2&-1&1 -1&2&-1 1&-1&2] Verify that A^3 - 6A^2 + 9A - 4I = O and…
- Let A be a non-singular square matrix of order 3 × 3. Then |adj A| is equal…
- If A is an invertible matrix of order 2, then det (A-1) is equal toA. det (A)…

**Exercise 4.6**

- x + 2y = 2 2x + 3y = 3 Examine the consistency of the system of equations.…
- 2x - y = 5 x + y = 4 Examine the consistency of the system of equations.…
- x + 3y = 5 2x + 6y = 8 Examine the consistency of the system of equations.…
- x + y + z = 1 2x + 3y + 2z = 2 ax + ay + 2az = 4 Examine the consistency of the…
- 3x-y - 2z = 2 2y - z = -1 3x - 5y = 3 Examine the consistency of the system of…
- 5x - y + 4z = 5 2x + 3y + 5z = 2 5x - 2y + 6z = -1 Examine the consistency of…
- 5x + 2y = 4 7x + 3y = 5 Solve system of linear equations, using matrix method.…
- 2x - y = -2 3x + 4y = 3 Solve system of linear equations, using matrix method.…
- 4x - 3y = 3 3x - 5y = 7 Solve system of linear equations, using matrix method.…
- 5x + 2y = 3 3x + 2y = 5 Solve system of linear equations, using matrix method.…
- 2x+y+z = 1 x-2y-z = 3/2 3y-5z = 9 Solve system of linear equations, using…
- x - y + z = 4 2x + y - 3z = 0 x + y + z = 2 Solve system of linear equations,…
- 2x + 3y +3 z = 5 x - 2y + z = -4 3x - y - 2z = 3 Solve system of linear…
- x - y + 2z = 7 3x + 4y - 5z = -5 2x - y + 3z = 12 Solve system of linear…
- If a = [ccc 2&-3&5 3&2&-4 1&1&-2] , find A-1. Using A-1 solve the system of…
- The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs. 60. The cost of 2 kg…

**Miscellaneous Exercise**

- Prove that the determinant | ccc x heta -sintegrate heta &-x&1 costheta &1 | is…
- Without expanding the determinant, prove that | ccc a& a^2 b& b^2 c& c^2 | = |…
- Evaluate | ccc cosalpha cosbeta sinbeta &-sinalpha -sinbeta &0 sinalpha cosbeta…
- If a, b and c are real numbers, and delta = | lll b+c+a+b c+a+b+c a+b+c+a | = 0…
- Solve the equation | ccc x+a x+a x+a | = 0 , a not equal 0
- Prove that | ccc a^2 & ac+c^2 a^2 + ab & b^2 ab& b^2 + bc & c^2 | = 4a^2b^2c^2…
- If a^-1 = [ccc 3&-1&1 -15&6&-5 5&-2&2] b = [ccc 1&2&-2 -1&3&0 0&-2&1] , find…
- Let a = [ccc 1&-2&1 -2&3&1 1&1&5] . Verify that [adj A]-1 = adj (A-1)…
- Let a = [ccc 1&-2&1 -2&3&1 1&1&5] . Verify that (A-1)-1 = A
- Evaluate | ccc x+y y+y x+y |
- Evaluate | ccc 1 1+y 1+y |
- Prove that | lll alpha & alpha^2 & beta + gamma beta & beta^2 & gamma + alpha…
- Prove that | lll x& x^2 & 1+px^3 y& y^2 & 1+py^3 z& z^2 & 1+pz^3 | = (1 = pxyz)…
- Prove that | ccc 3a&-a+b&-a+c -b+a&3b&-b+c -c+a&-c+b&3c | = 3 (a+b+c)…
- Prove that | ccc 1&1+p&1+p+q 2&3+2p&4+3p+2q 3&6+3p&10+6p+3q | = 1…
- Prove that | ccc sinalpha & cos (alpha + delta) sinbeta & cos (beta + delta)…
- Solve the system of equations 2/x + 3/y + 10/z = 4 4/x - 6/y + 5/z = 1 6/x +…
- If a, b, c, are in A.P, then the determinant | lll x+2+3+2a x+3+4+20 x+4+5+2c |…
- If x, y, z are nonzero real numbers, then the inverse of matrix a = [lll x&0&0…
- Let a = [ccc 1 heta &1 -sintegrate heta &1 heta -1&-sintegrate heta &1] , where…

###### Exercise 4.1

**Question 1.**Evaluate the determinants:

**Answer:**We know that determinant of A is calculated as

Now,

= 2(-1) – 4(-5)

= -2 – (-20)

= -2 + 20

= 18

__The determinant of the above matrix is 18.__

**Question 2.**Evaluate the determinants:

**Answer:**We know that determinant of A is calculated as

Now,

= cosÎ¸(cosÎ¸) - (-sinÎ¸)(sinÎ¸)

= cos^{2}Î¸ + sin^{2}Î¸

= 1 [∵ cos^{2}Î¸ + sin^{2}Î¸ = 1]

__The determinant of the above matrix is 1.__

**Question 3.**Evaluate the determinants:

**Answer:**We know that determinant of A is calculated as

Now,

= (x^{2} – x + 1)(x + 1) - (x - 1)(x + 1)

= (x^{3} - x^{2} + x + x^{2} – x + 1) - (x^{2} - 1)

= x^{3} + 1 - x^{2} + 1

= x^{3} - x^{2} + 2

__Ans. The determinant of the above matrix is x__^{3} - x^{2} + 2.

**Question 4.**If then show that |2A| = 4|A|.

**Answer:**|A| =

We know that determinant of A is calculated as

= 1(2) - 2(4)

= 2 - 8

|A| = -6

__LHS__: |2A|

= 2(4) - 4(8)

= 8 - 32 = -24

|2A| = -24 …LHS

__RHS__: 4|A|

4|A|= 4(-6)

= -24

4|A| = -24 …RHS

LHS = RHS

Hence proved.

**Question 5.**If then show that |3A| = 27|A|

**Answer:**

We know that a determinant of a 3 x 3 matrix is calculated as

= 1[1(4) - 2(0)] – 0 + 1[0-0]

= 1[4 - 0] – 0 + 0

= 4

|A|= 4

__LHS__: |3A|

= 3[3(12) - 0(6)] – 0 + 3[0 – 0]

= 3(36) – 0 + 0

= 108

|3A| = 108 ----LHS

__RHS__: 27|A|

27|A| = 27(4)

= 108

27|A| = 108 ----RHS

LHS=RHS

Hence proved.

**Question 6.**Evaluate the determinants

**Answer:**Now,

We know that a determinant of a 3x3 matrix is calculated as

= 3[0 – (-1)(-5)] +1[0 – (-1)(3)] – 2[0 – 0]

= 3(-5) + 1(3) – 0

= -15 + 3

= -12

__The determinant of the above matrix is -12__

**Question 7.**Evaluate the determinants

**Answer:**Now,

We know that a determinant of a 3 x 3 matrix is calculated as

= 3[1 – (-2)(3)] + 4[1 – (-2)(2)] + 5[3-2]

= 3[1 + 6] + 4[1 + 4] + 5[1]

= 3[7] + 4[5] + 5

= 21 + 20 + 5

= 46

__The determinant of the above matrix is 46.__

**Question 8.**Evaluate the determinants

**Answer:**Now,

We know that a determinant of a 3 x 3 matrix is calculated as

= 0 – 1[0 – (-3)(-2)] + 2[(-1)(3) – 0]

= 0 – 1[0 - 6] + 2[-3 – 0]

= 0 – 1[-6] + 2[-3]

= 0 + 6 – 6

= 0

__The determinant of the above matrix is 0.__

**Question 9.**Evaluate the determinants

**Answer:**Now,

We know that a determinant of a 3 x 3 matrix is calculated as

= 2[0 – (-1)(-5)] + 1[0 – (-1)(3)] – 2[0 – 3(2)]

= 2[0 – 5] + 1[0 + 3] – 2[-6]

= 2[-5] + 1[3] -2[-6]

= -10 + 3 + 12

= 5

__The determinant of the above matrix is 5.__

**Question 10.**If A = , find |A|.

**Answer:**GIVEN:

Now,

We know that a determinant of a 3 x 3 matrix is calculated as

.

= 1[-9 – (-3)(4)] – 1[2(-9) – (-3)(5)] – 2[2(4) – 1(5)]

= 1[-9 + 12] – 1[-18 + 15] – 2[8 – 5]

= 1[3] – 1[-3] – 2[3]

= 3 + 3 – 6

= 0

__Ans. |A| = 0__

**Question 11.**Find values of x, if

**Answer:**We have

We know that determinant of A is calculated as

⇒ 2(1) – 4(5) = 2x(x) – 4(6)

⇒ 2 – 20 = 2x^{2} - 24

⇒ -18 = 2x^{2} – 24

⇒ 2x^{2} = -24 + 18

⇒ 2x^{2} = 6

⇒ x^{2} = 6/2

⇒ x^{2} = 3

x = �√3

__Ans. The value of x is �√3__

**Question 12.**Evaluate the determinants

**Answer:**We know that determinant of A is calculated as

We have

⇒ 2(5) – 3(4) = x(5) – 3(2x)

⇒ 10 – 12 = 5x – 6x

⇒ -2 = -x

⇒ x = 2

__Ans. The value of x is 2.__

**Question 13.**If , then x is equal to

A. 6

B. ±6

C. –6

D. 0

**Answer:**We have

We know that determinant of A is calculated as

⇒ x(x) – 2(18) = 6(6) – 2(18)

⇒ x^{2} - 36 = 36 – 36

⇒ x^{2} =36 – 36 + 36

⇒ x^{2} = 36

⇒ x = ±6

**Question 1.**

Evaluate the determinants:

**Answer:**

We know that determinant of A is calculated as

Now,

= 2(-1) – 4(-5)

= -2 – (-20)

= -2 + 20

= 18

__The determinant of the above matrix is 18.__

**Question 2.**

Evaluate the determinants:

**Answer:**

We know that determinant of A is calculated as

Now,

= cosÎ¸(cosÎ¸) - (-sinÎ¸)(sinÎ¸)

= cos^{2}Î¸ + sin^{2}Î¸

= 1 [∵ cos^{2}Î¸ + sin^{2}Î¸ = 1]

__The determinant of the above matrix is 1.__

**Question 3.**

Evaluate the determinants:

**Answer:**

We know that determinant of A is calculated as

Now,

= (x^{2} – x + 1)(x + 1) - (x - 1)(x + 1)

= (x^{3} - x^{2} + x + x^{2} – x + 1) - (x^{2} - 1)

= x^{3} + 1 - x^{2} + 1

= x^{3} - x^{2} + 2

__Ans. The determinant of the above matrix is x ^{3} - x^{2} + 2.__

**Question 4.**

If then show that |2A| = 4|A|.

**Answer:**

|A| =

We know that determinant of A is calculated as

= 1(2) - 2(4)

= 2 - 8

|A| = -6

__LHS__: |2A|

= 2(4) - 4(8)

= 8 - 32 = -24

|2A| = -24 …LHS

__RHS__: 4|A|

4|A|= 4(-6)

= -24

4|A| = -24 …RHS

LHS = RHS

Hence proved.

**Question 5.**

If then show that |3A| = 27|A|

**Answer:**

We know that a determinant of a 3 x 3 matrix is calculated as

= 1[1(4) - 2(0)] – 0 + 1[0-0]

= 1[4 - 0] – 0 + 0

= 4

|A|= 4

__LHS__: |3A|

= 3[3(12) - 0(6)] – 0 + 3[0 – 0]

= 3(36) – 0 + 0

= 108

|3A| = 108 ----LHS

__RHS__: 27|A|

27|A| = 27(4)

= 108

27|A| = 108 ----RHS

LHS=RHS

Hence proved.

