Determinants Class 12th Mathematics Part I CBSE Solution

Class 12th 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…

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



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 1st column by addition of 1st and 2nd column)



C1→ C1 - C3 (i.e. Replacing 1st column by subtraction of 1st 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 2.

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




Answer:

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


C1→ C1 + C3 (i.e. Replacing 1st column by addition of 1st 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 3.

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




Answer:


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




C1→ C1 - C3 (i.e. Replacing 1st column by subtraction of 1st 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 3rd column by addition of 2nd and 3rd column)



Taking ab + bc + ac outside determinant



C1→ C1 - C3 (i.e. replace 1st column by subtraction of 1st 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 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 1st row with 3rd 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 1st row with 2nd row)




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



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



∴ LHS = RHS



Question 6.

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


Answer:

To Prove:



R1→ cR1 (i.e. replace 1st 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 1st row by subtraction of 1st row and b times 2nd row)



Taking a outside the determinant from 1st 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 1st,2nd and 3rd column respectively


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


.


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



panding the determinant along 1st 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 1st row by subtraction of 1st and 2nd row)

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



Since we know a2 - b2 = (a + b)(a - b)


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



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



Expanding the determinant along 1st 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 1st column by subtraction of 1st and 2nd column)

C2 → C2 - C3 (i.e. Replacing 2nd column by subtraction of 2nd and 3rd 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 1st and 2nd column respectively



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



As a2 - c2) = (a + c)(a - c) therefore taking (a - c) outside the determinant from 1st column we get


.


Expanding the determinant along 1st 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 1st row by subtraction of 1st and 2nd row)

R2 → R2 - R3 (i.e. Replacing 2nd row by subtraction of 2nd and 3rd 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 1st and 2nd row respectively



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



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



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



Expanding the determinant along 1st 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 1st row by addition of 1st, 2nd and 3rd row)


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



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


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



Expanding the determinant along 1st 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 1st row by addition of 1st, 2nd and 3rd row)


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



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


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



Expanding the determinant along 1st row



∴ LHS = (3y + k) k2 = RHS




Question 13.

By using properties of determinants, show that:



Answer:

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


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



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


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



Expanding the determinant along 1st row



∴ LHS = (a + b + c)3 = RHS




Question 14.

By using properties of determinants, show that:



Answer:

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


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



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


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



Expanding the determinant along 1st column



∴ LHS = 2(x + y + z)3 = RHS




Question 15.

By using properties of determinants, show that:



Answer:

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


Taking (1 - x3) outside the determinant from 1st row


.


Expanding the determinant along 1st row



HS = (1 - x3)2 = RHS




Question 16.

By using properties of determinants, show that:



Answer:

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

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


.


Taking both (1 + a2 + b2) outside the determinant from 1st and 2nd row



Expanding the determinant along 1st 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 1st, 2nd and 3rd row respectively.


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


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


.


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


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


→ cC3 (i.e. replace 3rd column by ‘c’ times 3rd 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 1st column by addition of 1st, 2nd and 3rd column)



Expanding determinant along 1st 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 1st, 2nd and 3rd 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.



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 aij of a determinant is the determinant obtained by removing ith row and jth column in which element aij lies. It is denoted by Mij.

Cofactor: Cofactor of an element aij, Aij = (-1)i+j Mij.


Minor of element aij = Mij


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


Here removing 1st row and 1st 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 aij, Aij = (-1)i+j × Mij


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 aij = Mij


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


Here removing 1st row and 1st 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 aij, Aij = (-1)i+j × Mij


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 aij = Mij

a11 = 1, Minor of element a11 = M11 =  = (1 × 1) – (0 × 0) = 1


Here removing 1st row and 1st 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 aij, Aij = (-1)i+j × Mij


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 aij = Mij


a11 = 1, Minor of element a11 = M11 =  = (5 × 2) – ((-1) × 1) = 10 + 1 = 11


Here removing 1st row and 1st 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 aij, Aij = (-1)i+j × Mij


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 Aij represents cofactors of aij 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 Aij = (-1)i+j × Mij, Mij = Minor of ith row & jth 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 Aij represents cofactors of aij 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 Aij = (-1)i+j × Mij, Mij = Minor of ith row & jth 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 Aij is Cofactors of aij, 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



Exercise 4.5
Question 1.

Find adjoint of each of the matrices.



Answer:

Adjoint of the matrix A = [aij]n×n is defined as the transpose of the matrix [Aij]n×n where Aij is the co-factor of the element aij.

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 = [aij]n×n is defined as the transpose of the matrix [Aij]n×n where Aij is the co-factor of the element aij.

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 = [aij]n×n is defined as the transpose of the matrix [Aij]n×n where Aij is the co-factor of the element aij.

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 = [aij]n×n is defined as the transpose of the matrix [Aij]n×n where Aij is the co-factor of the element aij.

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 = [aij]n×n is defined as the transpose of the matrix [Aij]n×n where Aij is the co-factor of the element aij.


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 = [aij]n×n is defined as the transpose of the matrix [Aij]n×n where Aij is the co-factor of the element aij.


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 = [aij]n×n is defined as the transpose of the matrix [Aij]n×n where Aij is the co-factor of the element aij.

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 = [aij]n×n is defined as the transpose of the matrix [Aij]n×n where Aij is the co-factor of the element aij.

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 = [aij]n×n is defined as the transpose of the matrix [Aij]n×n where Aij is the co-factor of the element aij.

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 = [aij]n×n is defined as the transpose of the matrix [Aij]n×n where Aij is the co-factor of the element aij.

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 = [aij]n×n is defined as the transpose of the matrix [Aij]n×n where Aij is the co-factor of the element aij.

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 =