Determinants Class 12th Mathematics Part I CBSE Solution

Class 12^th Mathematics Part I CBSE Solution

Exercise 4.1

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

Exercise 4.2

1. | lll x+a y+b x+c | = 0 Using the property of determinants and without expanding…
2. | lll a-b b-c c-a | = 0 Using the property of determinants and without expanding…
3. | lll 2&7&65 3&8&75 5&9&86 | = 0 Using the property of determinants and without…
4. Using the property of determinants and without expanding in prove that:…
5. | ccc b+c+r+z c+a+p+x a+b+q+y | = 2 | ccc a b c | Using the property of…
6. | ccc 0&-b -a&0&-c b&0 | = 0 Using the property of determinants and without…
7. | ccc - a^2 ba& - b^2 ca& - c^2 | = 4a^2b^2c^2 Using the property of…
8. | lll 1& a^2 1& b^2 1& c^2 | = (a-b) (b-c) (c-a) By using properties of…
9. | lll 1&1&1 a a^3 & b^3 & c^3 | = (a-b) (b-c) (c-a) (a+b+c) By using properties…
10. | lll x& x^2 y& y^2 z& z^2 | = (x-y) (y-z) (z-x) (xy+yz+zx) By using properties…
11. | ccc x+4&2x&2x 2x+4&2x 2x&2x+4 | = (5x+4) (4-x)^2 By using properties of…
12. | ccc y+k y+k y+k | = k^2 (3y+k) By using properties of determinants, show…
13. | ccc a-b-c&2a&2a 2b&2b 2c&2c | = (a+b+c)^3 By using properties of…
14. | ccc x+y+2z z+z+2x z+x+2y | = 2 (x+y+z)^3 By using properties of…
15. | ccc 1& x^2 x^2 &1 x& x^2 &1 | = (1-x^3)^2 By using properties of…
16. | 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…
17. | ccc a^2 + 1 ab& b^2 + 1 ca& c^2 + 1 | = 1+a^2 + b^2 + c^2 By using properties…
18. Let A be a square matrix of order 3 × 3, then | kA| is equal toA. k|A| B. k^2…
19. Which of the following is correctA. Determinant is a square matrix. B.…

Exercise 4.3

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

Exercise 4.4

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

Exercise 4.5

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

Exercise 4.6

1. x + 2y = 2 2x + 3y = 3 Examine the consistency of the system of equations.…
2. 2x - y = 5 x + y = 4 Examine the consistency of the system of equations.…
3. x + 3y = 5 2x + 6y = 8 Examine the consistency of the system of equations.…
4. x + y + z = 1 2x + 3y + 2z = 2 ax + ay + 2az = 4 Examine the consistency of the…
5. 3x-y - 2z = 2 2y - z = -1 3x - 5y = 3 Examine the consistency of the system of…
6. 5x - y + 4z = 5 2x + 3y + 5z = 2 5x - 2y + 6z = -1 Examine the consistency of…
7. 5x + 2y = 4 7x + 3y = 5 Solve system of linear equations, using matrix method.…
8. 2x - y = -2 3x + 4y = 3 Solve system of linear equations, using matrix method.…
9. 4x - 3y = 3 3x - 5y = 7 Solve system of linear equations, using matrix method.…
10. 5x + 2y = 3 3x + 2y = 5 Solve system of linear equations, using matrix method.…
11. 2x+y+z = 1 x-2y-z = 3/2 3y-5z = 9 Solve system of linear equations, using…
12. x - y + z = 4 2x + y - 3z = 0 x + y + z = 2 Solve system of linear equations,…
13. 2x + 3y +3 z = 5 x - 2y + z = -4 3x - y - 2z = 3 Solve system of linear…
14. x - y + 2z = 7 3x + 4y - 5z = -5 2x - y + 3z = 12 Solve system of linear…
15. If a = [ccc 2&-3&5 3&2&-4 1&1&-2] , find A-1. Using A-1 solve the system of…
16. The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs. 60. The cost of 2 kg…