**Question 6.**

Evaluate the determinants

**Answer:**

Now,

We know that a determinant of a 3x3 matrix is calculated as

= 3[0 – (-1)(-5)] +1[0 – (-1)(3)] – 2[0 – 0]

= 3(-5) + 1(3) – 0

= -15 + 3

= -12

__The determinant of the above matrix is -12__

**Question 7.**

Evaluate the determinants

**Answer:**

Now,

We know that a determinant of a 3 x 3 matrix is calculated as

= 3[1 – (-2)(3)] + 4[1 – (-2)(2)] + 5[3-2]

= 3[1 + 6] + 4[1 + 4] + 5[1]

= 3[7] + 4[5] + 5

= 21 + 20 + 5

= 46

__The determinant of the above matrix is 46.__

**Question 8.**

Evaluate the determinants

**Answer:**

Now,

We know that a determinant of a 3 x 3 matrix is calculated as

= 0 – 1[0 – (-3)(-2)] + 2[(-1)(3) – 0]

= 0 – 1[0 - 6] + 2[-3 – 0]

= 0 – 1[-6] + 2[-3]

= 0 + 6 – 6

= 0

__The determinant of the above matrix is 0.__

**Question 9.**

Evaluate the determinants

**Answer:**

Now,

We know that a determinant of a 3 x 3 matrix is calculated as

= 2[0 – (-1)(-5)] + 1[0 – (-1)(3)] – 2[0 – 3(2)]

= 2[0 – 5] + 1[0 + 3] – 2[-6]

= 2[-5] + 1[3] -2[-6]

= -10 + 3 + 12

= 5

__The determinant of the above matrix is 5.__

**Question 10.**

If A = , find |A|.

**Answer:**

GIVEN:

Now,

We know that a determinant of a 3 x 3 matrix is calculated as

.

= 1[-9 – (-3)(4)] – 1[2(-9) – (-3)(5)] – 2[2(4) – 1(5)]

= 1[-9 + 12] – 1[-18 + 15] – 2[8 – 5]

= 1[3] – 1[-3] – 2[3]

= 3 + 3 – 6

= 0

__Ans. |A| = 0__

**Question 11.**

Find values of x, if

**Answer:**

We have

We know that determinant of A is calculated as

⇒ 2(1) – 4(5) = 2x(x) – 4(6)

⇒ 2 – 20 = 2x^{2} - 24

⇒ -18 = 2x^{2} – 24

⇒ 2x^{2} = -24 + 18

⇒ 2x^{2} = 6

⇒ x^{2} = 6/2

⇒ x^{2} = 3

x = �√3

__Ans. The value of x is �√3__

**Question 12.**

Evaluate the determinants

**Answer:**

We know that determinant of A is calculated as

We have

⇒ 2(5) – 3(4) = x(5) – 3(2x)

⇒ 10 – 12 = 5x – 6x

⇒ -2 = -x

⇒ x = 2

__Ans. The value of x is 2.__

**Question 13.**

If , then x is equal to

A. 6

B. ±6

C. –6

D. 0

**Answer:**

We have

We know that determinant of A is calculated as

⇒ x(x) – 2(18) = 6(6) – 2(18)

⇒ x^{2} - 36 = 36 – 36

⇒ x^{2} =36 – 36 + 36

⇒ x^{2} = 36

⇒ x = ±6

###### Exercise 4.2

**Question 1.**Using the property of determinants and without expanding in prove that:

**Answer:**Applying Operations C_{1}→ C_{1} + C_{2} (i.e. Replacing 1^{st} column by addition of 1^{st} and 2^{nd} column)

C_{1}→ C_{1} - C_{3} (i.e. Replacing 1^{st} column by subtraction of 1^{st} and 3^{rd} column)

If any one of the rows or columns of a determinant is 0 then the value of that determinant is 0.

∴ LHS = 0 = RHS

**Question 2.**Using the property of determinants and without expanding in prove that:

**Answer:**Applying Operation C_{1}→ C_{1} + C_{2} (i.e. Replacing 1^{st} column by addition of 1^{st} and 2^{nd} column)

C_{1}→ C_{1} + C_{3} (i.e. Replacing 1^{st} column by addition of 1^{st} and 3^{rd} column)

If any one of the rows or columns of a determinant is 0 then the value of that determinant is 0.

∴ LHS = 0 = RHS

**Question 3.**Using the property of determinants and without expanding in prove that:

**Answer:**

Applying Operation C_{1}→ C_{1} + 9C_{2} (i.e. Replacing 1^{st} column by addition of 1^{st} column and 9 times second column)

C_{1}→ C_{1} - C_{3} (i.e. Replacing 1^{st} column by subtraction of 1^{st} and 3rd column)

If any one of the rows or columns of a determinant is 0 then the value of that determinant is 0.

∴ LHS = 0 = RHS

**Question 4.**Using the property of determinants and without expanding in prove that:

**Answer:**

C_{3}→ C_{2} + C_{3} (i.e. replace 3^{rd} column by addition of 2^{nd} and 3^{rd} column)

Taking ab + bc + ac outside determinant

C_{1}→ C_{1} - C_{3} (i.e. replace 1^{st} column by subtraction of 1^{st} and 3^{rd} column)

If any one of the rows or columns of a determinant is 0 then the value of that determinant is 0.

∴ LHS = 0 = RHS

∴

**Question 5.**Using the property of determinants and without expanding in prove that:

**Answer:**When two determinants are added each of the corresponding elements gets added.

Here we can split the LHS determinant as

For the determinant, u perform the following transformation R_{1}↔ R_{3} (i.e. interchange 1^{st} row with 3^{rd} row)

When two particular rows/columns of a determinant are interchanged the value becomes negative 1 times the original value.

R_{1}↔ R_{2} (i.e. interchange 1^{st} row with 2^{nd} row)

R1 ↔ R3 (i.e. interchange 1^{st} row with 3^{rd} row)

R1 ↔ R2 (i.e. interchange 1^{st} row with 2^{nd} row)

∴ LHS = RHS

∴

**Question 6.**Using the property of determinants and without expanding in prove that:

**Answer:**To Prove:

R_{1}→ cR_{1} (i.e. replace 1^{st} row by multiplying it with c)

As we are multiplying we should also divide c so that the original given determinant is not changed

R_{1}→ R_{1} - bR_{2} (i.e. replace 1^{st} row by subtraction of 1^{st} row and b times 2^{nd} row)

Taking a outside the determinant from 1^{st} row

If any two rows or columns of a determinant are identical then the value of that determinant is 0 because we get a row or column with all elements 0 when we when we subtract those particular rows/columns here the transformation is R1 → R1 - R3

∴LHS = 0 = RHS

∴

**Question 7.**Using the property of determinants and without expanding in prove that:

**Answer:**Taking ‘a’, ‘b’ and ‘c’ outside the determinant from 1^{st},2^{nd} and 3^{rd} column respectively

Taking ‘a’, ‘b’ and ‘c’ outside the determinant from 1^{st},2^{nd} and 3^{rd} row respectively

.

R1 → R1 + R2 (i.e. Replacing 1^{st} row by addition of 1^{st} and 2^{nd} row)

panding the determinant along 1^{st} row

∴ LHS = a^{2} b^{2} c^{2} × 2(1 - (-1)) = 4a^{2}b^{2}c^{2} = RHS

∴

**Question 8.**By using properties of determinants, show that:

**Answer:**R1 → R1 - R2 (i.e. Replacing 1^{st} row by subtraction of 1^{st} and 2^{nd} row)

R2 → R2 - R3 (i.e. Replacing 2^{nd} row by subtraction of 2^{nd} and 3^{rd} row)

Since we know a^{2} - b^{2} = (a + b)(a - b)

Therefore taking (a - b) and (b - c) outside the determinant from 1^{st} and 2^{nd} row respectively

R1 → R1 - R2 (i.e. Replacing 1^{st} row by subtraction of 1^{st} and 2^{nd} row)

Expanding the determinant along 1^{st} column

∴

∴LHS = (a - b)(b - c)(0 - (a - c))

∴LHS = (a - b)(b - c)(c - a) = RHS

∴

**Question 9.**By using properties of determinants, show that:

**Answer:**C1 → C1 - C2 (i.e. Replacing 1^{st} column by subtraction of 1^{st} and 2^{nd} column)

C2 → C2 - C3 (i.e. Replacing 2^{nd} column by subtraction of 2^{nd} and 3^{rd} column)

We have a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2}) and b^{3} - c^{3} = (b - c)(b^{2} + bc + c^{2})

Therefore taking (a - b) and (b - c) outside the determinant from 1^{st} and 2^{nd} column respectively

C1 → C1 - C2 (i.e. Replacing 1^{st} column by subtraction of 1^{st} and 2^{nd} column)

As a^{2} - c^{2}) = (a + c)(a - c) therefore taking (a - c) outside the determinant from 1^{st} column we get

.

Expanding the determinant along 1^{st} row

∴ LHS = (a - b)(b - c)(a - c)( - (a + b + c))

Adjusting the minus sign with (a - c)

∴LHS = (a - b)(b - c)(c - a)(a + b + c) = RHS

∴

**Question 10.**By using properties of determinants, show that:

**Answer:**R1 → R1 - R2 (i.e. Replacing 1^{st} row by subtraction of 1^{st} and 2^{nd} row)

R2 → R2 - R3 (i.e. Replacing 2^{nd} row by subtraction of 2^{nd} and 3^{rd} row)

We know x^{2} - y^{2} = (x + y)(x - y) and y^{2} - z^{2} = (y + z)(y - z)

Therefore taking (x - y) and (y - z) outside the determinant from 1^{st} and 2^{nd} row respectively

R1 → R1 - R2 (i.e. Replacing 1^{st} row by subtraction of 1^{st} and 2^{nd} row)

Taking (x - z) outside determinant from 1^{st} row

C2 → C2 - C3 (i.e. Replacing 2^{nd} column by subtraction of 2^{nd} and 3^{rd} column)

Expanding the determinant along 1^{st} row

∴ LHS = (x - y)(y - z)(x - z)(z^{2} - xy - xz - yz - z^{2})

Cancelling z^{2} and adjusting the negative sign with (x - z)

∴LHS = (x - y)(y - z)(z - x)(xy + yz + zx) = RHS

∴

**Question 11.**By using properties of determinants, show that:

**Answer:**R1 → R1 + R2 + R3 (i.e. replace 1^{st} row by addition of 1^{st}, 2^{nd} and 3^{rd} row)

Taking 5x + 4 outside the determinant from 1^{st} row

C2 → C2 - C1 (i.e. replace 2^{nd} column by subtraction of 2^{nd} and 1^{st} column)

C3 → C3 - C1 (i.e. replace 3^{rd} column by subtraction of 3^{rd} and 1^{st} column)

Expanding the determinant along 1^{st} row

∴ LHS = (5x - 4) (4 - x)^{2} = RHS

∴

**Question 12.**By using properties of determinants, show that:

**Answer:**R1 → R1 + R2 + R3 (i.e. replace 1^{st} row by addition of 1^{st}, 2^{nd} and 3^{rd} row)

Taking 3y + k outside the determinant from 1^{st} row

C2 → C2 - C1 (i.e. replace 2^{nd} column by subtraction of 2^{nd} and 1^{st} column)

C3 → C3 - C1 (i.e. replace 3^{rd} column by subtraction of 3^{rd} and 1^{st} column)

Expanding the determinant along 1^{st} row

∴ LHS = (3y + k) k^{2} = RHS

∴

**Question 13.**By using properties of determinants, show that:

**Answer:**R1 → R1 + R2 + R3 (i.e. replace 1^{st} row by addition of 1^{st}, 2^{nd} and 3^{rd} row)

Taking a + b + c outside the determinant from 1^{st} row

C2 → C2 - C1 (i.e. replace 2^{nd} column by subtraction of 2^{nd} and 1^{st} column)

C3 → C3 - C1 (i.e. replace 3^{rd} column by subtraction of 3^{rd} and 1^{st} column)

Expanding the determinant along 1^{st} row

∴ LHS = (a + b + c)^{3} = RHS

∴

**Question 14.**By using properties of determinants, show that:

**Answer:**C1 → C1 + C2 + C3 (i.e. replace 1^{st} column by addition of 1^{st}, 2^{nd} and 3^{rd} column)

Taking 2(x + y + z) outside the determinant from 1^{st} column

R2 → R2 - R1 (i.e. replace 2^{nd} row by subtraction of 2^{nd} and 1^{st} row)

R3 → R3 - R1 (i.e. replace 3^{rd} row by subtraction of 3^{rd} and 1^{st} row)

Expanding the determinant along 1^{st} column

∴ LHS = 2(x + y + z)^{3} = RHS

∴

**Question 15.**By using properties of determinants, show that:

**Answer:**R1 → R1 - xR2 (i.e. replace 1^{st} row by subtraction of 1^{st} row and ‘x’ times 2^{nd} row)

Taking (1 - x^{3}) outside the determinant from 1^{st} row

.

