Class 12 Maths NCERT Solutions for Chapter 4 Determinants Exercise 4.5

Determinants Exercise 4.5 Solutions

1. Find the adjoint of the matrix

Solution

Let A =
We have,
A₁₁ = 4, A₁₂ = -3, A₂₁ = -2, A₂₂ = 1

2. Find adjoint of each of the matrices.

Solution

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

Solution

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

Solution

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

Solution

Let A =

We have,
|A| = 6 + 8 = 14
Now,
A₁₁ = 3, A₁₂ = -4, A₂₁ = 2, A₂₂ = 2

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

Solution

Let A =

We have,
|A| = -2 + 15= 13
Now,

A₁₁ = 2, A₁₂ = 3, A₂₁ = -5, A₂₂ = -1

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

Solution

Let A =

We have,
|A| = 1(10 - 0) - 2(0 - 0) + 3(0 - 0) = 10
Now,
A₁₁ = 10 - 0 = 10 , A₁₂ = -(0 - 0) = 0, A₁₃ = 0 - 0 = 0,
A₂₁ = -(10 - 0) = -10, A₂₂ = 5 - 0 = 5, A₂₃ = -(0 - 0) = 0
A₃₁ = 8 - 6= 2 , A₃₂ = -(4 - 0) = -4, A₃₃ = 2 - 0 = 2

8. Find the inverse of each of the matrix (if it exists).

Solution

Let A =

We have,
|A| = 1(-3 - 0) - 0 + 0 = -3
Now,
A₁₁ = -3 - 0 = -3 , A₁₂ = -(-3 - 0) = 3, A₁₃ = 6 - 15 = -9,
A₂₁ = -(0 - 0) = 0, A₂₂ = -1 - 0 = -1, A₂₃ = -(2 - 0) = -2
A₃₁ = 0 - 0 = 0 , A₃₂ = -(0 - 0) = 0, A₃₃ = 3 - 0 = 3

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

Solution

Let A =

We have,
|A| = 2(-1 - 0) - 1(4 - 0) + 3(8 - 7)
= 2(-1) - 1(4) + 3(1)
= -2 - 4 + 3
= - 3
Now,
A₁₁ = -1 - 0 = -1 , A₁₂ = -(4 - 0) = -4, A₁₃ = 8 - 7 = 1
A₂₁ = -(1 - 6) = 5, A₂₂ = 2 + 21 = 23, A₂₃ = -(4 + 7) = -11
A₃₁ = 0 - 3 = 3 , A₃₂ = -(0 - 12) = 12, A₃₃ = -2 - 4 = -6

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

Solution

Let A =

By expanding along C₁ , we have :
|A| = 1(8 - 6) - 0 + 3(3 - 4) = 2 - 3 = -1
Now,
A₁₁ = 8 - 6 = 2 , A₁₂ = -(0 + 9) = -9, A₁₃ = 0 - 6 = -6
A₂₁ = -(-4 + 4) = 0, A₂₂ = 4 - 6 = -2, A₂₃ = -(-2 + 3) = -1
A₃₁ = 3 - 4 = -1 , A₃₂ = -(-3 - 0) = 3, A₃₃ = 2 - 0 = 2

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

Solution

Let A =

We have,
|A| = 1(−cos² α – sin²α) = −(cos²α + sin² α) = − 1
Now,
A₁₁ = −cos² α – sin²α = -1 , A₁₂ = 0, A₁₃ = 0
A₂₁ = 0, A₂₂ = −cosα , A₂₃ = – sinα
A₃₁ = 0 , A₃₂ = – sinα, A₃₃ = cosα

12. Let A =

and B =

. Verify that (AB)^-1 = B^-1 A^-1 .

Solution

Let A =

We have,
|A| = 15 - 14 = 1
Now,
A₁₁ = 5, A₁₂ = - 2, A₂₁ = -7, A₂₂ = 3

Now, Let B =

We have,
|B| = 54 - 56 = -2

From (1) and (2), we have :
(AB)^-1 = B^-1 A^-1
Hence, the given result is proved.

13. If A =

show that A² - 5A + 7I = 0. Hence find A^-1.

Solution

Hence, A² - 5A + 7I = 0.
∴ A . A - 5A = -7I
⇒ A.A(A^-1 ) - 5AA^-1 = -7IA^-1 [Post-multiplying by A^-1 as |A| ≠ 0]
⇒ A(AA^-1 ) - 5I = -7A^-1
⇒ AI -5I = -7A^-1
⇒ A^-1 = - (1/7)[A - 5I]
⇒ A^-1 = (1/7)[5I - A]

14. For the matrix A =

find the numbers a and b such that A² + aA + bI = 0.

Solution

Now,
A² + aA + bI = 0
⇒ (AA)A^-1 - aAA^-1 = bIA^-1 = 0 [Post-multiplying by A^-1 as |A| ≠ 0]
⇒ A(AA^-1 ) - aI + b(IA^-1 ) = 0
⇒ AI + aI + bA^-1 = 0
⇒ A + aI = -bA^-1
⇒ A^-1 = -(1/b)[A + aI]
Now,