Miscellaneous Exercise

1. Prove that the determinant | ccc x heta -sintegrate heta &-x&1 costheta &1 | is…
2. Without expanding the determinant, prove that | ccc a& a^2 b& b^2 c& c^2 | = |…
3. Evaluate | ccc cosalpha cosbeta sinbeta &-sinalpha -sinbeta &0 sinalpha cosbeta…
4. If a, b and c are real numbers, and delta = | lll b+c+a+b c+a+b+c a+b+c+a | = 0…
5. Solve the equation | ccc x+a x+a x+a | = 0 , a not equal 0
6. Prove that | ccc a^2 & ac+c^2 a^2 + ab & b^2 ab& b^2 + bc & c^2 | = 4a^2b^2c^2…
7. 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…
8. Let a = [ccc 1&-2&1 -2&3&1 1&1&5] . Verify that [adj A]-1 = adj (A-1)…
9. Let a = [ccc 1&-2&1 -2&3&1 1&1&5] . Verify that (A-1)-1 = A
10. Evaluate | ccc x+y y+y x+y |
11. Evaluate | ccc 1 1+y 1+y |
12. Prove that | lll alpha & alpha^2 & beta + gamma beta & beta^2 & gamma + alpha…
13. Prove that | lll x& x^2 & 1+px^3 y& y^2 & 1+py^3 z& z^2 & 1+pz^3 | = (1 = pxyz)…
14. Prove that | ccc 3a&-a+b&-a+c -b+a&3b&-b+c -c+a&-c+b&3c | = 3 (a+b+c)…
15. Prove that | ccc 1&1+p&1+p+q 2&3+2p&4+3p+2q 3&6+3p&10+6p+3q | = 1…
16. Prove that | ccc sinalpha & cos (alpha + delta) sinbeta & cos (beta + delta)…
17. Solve the system of equations 2/x + 3/y + 10/z = 4 4/x - 6/y + 5/z = 1 6/x +…
18. If a, b, c, are in A.P, then the determinant | lll x+2+3+2a x+3+4+20 x+4+5+2c |…
19. If x, y, z are nonzero real numbers, then the inverse of matrix a = [lll x&0&0…
20. Let a = [ccc 1 heta &1 -sintegrate heta &1 heta -1&-sintegrate heta &1] , where…

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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, 

= cosθ(cosθ) - (-sinθ)(sinθ)

= cos²θ + sin²θ

= 1 [∵ cos²θ + sin²θ = 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, 

= (x² – x + 1)(x + 1) - (x - 1)(x + 1)

= (x³ - x² + x + x² – x + 1) - (x² - 1)

= x³ + 1 - x² + 1

= x³ - x² + 2

Ans. The determinant of the above matrix is x³ - x² + 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 = 2x² - 24

⇒ -18 = 2x² – 24

⇒ 2x² = -24 + 18

⇒ 2x² = 6

⇒ x² = 6/2

⇒ x² = 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)

⇒ x² - 36 = 36 – 36

⇒ x² =36 – 36 + 36

⇒ x² = 36

⇒ x = ±6

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Exercise 4.2

Question 1.

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

Answer:

Applying Operations C₁→ C₁ + C₂ (i.e. Replacing 1^st column by addition of 1^st and 2^nd column)

C₁→ C₁ - C₃ (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 C₁→ C₁ + C₂ (i.e. Replacing 1^st column by addition of 1^st and 2^nd column)

C₁→ C₁ + C₃ (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 C₁→ C₁ + 9C₂ (i.e. Replacing 1^st column by addition of 1^st column and 9 times second column)

C₁→ C₁ - C₃ (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:

C₃→ C₂ + C₃ (i.e. replace 3^rd column by addition of 2^nd and 3^rd column)

Taking ab + bc + ac outside determinant

C₁→ C₁ - C₃ (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 R₁↔ R₃ (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.

R₁↔ R₂ (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:



R₁→ cR₁ (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

R₁→ R₁ - bR₂ (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 = a² b² c² × 2(1 - (-1)) = 4a²b²c² = 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 a² - b² = (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 a³ - b³ = (a - b)(a² + ab + b²) and b³ - c³ = (b - c)(b² + bc + c²)

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 a² - c²) = (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 x² - y² = (x + y)(x - y) and y² - z² = (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)(z² - xy - xz - yz - z²)

Cancelling z² 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) k² = 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 - x³) outside the determinant from 1^st row

.

Expanding the determinant along 1^st row

HS = (1 - x³)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 + a² + b²) outside the determinant from 1^st and 2^nd row

Expanding the determinant along 1^st row

∴ LHS = (1 + a² + b²)² [1 + a² - b² - (-2b²)]

∴ LHS = (1 + a² + b² )² (1 + a² + b²)

∴ LHS = (1 + a² + b² )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 + a² + b² + c²) = RHS

∴

Question 18.

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

B. k² |A|

C. k³ |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| = k³|A|

Therefore, answer is option (C) k³|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.