Expanding the determinant along 1^{st} row

HS = (1 - x^{3})^{2} = RHS

∴

**Question 16.**By using properties of determinants, show that:

**Answer:**R1 → R1 + bR3 (i.e. replace 1^{st} row by addition of 1^{st} row and b times 3^{rd} row)

R2 → R2 - aR3 (i.e. replace 2^{nd} row by subtraction of 2^{nd} row and a times 3^{rd} row)

.

Taking both (1 + a^{2} + b^{2}) outside the determinant from 1^{st} and 2^{nd} row

Expanding the determinant along 1^{st} row

∴ LHS = (1 + a^{2} + b^{2})^{2} [1 + a^{2} - b^{2} - (-2b^{2})]

∴ LHS = (1 + a^{2} + b^{2} )^{2} (1 + a^{2} + b^{2})

∴ LHS = (1 + a^{2} + b^{2} )^{3} = RHS

**Question 17.**By using properties of determinants, show that:

**Answer:**Taking out a, b and c from the determinant from 1^{st}, 2^{nd} and 3^{rd} row respectively.

R2 → R2 - R1 (i.e. replace 2^{nd} row by subtraction of 2^{nd} and 1^{st} row)

R3 → R3 - R1 (i.e. replace 3^{rd} row by subtraction of 3^{rd} and 1^{st} row)

.

C1 → aC1 (i.e. replace 1^{st} column by ‘a’ times 1^{st} column)

C2 → bC2 (i.e. replace 2^{nd} column by ‘b’ times 2^{nd} column)

→ cC3 (i.e. replace 3^{rd} column by ‘c’ times 3^{rd} column)

As we are multiplying by a, b and c we should also divide by a, b and c to keep the original determinant value unchanged.

.

C1 → C1 + C2 + C3 (i.e. replace 1^{st} column by addition of 1^{st}, 2^{nd} and 3^{rd} column)

Expanding determinant along 1^{st} column

∴ LHS = (1 + a^{2} + b^{2} + c^{2}) = RHS

∴

**Question 18.**Let A be a square matrix of order 3 × 3, then | kA| is equal to

A. k|A|

B. k^{2} |A|

C. k^{3} |A|

D. 3k |A|

**Answer:**Let A be any 3×3 matrix

∴

Taking out k from the determinant from 1^{st}, 2^{nd} and 3^{rd} row

∴

∴ |kA| = k^{3}|A|

Therefore, answer is option (C) k^{3}|A|

**Question 19.**Which of the following is correct

A. Determinant is a square matrix.

B. Determinant is a number associated to a matrix.

C. Determinant is a number associated to a square matrix.

D. None of these

**Answer:**Determinant is an operation which we perform on arranged numbers. A square matrix is set of arranged numbers. We perform some operations on a matrix and we get a value that value is called as determinant of that matrix hence determinant is a number associated to square matrix.

**Question 1.**

Using the property of determinants and without expanding in prove that:

**Answer:**

Applying Operations C_{1}→ C_{1} + C_{2} (i.e. Replacing 1^{st} column by addition of 1^{st} and 2^{nd} column)

C_{1}→ C_{1} - C_{3} (i.e. Replacing 1^{st} column by subtraction of 1^{st} and 3^{rd} column)

If any one of the rows or columns of a determinant is 0 then the value of that determinant is 0.

∴ LHS = 0 = RHS

**Question 2.**

Using the property of determinants and without expanding in prove that:

**Answer:**

Applying Operation C_{1}→ C_{1} + C_{2} (i.e. Replacing 1^{st} column by addition of 1^{st} and 2^{nd} column)

C_{1}→ C_{1} + C_{3} (i.e. Replacing 1^{st} column by addition of 1^{st} and 3^{rd} column)

If any one of the rows or columns of a determinant is 0 then the value of that determinant is 0.

∴ LHS = 0 = RHS

**Question 3.**

Using the property of determinants and without expanding in prove that:

**Answer:**

Applying Operation C_{1}→ C_{1} + 9C_{2} (i.e. Replacing 1^{st} column by addition of 1^{st} column and 9 times second column)

C_{1}→ C_{1} - C_{3} (i.e. Replacing 1^{st} column by subtraction of 1^{st} and 3rd column)

If any one of the rows or columns of a determinant is 0 then the value of that determinant is 0.

∴ LHS = 0 = RHS

**Question 4.**

Using the property of determinants and without expanding in prove that:

**Answer:**

C_{3}→ C_{2} + C_{3} (i.e. replace 3^{rd} column by addition of 2^{nd} and 3^{rd} column)

Taking ab + bc + ac outside determinant

C_{1}→ C_{1} - C_{3} (i.e. replace 1^{st} column by subtraction of 1^{st} and 3^{rd} column)

If any one of the rows or columns of a determinant is 0 then the value of that determinant is 0.

∴ LHS = 0 = RHS

∴

**Question 5.**

Using the property of determinants and without expanding in prove that:

**Answer:**

When two determinants are added each of the corresponding elements gets added.

Here we can split the LHS determinant as

For the determinant, u perform the following transformation R_{1}↔ R_{3} (i.e. interchange 1^{st} row with 3^{rd} row)

When two particular rows/columns of a determinant are interchanged the value becomes negative 1 times the original value.

R_{1}↔ R_{2} (i.e. interchange 1^{st} row with 2^{nd} row)

R1 ↔ R3 (i.e. interchange 1^{st} row with 3^{rd} row)

R1 ↔ R2 (i.e. interchange 1^{st} row with 2^{nd} row)

∴ LHS = RHS

∴

**Question 6.**

Using the property of determinants and without expanding in prove that:

**Answer:**

To Prove:

R_{1}→ cR_{1} (i.e. replace 1^{st} row by multiplying it with c)

As we are multiplying we should also divide c so that the original given determinant is not changed

R_{1}→ R_{1} - bR_{2} (i.e. replace 1^{st} row by subtraction of 1^{st} row and b times 2^{nd} row)

Taking a outside the determinant from 1^{st} row

If any two rows or columns of a determinant are identical then the value of that determinant is 0 because we get a row or column with all elements 0 when we when we subtract those particular rows/columns here the transformation is R1 → R1 - R3

∴LHS = 0 = RHS

∴

**Question 7.**

Using the property of determinants and without expanding in prove that:

**Answer:**

Taking ‘a’, ‘b’ and ‘c’ outside the determinant from 1^{st},2^{nd} and 3^{rd} column respectively

Taking ‘a’, ‘b’ and ‘c’ outside the determinant from 1^{st},2^{nd} and 3^{rd} row respectively

.

R1 → R1 + R2 (i.e. Replacing 1^{st} row by addition of 1^{st} and 2^{nd} row)

panding the determinant along 1^{st} row

∴ LHS = a^{2} b^{2} c^{2} × 2(1 - (-1)) = 4a^{2}b^{2}c^{2} = RHS

∴

**Question 8.**

By using properties of determinants, show that:

**Answer:**

R1 → R1 - R2 (i.e. Replacing 1^{st} row by subtraction of 1^{st} and 2^{nd} row)

R2 → R2 - R3 (i.e. Replacing 2^{nd} row by subtraction of 2^{nd} and 3^{rd} row)

Since we know a^{2} - b^{2} = (a + b)(a - b)

Therefore taking (a - b) and (b - c) outside the determinant from 1^{st} and 2^{nd} row respectively

R1 → R1 - R2 (i.e. Replacing 1^{st} row by subtraction of 1^{st} and 2^{nd} row)

Expanding the determinant along 1^{st} column

∴

∴LHS = (a - b)(b - c)(0 - (a - c))

∴LHS = (a - b)(b - c)(c - a) = RHS

∴

**Question 9.**

By using properties of determinants, show that:

**Answer:**

C1 → C1 - C2 (i.e. Replacing 1^{st} column by subtraction of 1^{st} and 2^{nd} column)

C2 → C2 - C3 (i.e. Replacing 2^{nd} column by subtraction of 2^{nd} and 3^{rd} column)

We have a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2}) and b^{3} - c^{3} = (b - c)(b^{2} + bc + c^{2})

Therefore taking (a - b) and (b - c) outside the determinant from 1^{st} and 2^{nd} column respectively

C1 → C1 - C2 (i.e. Replacing 1^{st} column by subtraction of 1^{st} and 2^{nd} column)

As a^{2} - c^{2}) = (a + c)(a - c) therefore taking (a - c) outside the determinant from 1^{st} column we get

.

Expanding the determinant along 1^{st} row

∴ LHS = (a - b)(b - c)(a - c)( - (a + b + c))

Adjusting the minus sign with (a - c)

∴LHS = (a - b)(b - c)(c - a)(a + b + c) = RHS

∴

**Question 10.**

By using properties of determinants, show that:

**Answer:**

R1 → R1 - R2 (i.e. Replacing 1^{st} row by subtraction of 1^{st} and 2^{nd} row)

R2 → R2 - R3 (i.e. Replacing 2^{nd} row by subtraction of 2^{nd} and 3^{rd} row)

We know x^{2} - y^{2} = (x + y)(x - y) and y^{2} - z^{2} = (y + z)(y - z)

Therefore taking (x - y) and (y - z) outside the determinant from 1^{st} and 2^{nd} row respectively

R1 → R1 - R2 (i.e. Replacing 1^{st} row by subtraction of 1^{st} and 2^{nd} row)

Taking (x - z) outside determinant from 1^{st} row

C2 → C2 - C3 (i.e. Replacing 2^{nd} column by subtraction of 2^{nd} and 3^{rd} column)

Expanding the determinant along 1^{st} row

∴ LHS = (x - y)(y - z)(x - z)(z^{2} - xy - xz - yz - z^{2})

Cancelling z^{2} and adjusting the negative sign with (x - z)

∴LHS = (x - y)(y - z)(z - x)(xy + yz + zx) = RHS

∴

**Question 11.**

By using properties of determinants, show that:

**Answer:**

R1 → R1 + R2 + R3 (i.e. replace 1^{st} row by addition of 1^{st}, 2^{nd} and 3^{rd} row)

Taking 5x + 4 outside the determinant from 1^{st} row

C2 → C2 - C1 (i.e. replace 2^{nd} column by subtraction of 2^{nd} and 1^{st} column)

C3 → C3 - C1 (i.e. replace 3^{rd} column by subtraction of 3^{rd} and 1^{st} column)

Expanding the determinant along 1^{st} row

∴ LHS = (5x - 4) (4 - x)^{2} = RHS

∴

**Question 12.**

By using properties of determinants, show that:

**Answer:**

R1 → R1 + R2 + R3 (i.e. replace 1^{st} row by addition of 1^{st}, 2^{nd} and 3^{rd} row)

Taking 3y + k outside the determinant from 1^{st} row

C2 → C2 - C1 (i.e. replace 2^{nd} column by subtraction of 2^{nd} and 1^{st} column)

C3 → C3 - C1 (i.e. replace 3^{rd} column by subtraction of 3^{rd} and 1^{st} column)

Expanding the determinant along 1^{st} row

∴ LHS = (3y + k) k^{2} = RHS

∴

**Question 13.**

By using properties of determinants, show that:

**Answer:**

R1 → R1 + R2 + R3 (i.e. replace 1^{st} row by addition of 1^{st}, 2^{nd} and 3^{rd} row)

Taking a + b + c outside the determinant from 1^{st} row

C2 → C2 - C1 (i.e. replace 2^{nd} column by subtraction of 2^{nd} and 1^{st} column)

C3 → C3 - C1 (i.e. replace 3^{rd} column by subtraction of 3^{rd} and 1^{st} column)

Expanding the determinant along 1^{st} row

∴ LHS = (a + b + c)^{3} = RHS

∴

**Question 14.**

By using properties of determinants, show that:

**Answer:**

C1 → C1 + C2 + C3 (i.e. replace 1^{st} column by addition of 1^{st}, 2^{nd} and 3^{rd} column)

Taking 2(x + y + z) outside the determinant from 1^{st} column

R2 → R2 - R1 (i.e. replace 2^{nd} row by subtraction of 2^{nd} and 1^{st} row)

R3 → R3 - R1 (i.e. replace 3^{rd} row by subtraction of 3^{rd} and 1^{st} row)

Expanding the determinant along 1^{st} column

∴ LHS = 2(x + y + z)^{3} = RHS

∴

**Question 15.**

By using properties of determinants, show that:

**Answer:**

R1 → R1 - xR2 (i.e. replace 1^{st} row by subtraction of 1^{st} row and ‘x’ times 2^{nd} row)

Taking (1 - x^{3}) outside the determinant from 1^{st} row

.

Expanding the determinant along 1^{st} row

HS = (1 - x^{3})^{2} = RHS

∴

**Question 16.**

By using properties of determinants, show that:

**Answer:**

R1 → R1 + bR3 (i.e. replace 1^{st} row by addition of 1^{st} row and b times 3^{rd} row)

R2 → R2 - aR3 (i.e. replace 2^{nd} row by subtraction of 2^{nd} row and a times 3^{rd} row)

.

Taking both (1 + a^{2} + b^{2}) outside the determinant from 1^{st} and 2^{nd} row

Expanding the determinant along 1^{st} row

∴ LHS = (1 + a^{2} + b^{2})^{2} [1 + a^{2} - b^{2} - (-2b^{2})]

∴ LHS = (1 + a^{2} + b^{2} )^{2} (1 + a^{2} + b^{2})

∴ LHS = (1 + a^{2} + b^{2} )^{3} = RHS

**Question 17.**

By using properties of determinants, show that:

**Answer:**

Taking out a, b and c from the determinant from 1^{st}, 2^{nd} and 3^{rd} row respectively.

R2 → R2 - R1 (i.e. replace 2^{nd} row by subtraction of 2^{nd} and 1^{st} row)

R3 → R3 - R1 (i.e. replace 3^{rd} row by subtraction of 3^{rd} and 1^{st} row)

.

C1 → aC1 (i.e. replace 1^{st} column by ‘a’ times 1^{st} column)

C2 → bC2 (i.e. replace 2^{nd} column by ‘b’ times 2^{nd} column)

→ cC3 (i.e. replace 3^{rd} column by ‘c’ times 3^{rd} column)

As we are multiplying by a, b and c we should also divide by a, b and c to keep the original determinant value unchanged.

.

C1 → C1 + C2 + C3 (i.e. replace 1^{st} column by addition of 1^{st}, 2^{nd} and 3^{rd} column)

Expanding determinant along 1^{st} column

∴ LHS = (1 + a^{2} + b^{2} + c^{2}) = RHS

∴

**Question 18.**

Let A be a square matrix of order 3 × 3, then | kA| is equal to

A. k|A|

B. k^{2} |A|

C. k^{3} |A|

D. 3k |A|

**Answer:**

Let A be any 3×3 matrix

∴

Taking out k from the determinant from 1^{st}, 2^{nd} and 3^{rd} row

∴

∴ |kA| = k^{3}|A|

Therefore, answer is option (C) k^{3}|A|

**Question 19.**

Which of the following is correct

A. Determinant is a square matrix.

B. Determinant is a number associated to a matrix.

C. Determinant is a number associated to a square matrix.

D. None of these

**Answer:**

Determinant is an operation which we perform on arranged numbers. A square matrix is set of arranged numbers. We perform some operations on a matrix and we get a value that value is called as determinant of that matrix hence determinant is a number associated to square matrix.

###### Exercise 4.3

**Question 1.**Find area of the triangle with vertices at the point given in each of the following:

(1, 0), (6, 0), (4, 3)

**Answer:**Given vertices of the triangle are (1, 0), (6, 0), (4, 3)

Let the vertices of the triangle be given by (x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})

Area of triangle is given by Î” =

Area of triangle = Î” =

Expanding the determinant along Row 1

Î” = 1/2 × [1 × (0 × 1 – 3 × 1) – 0 × (6 × 1 – 4 × 1) + 1 × (6 × 3 – 4 × 0)]

Î” = 1/2 × [1 × (0 – 3) – 0 + 1 × (18 – 0)]

Î” = 1/2 × (-3 + 18) = 15/2 sq. units

∴ Î” = 15/2 sq. units = 7.5 sq. units

**Question 2.**Find area of the triangle with vertices at the point given in each of the following:

(2, 7), (1, 1), (10, 8)

**Answer:**Given vertices of the triangle are (2, 7), (1, 1), (10, 8)

Let the vertices of the triangle be given by (x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})

Area of triangle is given by Î” =

Area of triangle = Î” =

Expanding the determinant along Row 1

Î” = 1/2 × [2 × (1 × 1 – 8 × 1) – 7 × (1 × 1 – 10 × 1) + 1 × (8 × 1 – 1 × 10)]

Î” = 1/2 × [2 × (1 – 8) – 7 × (1 – 10) + 1 × (8 – 10)]

Î” = 1/2 × [2 × (-7) – 7 × (-9) + 1 × (-2)] = 1/2 × (-14 + 63 – 2) sq. units

Î” = 1/2 × 47 sq. units = 47/2 sq. units

∴ Î” = 47/2 sq. units = 23.5 sq. units

**Question 3.**Find area of the triangle with vertices at the point given in each of the following:

(–2, –3), (3, 2), (–1, –8)

**Answer:**Given vertices of the triangle are (–2, –3), (3, 2), (–1, –8)

Let the vertices of the triangle be given by (x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})

Area of triangle is given by Î” =

Area of triangle = Î” =

Expanding the determinant along Row 1

Î” = 1/2 × |[(-2) × (2 × 1 – (-8) × 1) – (-3) × (3 × 1 – (-1) × 1) + 1 × (3 × (-8) – 2 × (-1))]|

Î” = 1/2 × |[(-2) × (2 + 8) + 3 × (3 + 1) + 1 × (-24 + 2)]|

Î” = 1/2 × |[(-2) × 10 + 3 × 4 + 1 × (-22)]| = 1/2 ×| (-20 + 12 – 22)| sq. units

Î” = 1/2 × |-30| sq. units = 30/2 sq. units

∴ Î” = 30/2 sq. units = 15 sq. units

**Question 4.**Show that points

A (a, b + c), B (b, c + a), C (c, a + b) are collinear.

**Answer:**Given vertices of the triangle are A (a, b + c), B (b, c + a), C (c, a + b)

Let the vertices of the triangle be given by (x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})

Area of triangle is given by Î” =

For points to be collinear area of triangle = Î” = 0

So, we have to show that area of triangle formed by ABC is 0

Area of triangle = Î” =

Expanding the determinant along Row 1

Î” = 1/2 × [a × {(c + a) × 1 – (a + b) × 1) – (b + c) × {b × 1 – c × 1} + 1 × {b × (a + b) – c × (c +a)}]

Î” = 1/2 × [a × (c + a – a – b) – (b + c) × (b – c) + 1 × (ab + b^{2} – c^{2} - ca)]

Î” = 1/2 × [a × (c – b) – (b^{2} – c^{2}) + 1 × (ab + b^{2} – c^{2} – ca)]

Î” = 1/2 × (ac – ab – b^{2} + c^{2} + ab + b^{2} – c^{2} – ca) sq units

Î” = 1/2 × 0 sq units

∴ Î” = 0

∴ Given vertices of the triangle are A (a, b + c), B (b, c + a), C (c, a + b) are collinear

**Question 5.**Find values of k if area of triangle is 4 sq. units and vertices are

(k, 0), (4, 0), (0, 2)

**Answer:**Given vertices of the triangle are (k, 0), (4, 0), (0, 2)

Let the vertices of the triangle be given by (x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})

Area of triangle is given by Î” =

Given, Area of triangle = Î” = 4 sq. units

= 4

⇒ 4 = 1/2 |[k × (0 × 1 – 2 × 1) – 0 × (4 × 1 – 0 × 1) + 1 × (4 × 2 – 0 × 0)]|

⇒ 4 = 1/2 × |[k × (0 – 2) – 0 + 1 × (8 – 0)]|

⇒ � 4 × 2 = -2k + 8

⇒ 8 = -2k + 8 and -8 = -2k + 8

⇒ 8 – 8 = -2k and 8 + 8 = 2k

⇒ 2k = 0 and 16 = 2k

⇒ k = 0 and k = 8

**Question 6.**Find values of k if area of triangle is 4 sq. units and vertices are

(–2, 0), (0, 4), (0, k)

**Answer:**Given vertices of the triangle are (–2, 0), (0, 4), (0, k)

Let the vertices of the triangle be given by (x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})

Area of triangle is given by Î” =

Given, Area of triangle = Î” = 4 sq. units

= 4

⇒ 4 = 1/2 |[(-2) × (4 × 1 – k × 1) – 0 × (0 × 1 – 0 × 1) + 1 × (0 × k – 0 × 4)]|

⇒ 4 = 1/2 × |[(-2) × (4 – k) – 0 + 1 × (0 – 0)]|

⇒ 4 × 2 = |(-8 + 2k)

⇒ � 8= 2k - 8

⇒ 8 = 2k - 8 and ⇒ -8 = 2k - 8

⇒ 8 + 8 = 2k and ⇒ 8 - 8 = 2k

⇒ k = 16/2 and ⇒ k = 0/2

⇒ k = 8 and ⇒ k = 0

**Question 7.**Find equation of line joining (1, 2) and (3, 6) using determinants.

**Answer:**Equation of line joining points (x_{1}, y_{1}) & (x_{2}, y_{2}) is given by = 0

Given points are (1, 2) and (3, 6)

Equation of line is given by = 0

⇒ 1/2 × [1 × (6 × 1 – y × 1) – 2 × (3 × 1 – x × 1) + 1 × (3 × y – x × 6)] = 0

⇒ [(6 – y) – 2 × (3 – x) + (3y – 6x)] = 0 × 2

⇒ (6 – y – 6 + 2x + 3y – 6x) = 0

⇒ 2y – 4x = 0

⇒ y – 2x = 0 ⇒ y = 2x

∴ Required Equation of line is y = 2x

**Question 8.**Find equation of line joining (3, 1) and (9, 3) using determinants.

**Answer:**Equation of line joining points (x_{1}, y_{1}) & (x_{2}, y_{2}) is given by = 0

Given points are (3, 1) and (9, 3)

Equation of line is given by = 0

⇒ 1/2 × [3 × (3 × 1 – y × 1) – 1 × (9 × 1 – x × 1) + 1 × (9 × y – x × 3)] = 0

⇒ [3 × (3 – y) – 1 × (9 – x) + (9y – 3x)] = 0 × 2

⇒ (9 – 3y – 9 + x + 9y – 3x) = 0

⇒ 6y – 2x = 0

⇒ 2x – 6y = 0 ⇒ x – 3y = 0

∴ Required Equation of line is x – 3y = 0

**Question 9.**If area of triangle is 35 sq units with vertices (2, –6), (5, 4) and (k, 4). Then k is

A. 12

B. –2

C. –12, –2

D. 12, –2

**Answer:**Given vertices of the triangle are (2, – 6), (5, 4) and (k, 4).

Let the vertices of the triangle be given by (x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})