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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 (x₁, y₁), (x₂, y₂), (x₃, y₃)

Area of triangle is given by Δ = 

Area of triangle = Δ = 

Expanding the determinant along Row 1

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

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

Δ = 1/2 × (-3 + 18) = 15/2 sq. units

∴ Δ = 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 (x₁, y₁), (x₂, y₂), (x₃, y₃)

Area of triangle is given by Δ = 

Area of triangle = Δ = 

Expanding the determinant along Row 1

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

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

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

Δ = 1/2 × 47 sq. units = 47/2 sq. units

∴ Δ = 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 (x₁, y₁), (x₂, y₂), (x₃, y₃)

Area of triangle is given by Δ = 

Area of triangle = Δ = 

Expanding the determinant along Row 1

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

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

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

Δ = 1/2 × |-30| sq. units = 30/2 sq. units

∴ Δ = 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 (x₁, y₁), (x₂, y₂), (x₃, y₃)

Area of triangle is given by Δ = 

For points to be collinear area of triangle = Δ = 0

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

Area of triangle = Δ = 

Expanding the determinant along Row 1

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

Δ = 1/2 × [a × (c + a – a – b) – (b + c) × (b – c) + 1 × (ab + b² – c² - ca)]

Δ = 1/2 × [a × (c – b) – (b² – c²) + 1 × (ab + b² – c² – ca)]

Δ = 1/2 × (ac – ab – b² + c² + ab + b² – c² – ca) sq units

Δ = 1/2 × 0 sq units

∴ Δ = 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 (x₁, y₁), (x₂, y₂), (x₃, y₃)

Area of triangle is given by Δ = 

Given, Area of triangle = Δ = 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 (x₁, y₁), (x₂, y₂), (x₃, y₃)

Area of triangle is given by Δ = 

Given, Area of triangle = Δ = 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 (x₁, y₁) & (x₂, y₂) 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 (x₁, y₁) & (x₂, y₂) 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 (x₁, y₁), (x₂, y₂), (x₃, y₃)

Area of triangle is given by Δ = 

Given, Area of triangle = Δ = 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

a₁₁ = 2, Minor of element a₁₁ = M₁₁ = 3

Here removing 1^st row and 1^st column from the determinant we are left out with 3 so M₁₁ = 3.

Similarly, finding other Minors of the determinant

a₁₂ = -4, Minor of element a₁₂ = M₁₂ = 0

a₂₁ = 0, Minor of element a₂₁ = M₂₁ = -4

a₂₂ = 3, Minor of element a₂₂ = M₂₂ = 2

Cofactor of an element a_ij, A_ij = (-1)^i+j × M_ij

A₁₁ = (-1)¹⁺¹ × M₁₁ = 1 × 3 = 3

A₁₂ = (-1)¹⁺² × M₁₂ = (-1) × 0 = 0

A₂₁ = (-1)²⁺¹ × M₁₁ = (-1) × (-4) = 4

A₂₂ = (-1)²⁺² × M₂₂ = 1 × 2 = 2

Question 2.

Write Minors and Cofactors of the elements of following determinants:

Answer:

Minor of an element a_ij = M_ij

a₁₁ = a, Minor of element a₁₁ = M₁₁ = d

Here removing 1^st row and 1^st column from the determinant we are left out with d so M₁₁ = d.

Similarly, finding other Minors of the determinant

a₁₂ = c, Minor of element a₁₂ = M₁₂ = b

a₂₁ = b, Minor of element a₂₁ = M₂₁ = c

a₂₂ = d, Minor of element a₂₂ = M₂₂ = a

Cofactor of an element a_ij, A_ij = (-1)^i+j × M_ij

A₁₁ = (-1)¹⁺¹ × M₁₁ = 1 × d = d

A₁₂ = (-1)¹⁺² × M₁₂ = (-1) × b = -b

A₂₁ = (-1)²⁺¹ × M₁₁ = (-1) × c = -c

A₂₂ = (-1)²⁺² × M₂₂ = 1 × a = a

Question 3.

Write Minors and Cofactors of the elements of following determinants:

Answer:

Minor of an element a_ij = M_ij

a₁₁ = 1, Minor of element a₁₁ = M₁₁ =  = (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 M₁₁ = 1