Area of triangle is given by Î” =

Given, Area of triangle = Î” = 35 sq. units

= 35

⇒ � 35 = 1/2 × [2 × (4 × 1 – 4 × 1) – (-6) × (5 × 1 – k × 1) + 1 × (5 × 4 – k × 4)]

⇒ � 35 = 1/2 × [2 × (4 – 4) + 6 × (5 – k) + 1 × (20 – 4k)]

⇒ � 35 × 2 = (2 × 0 + 30 – 6k + 20 – 4k)

⇒ � 70 = 30 – 6k + 20 – 4k

⇒ � 70 = 50 – 10k

⇒ 70 – 50 = -10k and ⇒ -70 – 50 = -10k

⇒ 20 = -10k and ⇒ -120 = -10k

⇒ k = -20/10 and ⇒ k = 120/10

⇒ k = -2 and ⇒ k = 12

**Question 1.**

Find area of the triangle with vertices at the point given in each of the following:

(1, 0), (6, 0), (4, 3)

**Answer:**

Given vertices of the triangle are (1, 0), (6, 0), (4, 3)

Let the vertices of the triangle be given by (x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})

Area of triangle is given by Î” =

Area of triangle = Î” =

Expanding the determinant along Row 1

Î” = 1/2 × [1 × (0 × 1 – 3 × 1) – 0 × (6 × 1 – 4 × 1) + 1 × (6 × 3 – 4 × 0)]

Î” = 1/2 × [1 × (0 – 3) – 0 + 1 × (18 – 0)]

Î” = 1/2 × (-3 + 18) = 15/2 sq. units

∴ Î” = 15/2 sq. units = 7.5 sq. units

**Question 2.**

Find area of the triangle with vertices at the point given in each of the following:

(2, 7), (1, 1), (10, 8)

**Answer:**

Given vertices of the triangle are (2, 7), (1, 1), (10, 8)

Let the vertices of the triangle be given by (x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})

Area of triangle is given by Î” =

Area of triangle = Î” =

Expanding the determinant along Row 1

Î” = 1/2 × [2 × (1 × 1 – 8 × 1) – 7 × (1 × 1 – 10 × 1) + 1 × (8 × 1 – 1 × 10)]

Î” = 1/2 × [2 × (1 – 8) – 7 × (1 – 10) + 1 × (8 – 10)]

Î” = 1/2 × [2 × (-7) – 7 × (-9) + 1 × (-2)] = 1/2 × (-14 + 63 – 2) sq. units

Î” = 1/2 × 47 sq. units = 47/2 sq. units

∴ Î” = 47/2 sq. units = 23.5 sq. units

**Question 3.**

Find area of the triangle with vertices at the point given in each of the following:

(–2, –3), (3, 2), (–1, –8)

**Answer:**

Given vertices of the triangle are (–2, –3), (3, 2), (–1, –8)

Let the vertices of the triangle be given by (x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})

Area of triangle is given by Î” =

Area of triangle = Î” =

Expanding the determinant along Row 1

Î” = 1/2 × |[(-2) × (2 × 1 – (-8) × 1) – (-3) × (3 × 1 – (-1) × 1) + 1 × (3 × (-8) – 2 × (-1))]|

Î” = 1/2 × |[(-2) × (2 + 8) + 3 × (3 + 1) + 1 × (-24 + 2)]|

Î” = 1/2 × |[(-2) × 10 + 3 × 4 + 1 × (-22)]| = 1/2 ×| (-20 + 12 – 22)| sq. units

Î” = 1/2 × |-30| sq. units = 30/2 sq. units

∴ Î” = 30/2 sq. units = 15 sq. units

**Question 4.**

Show that points

A (a, b + c), B (b, c + a), C (c, a + b) are collinear.

**Answer:**

Given vertices of the triangle are A (a, b + c), B (b, c + a), C (c, a + b)

_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})

Area of triangle is given by Î” =

For points to be collinear area of triangle = Î” = 0

So, we have to show that area of triangle formed by ABC is 0

Area of triangle = Î” =

Expanding the determinant along Row 1

Î” = 1/2 × [a × {(c + a) × 1 – (a + b) × 1) – (b + c) × {b × 1 – c × 1} + 1 × {b × (a + b) – c × (c +a)}]

Î” = 1/2 × [a × (c + a – a – b) – (b + c) × (b – c) + 1 × (ab + b^{2} – c^{2} - ca)]

Î” = 1/2 × [a × (c – b) – (b^{2} – c^{2}) + 1 × (ab + b^{2} – c^{2} – ca)]

Î” = 1/2 × (ac – ab – b^{2} + c^{2} + ab + b^{2} – c^{2} – ca) sq units

Î” = 1/2 × 0 sq units

∴ Î” = 0

∴ Given vertices of the triangle are A (a, b + c), B (b, c + a), C (c, a + b) are collinear

**Question 5.**

Find values of k if area of triangle is 4 sq. units and vertices are

(k, 0), (4, 0), (0, 2)

**Answer:**

Given vertices of the triangle are (k, 0), (4, 0), (0, 2)

_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})

Area of triangle is given by Î” =

Given, Area of triangle = Î” = 4 sq. units

= 4

⇒ 4 = 1/2 |[k × (0 × 1 – 2 × 1) – 0 × (4 × 1 – 0 × 1) + 1 × (4 × 2 – 0 × 0)]|

⇒ 4 = 1/2 × |[k × (0 – 2) – 0 + 1 × (8 – 0)]|

⇒ � 4 × 2 = -2k + 8

⇒ 8 = -2k + 8 and -8 = -2k + 8

⇒ 8 – 8 = -2k and 8 + 8 = 2k

⇒ 2k = 0 and 16 = 2k

⇒ k = 0 and k = 8

**Question 6.**

Find values of k if area of triangle is 4 sq. units and vertices are

(–2, 0), (0, 4), (0, k)

**Answer:**

Given vertices of the triangle are (–2, 0), (0, 4), (0, k)

_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})

Area of triangle is given by Î” =

Given, Area of triangle = Î” = 4 sq. units

= 4

⇒ 4 = 1/2 |[(-2) × (4 × 1 – k × 1) – 0 × (0 × 1 – 0 × 1) + 1 × (0 × k – 0 × 4)]|

⇒ 4 = 1/2 × |[(-2) × (4 – k) – 0 + 1 × (0 – 0)]|

⇒ 4 × 2 = |(-8 + 2k)

⇒ � 8= 2k - 8

⇒ 8 = 2k - 8 and ⇒ -8 = 2k - 8

⇒ 8 + 8 = 2k and ⇒ 8 - 8 = 2k

⇒ k = 16/2 and ⇒ k = 0/2

⇒ k = 8 and ⇒ k = 0

**Question 7.**

Find equation of line joining (1, 2) and (3, 6) using determinants.

**Answer:**

Equation of line joining points (x_{1}, y_{1}) & (x_{2}, y_{2}) is given by = 0

Given points are (1, 2) and (3, 6)

Equation of line is given by = 0

⇒ 1/2 × [1 × (6 × 1 – y × 1) – 2 × (3 × 1 – x × 1) + 1 × (3 × y – x × 6)] = 0

⇒ [(6 – y) – 2 × (3 – x) + (3y – 6x)] = 0 × 2

⇒ (6 – y – 6 + 2x + 3y – 6x) = 0

⇒ 2y – 4x = 0

⇒ y – 2x = 0 ⇒ y = 2x

∴ Required Equation of line is y = 2x

**Question 8.**

Find equation of line joining (3, 1) and (9, 3) using determinants.

**Answer:**

Equation of line joining points (x_{1}, y_{1}) & (x_{2}, y_{2}) is given by = 0

Given points are (3, 1) and (9, 3)

Equation of line is given by = 0

⇒ 1/2 × [3 × (3 × 1 – y × 1) – 1 × (9 × 1 – x × 1) + 1 × (9 × y – x × 3)] = 0

⇒ [3 × (3 – y) – 1 × (9 – x) + (9y – 3x)] = 0 × 2

⇒ (9 – 3y – 9 + x + 9y – 3x) = 0

⇒ 6y – 2x = 0

⇒ 2x – 6y = 0 ⇒ x – 3y = 0

∴ Required Equation of line is x – 3y = 0

**Question 9.**

If area of triangle is 35 sq units with vertices (2, –6), (5, 4) and (k, 4). Then k is

A. 12

B. –2

C. –12, –2

D. 12, –2

**Answer:**

Given vertices of the triangle are (2, – 6), (5, 4) and (k, 4).

_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})

Area of triangle is given by Î” =

Given, Area of triangle = Î” = 35 sq. units

= 35

⇒ � 35 = 1/2 × [2 × (4 × 1 – 4 × 1) – (-6) × (5 × 1 – k × 1) + 1 × (5 × 4 – k × 4)]

⇒ � 35 = 1/2 × [2 × (4 – 4) + 6 × (5 – k) + 1 × (20 – 4k)]

⇒ � 35 × 2 = (2 × 0 + 30 – 6k + 20 – 4k)

⇒ � 70 = 30 – 6k + 20 – 4k

⇒ � 70 = 50 – 10k

⇒ 70 – 50 = -10k and ⇒ -70 – 50 = -10k

⇒ 20 = -10k and ⇒ -120 = -10k

⇒ k = -20/10 and ⇒ k = 120/10

⇒ k = -2 and ⇒ k = 12

###### Exercise 4.4

**Question 1.**Write Minors and Cofactors of the elements of following determinants:

**Answer:**Minor: Minor of an element a_{ij} of a determinant is the determinant obtained by removing i^{th} row and j^{th} column in which element a_{ij} lies. It is denoted by M_{ij}.

Cofactor: Cofactor of an element a_{ij}, A_{ij} = (-1)^{i+j} M_{ij}.

Minor of element a_{ij} = M_{ij}

a_{11} = 2, Minor of element a_{11} = M_{11} = 3

Here removing 1^{st} row and 1^{st} column from the determinant we are left out with 3 so M_{11} = 3.

Similarly, finding other Minors of the determinant

a_{12} = -4, Minor of element a_{12} = M_{12} = 0

a_{21} = 0, Minor of element a_{21} = M_{21} = -4

a_{22} = 3, Minor of element a_{22} = M_{22} = 2

Cofactor of an element a_{ij}, A_{ij} = (-1)^{i+j} × M_{ij}

A_{11} = (-1)^{1+1} × M_{11} = 1 × 3 = 3

A_{12} = (-1)^{1+2} × M_{12} = (-1) × 0 = 0

A_{21} = (-1)^{2+1} × M_{11} = (-1) × (-4) = 4

A_{22} = (-1)^{2+2} × M_{22} = 1 × 2 = 2

**Question 2.**Write Minors and Cofactors of the elements of following determinants:

**Answer:**

Minor of an element a_{ij} = M_{ij}

a_{11} = a, Minor of element a_{11} = M_{11} = d

Here removing 1^{st} row and 1^{st} column from the determinant we are left out with d so M_{11} = d.