Similarly, finding other Minors of the determinant

a₁₂ = 0, Minor of element a₁₂ = M₁₂ =  = (0 × 1) – (0 × 0) = 0

a₁₃ = 0, Minor of element a₁₃ = M₁₃ =  = (0 × 0) - (1 × 0) = 0

a₂₁ = 0, Minor of element a₂₁ = M₂₁ =  = (0 × 1) – (0 × 0) = 0

a₂₂ = 1, Minor of element a₂₂ = M₂₂ =  = (1 × 1) – (0 × 0) = 1

a₂₃ = 0, Minor of element a₂₃ = M₂₃ =  = (1 × 0) – (0 × 0) = 0

a₃₁ = 0, Minor of element a₃₁ = M₃₁ =  = (0 × 0) – (0 × 1) = 0

a₃₂ = 0, Minor of element a₃₂ = M₃₂ =  = (1 × 0) – (0 × 0) = 0

a₃₃ = 1, Minor of element a₃₃ = M₃₃ =  = (1 × 1) – (0 × 0) = 1

Cofactor of an element a_ij, A_ij = (-1)^i+j × M_ij

A₁₁ = (-1)¹⁺¹ × M₁₁ = 1 × 1 = 1

A₁₂ = (-1)¹⁺² × M₁₂ = (-1) × 0 = 0

A₁₃ = (-1)¹⁺³ × M₁₃ = 1 × 0 = 0

A₂₁ = (-1)²⁺¹ × M₂₁ = (-1) × 0 = 0

A₂₂ = (-1)²⁺² × M₂₂ = 1 × 1 = 1

A₂₃ = (-1)²⁺³ × M₂₃ = (-1) × 0 = 0

A₃₁ = (-1)³⁺¹ × M₃₁ = 1 × 0 = 0

A₃₂ = (-1)³⁺² × M₃₂ = (-1) × 0 = 0

A₃₃ = (-1)³⁺³ × M₃₃ = 1 × 1 = 1

Question 4.

Write Minors and Cofactors of the elements of following determinants:

Answer:

Minor of an element a_ij = M_ij

a₁₁ = 1, Minor of element a₁₁ = M₁₁ =  = (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 M₁₁ = 11

Similarly, finding other Minors of the determinant

a₁₂ = 0, Minor of element a₁₂ = M₁₂ = = (3 × 2) – ((-1) × 0) = (6 - 0) = 6

a₁₃ = 4, Minor of element a₁₃ = M₁₃ =  = (3 × 1) – (5 × 0) = 3 - 0 = 3

a₂₁ = 3, Minor of element a₂₁ = M₂₁ =  = (0 × 2) – (4 × 1) = 0 – 4 = -4

a₂₂ = 5, Minor of element a₂₂ = M₂₂ =  = (1 × 2) – (4 × 0) = 2 – 0 = 2

a₂₃ = -1, Minor of element a₂₃ = M₂₃ =  = (1 × 1) – (0 × 0) = 1

a₃₁ = 0, Minor of element a₃₁ = M₃₁ =  = (0 × (-1)) – (4 × 5) = 0 – 20 = -20

a₃₂ = 1, Minor of element a₃₂ = M₃₂ =  = (1 × (-1)) – (4 × 3) = -1 – 12 = -13

a₃₃ = 2, Minor of element a₃₃ = M₃₃ = = (1 × 5) – (0 × 3) = (5 – 0) = 5

Cofactor of an element a_ij, A_ij = (-1)^i+j × M_ij

A₁₁ = (-1)¹⁺¹ × M₁₁ = 1 × 11 = 11

A₁₂ = (-1)¹⁺² × M₁₂ = (-1) × 6 = -6

A₁₃ = (-1)¹⁺³ × M₁₃ = 1 × 3 = 3

A₂₁ = (-1)²⁺¹ × M₂₁ = (-1) × (-4) = 4

A₂₂ = (-1)²⁺² × M₂₂ = 1 × 2 = 2

A₂₃ = (-1)²⁺³ × M₂₃ = (-1) × 1 = -1

A₃₁ = (-1)³⁺¹ × M₃₁ = 1 × (-20) = -20

A₃₂ = (-1)³⁺² × M₃₂ = (-1) × (-13) = 13

A₃₃ = (-1)³⁺³ × M₃₃ = 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 = a₁₁ A₁₁ + a₁₂ A₁₂ + a₁₃ A₁₃

[Where A_ij represents cofactors of a_ij of determinant B.]

B = Sum of product of elements of R₁ 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

A₂₁ = (-1)²⁺¹ × M₂₁ = (-1) ×  = (-1) × (3 × 3 – 8 × 2) = (-1) × (-7) = 7

A₂₂ = (-1)²⁺² × M₂₂ = 1 × = (5 × 3 – 8 × 1) = 7

A₂₃ = (-1)²⁺³ × M₂₃ = (-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,

Δ = a₂₁A₂₁ + a₂₂A₂₂ + a₂₃A₂₃

Δ = 2 × 7 + 1 × (-7) = 14 - 7 = 7

Ans: Δ = 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 = a₁₁ A₁₁ + a₁₂ A₁₂ + a₁₃ A₁₃

[Where A_ij represents cofactors of a_ij of determinant B.]