Similarly, finding other Minors of the determinant

a_{12} = c, Minor of element a_{12} = M_{12} = b

a_{21} = b, Minor of element a_{21} = M_{21} = c

a_{22} = d, Minor of element a_{22} = M_{22} = a

Cofactor of an element a_{ij}, A_{ij} = (-1)^{i+j} × M_{ij}

A_{11} = (-1)^{1+1} × M_{11} = 1 × d = d

A_{12} = (-1)^{1+2} × M_{12} = (-1) × b = -b

A_{21} = (-1)^{2+1} × M_{11} = (-1) × c = -c

A_{22} = (-1)^{2+2} × M_{22} = 1 × a = a

**Question 3.**Write Minors and Cofactors of the elements of following determinants:

**Answer:**Minor of an element a_{ij} = M_{ij}

a_{11} = 1, Minor of element a_{11} = M_{11} = = (1 × 1) – (0 × 0) = 1

Here removing 1^{st} row and 1^{st} column from the determinant we are left out with the determinant. Solving this we get M_{11} = 1

Similarly, finding other Minors of the determinant

a_{12} = 0, Minor of element a_{12} = M_{12} = = (0 × 1) – (0 × 0) = 0

a_{13} = 0, Minor of element a_{13} = M_{13} = = (0 × 0) - (1 × 0) = 0

a_{21} = 0, Minor of element a_{21} = M_{21} = = (0 × 1) – (0 × 0) = 0

a_{22} = 1, Minor of element a_{22} = M_{22} = = (1 × 1) – (0 × 0) = 1

a_{23} = 0, Minor of element a_{23} = M_{23} = = (1 × 0) – (0 × 0) = 0

a_{31} = 0, Minor of element a_{31} = M_{31} = = (0 × 0) – (0 × 1) = 0

a_{32} = 0, Minor of element a_{32} = M_{32} = = (1 × 0) – (0 × 0) = 0

a_{33} = 1, Minor of element a_{33} = M_{33} = = (1 × 1) – (0 × 0) = 1

Cofactor of an element a_{ij}, A_{ij} = (-1)^{i+j} × M_{ij}

A_{11} = (-1)^{1+1} × M_{11} = 1 × 1 = 1

A_{12} = (-1)^{1+2} × M_{12} = (-1) × 0 = 0

A_{13} = (-1)^{1+3} × M_{13} = 1 × 0 = 0

A_{21} = (-1)^{2+1} × M_{21} = (-1) × 0 = 0

A_{22} = (-1)^{2+2} × M_{22} = 1 × 1 = 1

A_{23} = (-1)^{2+3} × M_{23} = (-1) × 0 = 0

A_{31} = (-1)^{3+1} × M_{31} = 1 × 0 = 0

A_{32} = (-1)^{3+2} × M_{32} = (-1) × 0 = 0

A_{33} = (-1)^{3+3} × M_{33} = 1 × 1 = 1

**Question 4.**Write Minors and Cofactors of the elements of following determinants:

**Answer:**

Minor of an element a_{ij} = M_{ij}

a_{11} = 1, Minor of element a_{11} = M_{11} = = (5 × 2) – ((-1) × 1) = 10 + 1 = 11

Here removing 1^{st} row and 1^{st} column from the determinant we are left out with the determinant. Solving this we get M_{11} = 11

Similarly, finding other Minors of the determinant

a_{12} = 0, Minor of element a_{12} = M_{12} = = (3 × 2) – ((-1) × 0) = (6 - 0) = 6

a_{13} = 4, Minor of element a_{13} = M_{13} = = (3 × 1) – (5 × 0) = 3 - 0 = 3

a_{21} = 3, Minor of element a_{21} = M_{21} = = (0 × 2) – (4 × 1) = 0 – 4 = -4

a_{22} = 5, Minor of element a_{22} = M_{22} = = (1 × 2) – (4 × 0) = 2 – 0 = 2

a_{23} = -1, Minor of element a_{23} = M_{23} = = (1 × 1) – (0 × 0) = 1

a_{31} = 0, Minor of element a_{31} = M_{31} = = (0 × (-1)) – (4 × 5) = 0 – 20 = -20

a_{32} = 1, Minor of element a_{32} = M_{32} = = (1 × (-1)) – (4 × 3) = -1 – 12 = -13

a_{33} = 2, Minor of element a_{33} = M_{33} = = (1 × 5) – (0 × 3) = (5 – 0) = 5

Cofactor of an element a_{ij}, A_{ij} = (-1)^{i+j} × M_{ij}

A_{11} = (-1)^{1+1} × M_{11} = 1 × 11 = 11

A_{12} = (-1)^{1+2} × M_{12} = (-1) × 6 = -6

A_{13} = (-1)^{1+3} × M_{13} = 1 × 3 = 3

A_{21} = (-1)^{2+1} × M_{21} = (-1) × (-4) = 4

A_{22} = (-1)^{2+2} × M_{22} = 1 × 2 = 2

A_{23} = (-1)^{2+3} × M_{23} = (-1) × 1 = -1

A_{31} = (-1)^{3+1} × M_{31} = 1 × (-20) = -20

A_{32} = (-1)^{3+2} × M_{32} = (-1) × (-13) = 13

A_{33} = (-1)^{3+3} × M_{33} = 1 × 5 = 5

**Question 5.**Using Cofactors of elements of second row, evaluate

**Answer:**To evaluate a determinant using cofactors, Let

B =

Expanding along Row 1

B =

B = a_{11} A_{11} + a_{12} A_{12} + a_{13} A_{13}

[Where A_{ij} represents cofactors of a_{ij} of determinant B.]

B = Sum of product of elements of R_{1} with their corresponding cofactors

Similarly, the determinant can be solved by expanding along column

So, B = sum of product of elements of any row or column with their corresponding cofactors

Cofactors of second row

A_{21} = (-1)^{2+1} × M_{21} = (-1) × = (-1) × (3 × 3 – 8 × 2) = (-1) × (-7) = 7

A_{22} = (-1)^{2+2} × M_{22} = 1 × = (5 × 3 – 8 × 1) = 7

A_{23} = (-1)^{2+3} × M_{23} = (-1) × = (-1) × (5 × 2 – 3 × 1) = (-1) × 7 = -7

[Where A_{ij} = (-1)^{i+j} × M_{ij}, M_{ij} = Minor of i^{th} row & j^{th} column]

Therefore,

Î” = a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23}

Î” = 2 × 7 + 1 × (-7) = 14 - 7 = 7

Ans: Î” = 7

**Question 6.**Using Cofactors of elements of third column, evaluate.

**Answer:**To evaluate a determinant using cofactors, Let

B =

Expanding along Row 1

B =

B = a_{11} A_{11} + a_{12} A_{12} + a_{13} A_{13}

[Where A_{ij} represents cofactors of a_{ij} of determinant B.]

B = Sum of product of elements of R_{1} with their corresponding cofactors

Similarly, the determinant can be solved by expanding along column

So, B = sum of product of elements of any row or column with their corresponding cofactors

Cofactors of third column

A_{13} = (-1)^{1+3} × M_{13} = 1 × = 1 × (1 × z – 1 × y) = (z – y)

A_{23} = (-1)^{2+3} × M_{23} = (-1) × = (-1) × (1 × z – 1 × x) = - (z - x) = (x - z)

A_{33} = (-1)^{3+3} × M_{33} = 1 × = 1 × (1 × y – 1 × x) = (y – x)

[Where A_{ij} = (-1)^{i+j} × M_{ij}, M_{ij} = Minor of i^{th} row & j^{th} column]

Therefore,

Î” = a_{13}A_{13} + a_{23}A_{23} + a_{33}A_{33}

Î” = yz (z - y) + zx (x - z) + xy (y - x) = z [y (z - y) + x (x - z)] + xy (y - x)

Î” = z (yz - y^{2} + x^{2} - xz) + xy (y - x) = z [(yz - xz) + (x^{2} - y^{2})] + xy (y - x)

Î” = z [z × (y - x) + (x + y) × (x - y)] + xy (y - x)

Î” = z × (y - x) × (z – x - y) + xy (y - x)

Î” = (y - x) × (z^{2} – xz – yz + xy)

Î” = (y - x) × [z (z - x) – y (z - x)] = (y - x) × (z - y) × (z - x)

Î” = (x - y) (y - z) (z - x)

Ans: Î” = (x - y) (y - z) (z - x)

**Question 7.**If and A_{ij} is Cofactors of a_{ij}, then value of Î” is given by

A. a_{11} A_{31}+ a_{12} A_{32} + a_{13} A_{33}

B. a_{11} A_{11}+ a_{12} A_{21} + a_{13} A_{31}

C. a_{21} A_{11}+ a_{22} A_{12} + a_{23} A_{13}

D. a_{11} A_{11}+ a_{21} A_{21} + a_{31} A_{31}

**Answer:**Î” =

Expanding along Column 1

Î” =

Î” = a_{11}A_{11} + a_{21}A_{21} + a_{31}A_{31}

**Question 1.**

Write Minors and Cofactors of the elements of following determinants:

**Answer:**

Minor: Minor of an element a_{ij} of a determinant is the determinant obtained by removing i^{th} row and j^{th} column in which element a_{ij} lies. It is denoted by M_{ij}.

Cofactor: Cofactor of an element a_{ij}, A_{ij} = (-1)^{i+j} M_{ij}.

Minor of element a_{ij} = M_{ij}

a_{11} = 2, Minor of element a_{11} = M_{11} = 3

Here removing 1^{st} row and 1^{st} column from the determinant we are left out with 3 so M_{11} = 3.

Similarly, finding other Minors of the determinant

a_{12} = -4, Minor of element a_{12} = M_{12} = 0

a_{21} = 0, Minor of element a_{21} = M_{21} = -4

a_{22} = 3, Minor of element a_{22} = M_{22} = 2

Cofactor of an element a_{ij}, A_{ij} = (-1)^{i+j} × M_{ij}

A_{11} = (-1)^{1+1} × M_{11} = 1 × 3 = 3

A_{12} = (-1)^{1+2} × M_{12} = (-1) × 0 = 0

A_{21} = (-1)^{2+1} × M_{11} = (-1) × (-4) = 4

A_{22} = (-1)^{2+2} × M_{22} = 1 × 2 = 2

**Question 2.**

Write Minors and Cofactors of the elements of following determinants:

**Answer:**

Minor of an element a_{ij} = M_{ij}

a_{11} = a, Minor of element a_{11} = M_{11} = d

Here removing 1^{st} row and 1^{st} column from the determinant we are left out with d so M_{11} = d.

Similarly, finding other Minors of the determinant

a_{12} = c, Minor of element a_{12} = M_{12} = b

a_{21} = b, Minor of element a_{21} = M_{21} = c

a_{22} = d, Minor of element a_{22} = M_{22} = a

Cofactor of an element a_{ij}, A_{ij} = (-1)^{i+j} × M_{ij}

A_{11} = (-1)^{1+1} × M_{11} = 1 × d = d

A_{12} = (-1)^{1+2} × M_{12} = (-1) × b = -b

A_{21} = (-1)^{2+1} × M_{11} = (-1) × c = -c

A_{22} = (-1)^{2+2} × M_{22} = 1 × a = a

**Question 3.**

Write Minors and Cofactors of the elements of following determinants:

**Answer:**

Minor of an element a_{ij} = M_{ij}

a_{11} = 1, Minor of element a_{11} = M_{11} = = (1 × 1) – (0 × 0) = 1

Here removing 1^{st} row and 1^{st} column from the determinant we are left out with the determinant. Solving this we get M_{11} = 1

Similarly, finding other Minors of the determinant

a_{12} = 0, Minor of element a_{12} = M_{12} = = (0 × 1) – (0 × 0) = 0

a_{13} = 0, Minor of element a_{13} = M_{13} = = (0 × 0) - (1 × 0) = 0

a_{21} = 0, Minor of element a_{21} = M_{21} = = (0 × 1) – (0 × 0) = 0

a_{22} = 1, Minor of element a_{22} = M_{22} = = (1 × 1) – (0 × 0) = 1

a_{23} = 0, Minor of element a_{23} = M_{23} = = (1 × 0) – (0 × 0) = 0

a_{31} = 0, Minor of element a_{31} = M_{31} = = (0 × 0) – (0 × 1) = 0

a_{32} = 0, Minor of element a_{32} = M_{32} = = (1 × 0) – (0 × 0) = 0

a_{33} = 1, Minor of element a_{33} = M_{33} = = (1 × 1) – (0 × 0) = 1

Cofactor of an element a_{ij}, A_{ij} = (-1)^{i+j} × M_{ij}

A_{11} = (-1)^{1+1} × M_{11} = 1 × 1 = 1

A_{12} = (-1)^{1+2} × M_{12} = (-1) × 0 = 0

A_{13} = (-1)^{1+3} × M_{13} = 1 × 0 = 0

A_{21} = (-1)^{2+1} × M_{21} = (-1) × 0 = 0

A_{22} = (-1)^{2+2} × M_{22} = 1 × 1 = 1

A_{23} = (-1)^{2+3} × M_{23} = (-1) × 0 = 0

A_{31} = (-1)^{3+1} × M_{31} = 1 × 0 = 0

A_{32} = (-1)^{3+2} × M_{32} = (-1) × 0 = 0

A_{33} = (-1)^{3+3} × M_{33} = 1 × 1 = 1

**Question 4.**

Write Minors and Cofactors of the elements of following determinants:

**Answer:**

Minor of an element a_{ij} = M_{ij}

a_{11} = 1, Minor of element a_{11} = M_{11} = = (5 × 2) – ((-1) × 1) = 10 + 1 = 11

Here removing 1^{st} row and 1^{st} column from the determinant we are left out with the determinant. Solving this we get M_{11} = 11

Similarly, finding other Minors of the determinant

a_{12} = 0, Minor of element a_{12} = M_{12} = = (3 × 2) – ((-1) × 0) = (6 - 0) = 6

a_{13} = 4, Minor of element a_{13} = M_{13} = = (3 × 1) – (5 × 0) = 3 - 0 = 3

a_{21} = 3, Minor of element a_{21} = M_{21} = = (0 × 2) – (4 × 1) = 0 – 4 = -4

a_{22} = 5, Minor of element a_{22} = M_{22} = = (1 × 2) – (4 × 0) = 2 – 0 = 2

a_{23} = -1, Minor of element a_{23} = M_{23} = = (1 × 1) – (0 × 0) = 1

a_{31} = 0, Minor of element a_{31} = M_{31} = = (0 × (-1)) – (4 × 5) = 0 – 20 = -20

a_{32} = 1, Minor of element a_{32} = M_{32} = = (1 × (-1)) – (4 × 3) = -1 – 12 = -13

a_{33} = 2, Minor of element a_{33} = M_{33} = = (1 × 5) – (0 × 3) = (5 – 0) = 5

Cofactor of an element a_{ij}, A_{ij} = (-1)^{i+j} × M_{ij}

A_{11} = (-1)^{1+1} × M_{11} = 1 × 11 = 11

A_{12} = (-1)^{1+2} × M_{12} = (-1) × 6 = -6

A_{13} = (-1)^{1+3} × M_{13} = 1 × 3 = 3

A_{21} = (-1)^{2+1} × M_{21} = (-1) × (-4) = 4

A_{22} = (-1)^{2+2} × M_{22} = 1 × 2 = 2

A_{23} = (-1)^{2+3} × M_{23} = (-1) × 1 = -1

A_{31} = (-1)^{3+1} × M_{31} = 1 × (-20) = -20

A_{32} = (-1)^{3+2} × M_{32} = (-1) × (-13) = 13

A_{33} = (-1)^{3+3} × M_{33} = 1 × 5 = 5

**Question 5.**

Using Cofactors of elements of second row, evaluate

**Answer:**

To evaluate a determinant using cofactors, Let

B =

Expanding along Row 1

B =

B = a_{11} A_{11} + a_{12} A_{12} + a_{13} A_{13}

[Where A_{ij} represents cofactors of a_{ij} of determinant B.]

B = Sum of product of elements of R_{1} with their corresponding cofactors

Similarly, the determinant can be solved by expanding along column

So, B = sum of product of elements of any row or column with their corresponding cofactors

Cofactors of second row

A_{21} = (-1)^{2+1} × M_{21} = (-1) × = (-1) × (3 × 3 – 8 × 2) = (-1) × (-7) = 7

A_{22} = (-1)^{2+2} × M_{22} = 1 × = (5 × 3 – 8 × 1) = 7

A_{23} = (-1)^{2+3} × M_{23} = (-1) × = (-1) × (5 × 2 – 3 × 1) = (-1) × 7 = -7

[Where A_{ij} = (-1)^{i+j} × M_{ij}, M_{ij} = Minor of i^{th} row & j^{th} column]

Therefore,

Î” = a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23}

Î” = 2 × 7 + 1 × (-7) = 14 - 7 = 7

Ans: Î” = 7

**Question 6.**

Using Cofactors of elements of third column, evaluate.

**Answer:**

To evaluate a determinant using cofactors, Let

B =

Expanding along Row 1

B =

B = a_{11} A_{11} + a_{12} A_{12} + a_{13} A_{13}

[Where A_{ij} represents cofactors of a_{ij} of determinant B.]

B = Sum of product of elements of R_{1} with their corresponding cofactors

Similarly, the determinant can be solved by expanding along column

So, B = sum of product of elements of any row or column with their corresponding cofactors

Cofactors of third column

A_{13} = (-1)^{1+3} × M_{13} = 1 × = 1 × (1 × z – 1 × y) = (z – y)

A_{23} = (-1)^{2+3} × M_{23} = (-1) × = (-1) × (1 × z – 1 × x) = - (z - x) = (x - z)

A_{33} = (-1)^{3+3} × M_{33} = 1 × = 1 × (1 × y – 1 × x) = (y – x)

[Where A_{ij} = (-1)^{i+j} × M_{ij}, M_{ij} = Minor of i^{th} row & j^{th} column]

Therefore,

Î” = a_{13}A_{13} + a_{23}A_{23} + a_{33}A_{33}

Î” = yz (z - y) + zx (x - z) + xy (y - x) = z [y (z - y) + x (x - z)] + xy (y - x)

Î” = z (yz - y^{2} + x^{2} - xz) + xy (y - x) = z [(yz - xz) + (x^{2} - y^{2})] + xy (y - x)

Î” = z [z × (y - x) + (x + y) × (x - y)] + xy (y - x)

Î” = z × (y - x) × (z – x - y) + xy (y - x)

Î” = (y - x) × (z^{2} – xz – yz + xy)

Î” = (y - x) × [z (z - x) – y (z - x)] = (y - x) × (z - y) × (z - x)

Î” = (x - y) (y - z) (z - x)

Ans: Î” = (x - y) (y - z) (z - x)

**Question 7.**

If and A_{ij} is Cofactors of a_{ij}, then value of Î” is given by

A. a_{11} A_{31}+ a_{12} A_{32} + a_{13} A_{33}

B. a_{11} A_{11}+ a_{12} A_{21} + a_{13} A_{31}

C. a_{21} A_{11}+ a_{22} A_{12} + a_{23} A_{13}

D. a_{11} A_{11}+ a_{21} A_{21} + a_{31} A_{31}

**Answer:**

Î” =

Expanding along Column 1

Î” =

Î” = a_{11}A_{11} + a_{21}A_{21} + a_{31}A_{31}

###### Exercise 4.5

**Question 1.**Find adjoint of each of the matrices.

**Answer:**Adjoint of the matrix A = [a_{ij}]_{n×n} is defined as the transpose of the matrix [A_{ij}]_{n×n} where A_{ij} is the co-factor of the element a_{ij}.

Let’s find the cofactors for all the positions first-

Here, A_{11} = 4, A_{12} = -3, A_{21} = -2, A_{22} = 1.

∴ Adj A =

=

**Question 2.**Find adjoint of each of the matrices.

**Answer:**Adjoint of the matrix A = [a_{ij}]_{n×n} is defined as the transpose of the matrix [A_{ij}]_{n×n} where A_{ij} is the co-factor of the element a_{ij}.

Let’s find the cofactors for all the positions first-

Here, A_{11} = 1{(3×1-0×5)} = 3

Similarly,

A_{12} = -12, A_{13} = 6, A_{21} = 1, A_{22} = 5, A_{23} = 2, A_{31} = -11, A_{32} = -1, A_{33} = 5.

**Question 3.**Verify A (adj A) = (adj A) A = |A|

**Answer:**Adjoint of the matrix A = [a_{ij}]_{n×n} is defined as the transpose of the matrix [A_{ij}]_{n×n} where A_{ij} is the co-factor of the element a_{ij}.

Let’s find the cofactors for all the positions first-

Here, A_{11} = -6, A_{12} = 4, A_{21} = -3, A_{22} = 2.

∴ Adj A =

=

So LHS = A(AdjA) =

Also AdjA(A) =

Determinant of A = |A| = 2(-6)-(3)(-4) = 0

So RHS = |A|I = 0

Hence A(AdjA) = AdjA(A) = |A|I = 0 {hence proved}

**Question 4.**Verify A (adj A) = (adj A) A = |A|

**Answer:**Adjoint of the matrix A = [a_{ij}]_{n×n} is defined as the transpose of the matrix [A_{ij}]_{n×n} where A_{ij} is the co-factor of the element a_{ij}.

Let’s find the cofactors for all the positions first-

Here, A_{11} = 0, A_{12} = -11, A_{13} = 0, A_{21} = 3, A_{22} = 1, A_{23} = -1, A_{31} = 2, A_{32} = 8, A_{33} = 3.

∴ Adj A =

=

So, LHS = A(AdjA) =

Also AdjA(A) =

Determinant of A = |A| = 11

So RHS = |A|I = .

Hence A(AdjA) = AdjA(A) = |A|I = {hence proved}

**Question 5.**Find the inverse of each of the matrices (if it exists)

**Answer:**We know that

Adjoint of the matrix A = [a_{ij}]_{n×n} is defined as the transpose of the matrix [A_{ij}]_{n×n} where A_{ij} is the co-factor of the element a_{ij}.

Let’s find the cofactors for all the positions first-

Here, A_{11} = 3, A_{12} = -4, A_{21} = 2, A_{22} = 2.

∴ Adj A =

=

And |A| = 2(3)-(-2)(4) = 14

So .

**Question 6.**Find the inverse of each of the matrices (if it exists)

**Answer:**We know that

Adjoint of the matrix A = [a_{ij}]_{n×n} is defined as the transpose of the matrix [A_{ij}]_{n×n} where A_{ij} is the co-factor of the element a_{ij}.

Let’s find the cofactors for all the positions first-

Here, A_{11} = 2, A_{12} = 3, A_{21} = -5, A_{22} = -1.

∴ Adj A =

=

And |A| = -1(2)-(-3)(5) = 13

So

**Question 7.**Find the inverse of each of the matrices (if it exists)

**Answer:**Adjoint of the matrix A = [a_{ij}]_{n×n} is defined as the transpose of the matrix [A_{ij}]_{n×n} where A_{ij} is the co-factor of the element a_{ij}.

Let’s find the cofactors for all the positions first-

Here, A_{11} = 10, A_{12} = 0, A_{13} = 0, A_{21} = -10, A_{22} = 5, A_{23} = 0, A_{31} = 2, A_{32} = -4, A_{33} = 2.

∴ Adj A =

And |A| = 10.

**Question 8.**Find the inverse of each of the matrices (if it exists)

**Answer:**Adjoint of the matrix A = [a_{ij}]_{n×n} is defined as the transpose of the matrix [A_{ij}]_{n×n} where A_{ij} is the co-factor of the element a_{ij}.

Let’s find the cofactors for all the positions first-

Here, A_{11} = -3, A_{12} = 3, A_{13} = -9, A_{21} = 0, A_{22} = -1, A_{23} = -2, A_{31} = 0, A_{32} = 0, A_{33} = 3.

∴ Adj A =

And |A| = -3.

.

**Question 9.**Find the inverse of each of the matrices (if it exists)

**Answer:**Adjoint of the matrix A = [a_{ij}]_{n×n} is defined as the transpose of the matrix [A_{ij}]_{n×n} where A_{ij} is the co-factor of the element a_{ij}.

Let’s find the cofactors for all the positions first-

Here, A_{11} = -1, A_{12} = -4, A_{13} = 1, A_{21} = 5, A_{22} = 23, A_{23} = -11, A_{31} = 3, A_{32} = 12, A_{33} = -6.

∴ Adj A =

= .

And |A| = -3.

.

**Question 10.**Find the inverse of each of the matrices (if it exists)

**Answer:**Adjoint of the matrix A = [a_{ij}]_{n×n} is defined as the transpose of the matrix [A_{ij}]_{n×n} where A_{ij} is the co-factor of the element a_{ij}.