B = Sum of product of elements of R₁ 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

A₁₃ = (-1)¹⁺³ × M₁₃ = 1 ×  = 1 × (1 × z – 1 × y) = (z – y)

A₂₃ = (-1)²⁺³ × M₂₃ = (-1) ×  = (-1) × (1 × z – 1 × x) = - (z - x) = (x - z)

A₃₃ = (-1)³⁺³ × M₃₃ = 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,

Δ = a₁₃A₁₃ + a₂₃A₂₃ + a₃₃A₃₃

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

Δ = z (yz - y² + x² - xz) + xy (y - x) = z [(yz - xz) + (x² - y²)] + xy (y - x)

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

Δ = z × (y - x) × (z – x - y) + xy (y - x)

Δ = (y - x) × (z² – xz – yz + xy)

Δ = (y - x) × [z (z - x) – y (z - x)] = (y - x) × (z - y) × (z - x)

Δ = (x - y) (y - z) (z - x)

Ans: Δ = (x - y) (y - z) (z - x)

Question 7.

If  and A_ij is Cofactors of a_ij, then value of Δ is given by
A. a₁₁ A₃₁+ a₁₂ A₃₂ + a₁₃ A₃₃

B. a₁₁ A₁₁+ a₁₂ A₂₁ + a₁₃ A₃₁

C. a₂₁ A₁₁+ a₂₂ A₁₂ + a₂₃ A₁₃

D. a₁₁ A₁₁+ a₂₁ A₂₁ + a₃₁ A₃₁

Answer:

Δ = 

Expanding along Column 1

Δ = 

Δ = a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁

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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, A₁₁ = 4, A₁₂ = -3, A₂₁ = -2, A₂₂ = 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, A₁₁ = 1{(3×1-0×5)} = 3

Similarly,

A₁₂ = -12, A₁₃ = 6, A₂₁ = 1, A₂₂ = 5, A₂₃ = 2, A₃₁ = -11, A₃₂ = -1, A₃₃ = 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, A₁₁ = -6, A₁₂ = 4, A₂₁ = -3, A₂₂ = 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, A₁₁ = 0, A₁₂ = -11, A₁₃ = 0, A₂₁ = 3, A₂₂ = 1, A₂₃ = -1, A₃₁ = 2, A₃₂ = 8, A₃₃ = 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, A₁₁ = 3, A₁₂ = -4, A₂₁ = 2, A₂₂ = 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, A₁₁ = 2, A₁₂ = 3, A₂₁ = -5, A₂₂ = -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, A₁₁ = 10, A₁₂ = 0, A₁₃ = 0, A₂₁ = -10, A₂₂ = 5, A₂₃ = 0, A₃₁ = 2, A₃₂ = -4, A₃₃ = 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, A₁₁ = -3, A₁₂ = 3, A₁₃ = -9, A₂₁ = 0, A₂₂ = -1, A₂₃ = -2, A₃₁ = 0, A₃₂ = 0, A₃₃ = 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, A₁₁ = -1, A₁₂ = -4, A₁₃ = 1, A₂₁ = 5, A₂₂ = 23, A₂₃ = -11, A₃₁ = 3, A₃₂ = 12, A₃₃ = -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, A₁₁ = 2, A₁₂ = -9, A₁₃ = -6, A₂₁ = 0, A₂₂ = -2, A₂₃ = -1, A₃₁ = -1, A₃₂ = 3, A₃₃ = 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, A₁₁ = -1, A₁₂ = 0, A₁₃ = 0, A₂₁ = 0, A₂₂ = -cosα, A₂₃ = -sinα, A₃₁ = 0, A₃₂ = -sinα, A₃₃ = cosα.

∴ 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 A² – 5A + 7I = O. Hence find A^–1.

Answer:

We have A² = A.A = .

So A² – 5A + 7I = 

Hence A² – 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 A² + aA + bI = O.

Answer:

We have A² = A.A = 

Since A² + aA + bI = 

So A² + 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 A³– 6A² + 5A + 11 I = O. Hence, find A^–1.

Answer:

Here A² = A.A =

And hence A³ = A. A² =

∴ A³– 6A² + 5A + 11 I =

Thus, A³– 6A² + 5A + 11 I = 0

Now, A³– 6A² + 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 A³ – 6A² + 9A – 4I = O and hence find A^-1.

Answer:

Here A² = A.A = 

And hence A³ = A. A² = 

∴ A³– 6A² + 9A -4I