Let’s find the cofactors for all the positions first-

Here, A_{11} = 2, A_{12} = -9, A_{13} = -6, A_{21} = 0, A_{22} = -2, A_{23} = -1, A_{31} = -1, A_{32} = 3, A_{33} = 2.

∴ Adj A =

=

And |A| = -1.

**Question 11.**Find the inverse of each of the matrices (if it exists)

**Answer:**Adjoint of the matrix A = [a_{ij}]_{n×n} is defined as the transpose of the matrix [A_{ij}]_{n×n} where A_{ij} is the co-factor of the element a_{ij}.

Let’s find the cofactors for all the positions first-

Here, A_{11} = -1, A_{12} = 0, A_{13} = 0, A_{21} = 0, A_{22} = -cosÎ±, A_{23} = -sinÎ±, A_{31} = 0, A_{32} = -sinÎ±, A_{33} = cosÎ±.

∴ Adj A =

And |A| = 1.

.

**Question 12.**Let . Verify that (AB)^{–1} = B^{–1} A^{–1}.

**Answer:**We have AB = = (61)(67)-(47)(87) = -2

Here determinant of matrix = |AB|≠ 0 hence (AB)^{-1} exists.

Also |A| = 1 ≠ 0 and |B| = -2 ≠ 0.

∴ A^{-1} and B^{-1} will also exist and are given by-

And hence,

{Hence proved}

**Question 13.**If , show that A^{2} – 5A + 7I = O. Hence find A^{–1}.

**Answer:**We have A^{2} = A.A = .

So A^{2} – 5A + 7I =

Hence A^{2} – 5A + 7I = 0

∴ A.A – 5A = -7I

Now post multiply with A^{-1}

So A.A.A^{-1}-5A.A^{-1} = -7I.A^{-1}

→ A.I – 5I = -7I.A^{-1} {since A.A^{-1} = I}

A – 5I = -7A^{-1} {since X.I = X}

**Question 14.**For the matrix , find the numbers a and b such that A^{2} + aA + bI = O.

**Answer:**We have A^{2} = A.A =

Since A^{2} + aA + bI =

So A^{2} + aA + bI =

Hence 10+3a+b = 0 …(i)

5+a = 0 …(ii)

5+2a+b = 0 …(iii)

From (ii) a = -5

Putting a in (iii) we get b = 5

So a = -5 and b = 5 satisfy the equation.

**Question 15.**For the matrix

Show that A^{3}– 6A^{2} + 5A + 11 I = O. Hence, find A^{–1}.

**Answer:**Here A^{2} = A.A =

And hence A^{3} = A. A^{2} =

∴ A^{3}– 6A^{2} + 5A + 11 I =

Thus, A^{3}– 6A^{2} + 5A + 11 I = 0

Now, A^{3}– 6A^{2} + 5A + 11 I = 0,

→ (A.A.A)- 6 (A.A) +5A = -11I

Post-multiply with A^{-1} on both sides-

→ (A.A.A.A^{-1})- 6 (A.A.A^{-1}) +5A.A^{-1} = -11I. A^{-1}

→ (A.A.I) – 6(A.I) + 5I = -11I. A^{-1} {since A.A^{-1} = I}

→ (A.A) – 6A +5I = -11A^{-1} {since X.I = X}

Hence

**Question 16.**If Verify that A^{3} – 6A^{2} + 9A – 4I = O and hence find A^{-1}.

**Answer:**Here A^{2} = A.A =

And hence A^{3} = A. A^{2} =

**Question 1.**

Find adjoint of each of the matrices.

**Answer:**

Adjoint of the matrix A = [a_{ij}]_{n×n} is defined as the transpose of the matrix [A_{ij}]_{n×n} where A_{ij} is the co-factor of the element a_{ij}.

Let’s find the cofactors for all the positions first-

Here, A_{11} = 4, A_{12} = -3, A_{21} = -2, A_{22} = 1.

∴ Adj A =

=

**Question 2.**

Find adjoint of each of the matrices.

**Answer:**

Adjoint of the matrix A = [a_{ij}]_{n×n} is defined as the transpose of the matrix [A_{ij}]_{n×n} where A_{ij} is the co-factor of the element a_{ij}.

Let’s find the cofactors for all the positions first-

Here, A_{11} = 1{(3×1-0×5)} = 3

Similarly,

A_{12} = -12, A_{13} = 6, A_{21} = 1, A_{22} = 5, A_{23} = 2, A_{31} = -11, A_{32} = -1, A_{33} = 5.

**Question 3.**

Verify A (adj A) = (adj A) A = |A|

**Answer:**

Adjoint of the matrix A = [a_{ij}]_{n×n} is defined as the transpose of the matrix [A_{ij}]_{n×n} where A_{ij} is the co-factor of the element a_{ij}.

Let’s find the cofactors for all the positions first-

Here, A_{11} = -6, A_{12} = 4, A_{21} = -3, A_{22} = 2.

∴ Adj A =

=

So LHS = A(AdjA) =

Also AdjA(A) =

Determinant of A = |A| = 2(-6)-(3)(-4) = 0

So RHS = |A|I = 0

Hence A(AdjA) = AdjA(A) = |A|I = 0 {hence proved}

**Question 4.**

Verify A (adj A) = (adj A) A = |A|

**Answer:**

_{ij}]_{n×n} is defined as the transpose of the matrix [A_{ij}]_{n×n} where A_{ij} is the co-factor of the element a_{ij}.

Let’s find the cofactors for all the positions first-

Here, A_{11} = 0, A_{12} = -11, A_{13} = 0, A_{21} = 3, A_{22} = 1, A_{23} = -1, A_{31} = 2, A_{32} = 8, A_{33} = 3.

∴ Adj A =

=

So, LHS = A(AdjA) =

Also AdjA(A) =

Determinant of A = |A| = 11

So RHS = |A|I = .

Hence A(AdjA) = AdjA(A) = |A|I = {hence proved}

**Question 5.**

Find the inverse of each of the matrices (if it exists)

**Answer:**

We know that

_{ij}]_{n×n} is defined as the transpose of the matrix [A_{ij}]_{n×n} where A_{ij} is the co-factor of the element a_{ij}.

Let’s find the cofactors for all the positions first-

Here, A_{11} = 3, A_{12} = -4, A_{21} = 2, A_{22} = 2.

∴ Adj A =

=

And |A| = 2(3)-(-2)(4) = 14

So .

**Question 6.**

Find the inverse of each of the matrices (if it exists)

**Answer:**

We know that

_{ij}]_{n×n} is defined as the transpose of the matrix [A_{ij}]_{n×n} where A_{ij} is the co-factor of the element a_{ij}.

Let’s find the cofactors for all the positions first-

Here, A_{11} = 2, A_{12} = 3, A_{21} = -5, A_{22} = -1.

∴ Adj A =

=

And |A| = -1(2)-(-3)(5) = 13

So

**Question 7.**

Find the inverse of each of the matrices (if it exists)

**Answer:**

_{ij}]_{n×n} is defined as the transpose of the matrix [A_{ij}]_{n×n} where A_{ij} is the co-factor of the element a_{ij}.

Let’s find the cofactors for all the positions first-

Here, A_{11} = 10, A_{12} = 0, A_{13} = 0, A_{21} = -10, A_{22} = 5, A_{23} = 0, A_{31} = 2, A_{32} = -4, A_{33} = 2.

∴ Adj A =

And |A| = 10.

**Question 8.**

Find the inverse of each of the matrices (if it exists)

**Answer:**

_{ij}]_{n×n} is defined as the transpose of the matrix [A_{ij}]_{n×n} where A_{ij} is the co-factor of the element a_{ij}.

Let’s find the cofactors for all the positions first-

Here, A_{11} = -3, A_{12} = 3, A_{13} = -9, A_{21} = 0, A_{22} = -1, A_{23} = -2, A_{31} = 0, A_{32} = 0, A_{33} = 3.

∴ Adj A =

And |A| = -3.

.

**Question 9.**

Find the inverse of each of the matrices (if it exists)

**Answer:**

_{ij}]_{n×n} is defined as the transpose of the matrix [A_{ij}]_{n×n} where A_{ij} is the co-factor of the element a_{ij}.

Let’s find the cofactors for all the positions first-

Here, A_{11} = -1, A_{12} = -4, A_{13} = 1, A_{21} = 5, A_{22} = 23, A_{23} = -11, A_{31} = 3, A_{32} = 12, A_{33} = -6.

∴ Adj A =

= .

And |A| = -3.

.

**Question 10.**

Find the inverse of each of the matrices (if it exists)

**Answer:**

_{ij}]_{n×n} is defined as the transpose of the matrix [A_{ij}]_{n×n} where A_{ij} is the co-factor of the element a_{ij}.

Let’s find the cofactors for all the positions first-

Here, A_{11} = 2, A_{12} = -9, A_{13} = -6, A_{21} = 0, A_{22} = -2, A_{23} = -1, A_{31} = -1, A_{32} = 3, A_{33} = 2.

∴ Adj A =

=

And |A| = -1.

**Question 11.**

Find the inverse of each of the matrices (if it exists)

**Answer:**

_{ij}]_{n×n} is defined as the transpose of the matrix [A_{ij}]_{n×n} where A_{ij} is the co-factor of the element a_{ij}.

Let’s find the cofactors for all the positions first-

Here, A_{11} = -1, A_{12} = 0, A_{13} = 0, A_{21} = 0, A_{22} = -cosÎ±, A_{23} = -sinÎ±, A_{31} = 0, A_{32} = -sinÎ±, A_{33} = cosÎ±.

∴ Adj A =

And |A| = 1.

.

**Question 12.**

Let . Verify that (AB)^{–1} = B^{–1} A^{–1}.

**Answer:**

We have AB = = (61)(67)-(47)(87) = -2

Here determinant of matrix = |AB|≠ 0 hence (AB)^{-1} exists.

Also |A| = 1 ≠ 0 and |B| = -2 ≠ 0.

∴ A^{-1} and B^{-1} will also exist and are given by-

And hence,

{Hence proved}

**Question 13.**

If , show that A^{2} – 5A + 7I = O. Hence find A^{–1}.

**Answer:**

We have A^{2} = A.A = .

So A^{2} – 5A + 7I =

Hence A^{2} – 5A + 7I = 0

∴ A.A – 5A = -7I

Now post multiply with A^{-1}

So A.A.A^{-1}-5A.A^{-1} = -7I.A^{-1}

→ A.I – 5I = -7I.A^{-1} {since A.A^{-1} = I}

A – 5I = -7A^{-1} {since X.I = X}

**Question 14.**

For the matrix , find the numbers a and b such that A^{2} + aA + bI = O.

**Answer:**

We have A^{2} = A.A =

Since A^{2} + aA + bI =

So A^{2} + aA + bI =

Hence 10+3a+b = 0 …(i)

5+a = 0 …(ii)

5+2a+b = 0 …(iii)

From (ii) a = -5

Putting a in (iii) we get b = 5

So a = -5 and b = 5 satisfy the equation.

**Question 15.**

For the matrix

Show that A^{3}– 6A^{2} + 5A + 11 I = O. Hence, find A^{–1}.

**Answer:**

Here A^{2} = A.A =

And hence A^{3} = A. A^{2} =

∴ A^{3}– 6A^{2} + 5A + 11 I =

Thus, A^{3}– 6A^{2} + 5A + 11 I = 0

Now, A^{3}– 6A^{2} + 5A + 11 I = 0,

→ (A.A.A)- 6 (A.A) +5A = -11I

Post-multiply with A^{-1} on both sides-

→ (A.A.A.A^{-1})- 6 (A.A.A^{-1}) +5A.A^{-1} = -11I. A^{-1}

→ (A.A.I) – 6(A.I) + 5I = -11I. A^{-1} {since A.A^{-1} = I}

→ (A.A) – 6A +5I = -11A^{-1} {since X.I = X}

Hence

**Question 16.**

If Verify that A^{3} – 6A^{2} + 9A – 4I = O and hence find A^{-1}.

**Answer:**

Here A^{2} = A.A =

And hence A^{3} = A. A^{2} =