Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Ex 2.2

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 2 Matrices Ex 2.2 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 2 Matrices Ex 2.2

Question 1.
Find the co-factors of the elements of the following matrices
(i) $\left[\begin{array}{cc} -1 & 2 \\ -3 & 4 \end{array}\right]$
Solution:
Let A = $\left[\begin{array}{cc} -1 & 2 \\ -3 & 4 \end{array}\right]$
Here, a₁₁ = -11, M₁₁ = 4
∴ A₁₁ = (-1)¹⁺¹(4) = 4
a₁₂ = 2, M₁₂ = -3
∴ A₁₂ = (-1)¹⁺²(- 3) = 3
a₂₁ = – 3, M₂₁ = -2
∴ A₂₁ = (- 1)²⁺¹(2) = -2
a₂₂ = 4, M₂₂ = -1
∴ A₂₂ = (-1)²⁺²(-1) = -1.

(ii) $\left[\begin{array}{ccc} 1 & -1 & 2 \\ -2 & 3 & 5 \\ -2 & 0 & -1 \end{array}\right]$
Solution:
Let A = $\left[\begin{array}{ccc} 1 & -1 & 2 \\ -2 & 3 & 5 \\ -2 & 0 & -1 \end{array}\right]$
The co-factor of a_ij is given by A_ij = (-1)^i+jM_ij

Question 2.
Find the matrix of co-factors for the following matrices
(i) $\left[\begin{array}{rr} 1 & 3 \\ 4 & -1 \end{array}\right]$
Solution:
Let A = $\left[\begin{array}{rr} 1 & 3 \\ 4 & -1 \end{array}\right]$
Here, a₁₁ = 1, M₁₁ = -1
∴ A₁₁ = (-1)¹⁺¹(-1) = -1
a₁₂ = 3, M₁₂ = 4
∴ A₁₂ = (-1)¹⁺²(4) = -4
a₂₁ = 4, M₂₁ = 3
∴ A₂₁ = (-1)²⁺¹(3) = -3
a₂₂ = -1, M₂₂ = 1
∴ A₂₂ = (-1)²⁺¹(1) = 1
∴ the co-factor matrix = $\left[\begin{array}{ll} A_{11} & A_{12} \\ A_{21} & A_{22} \end{array}\right]$
= $\left(\begin{array}{rr} -1 & -4 \\ -3 & 1 \end{array}\right)$

(ii) $\left[\begin{array}{rrr} 1 & 0 & 2 \\ -2 & 1 & 3 \\ 0 & 3 & -5 \end{array}\right]$
Solution:
Let A = $\left[\begin{array}{rrr} 1 & 0 & 2 \\ -2 & 1 & 3 \\ 0 & 3 & -5 \end{array}\right]$



A₁₁ = -14, A₁₂ = -10, A₁₃ = -6,
A₂₁ = 6, A₂₂ = -5, A₂₃ = -3,
A₃₁ = -2, A₃₂ = -7, A₃₃ = 1.
∴ the co-factor matrix
= $\left[\begin{array}{lll} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{array}\right]$ = $\left[\begin{array}{rrr} -14 & -10 & -6 \\ 6 & -5 & -3 \\ -2 & -7 & 1 \end{array}\right]$

Question 3.
Find the adjoint of the following matrices.
(i) $\left[\begin{array}{cc} 2 & -3 \\ 3 & 5 \end{array}\right]$
Solution:
Let A = $\left[\begin{array}{cc} 2 & -3 \\ 3 & 5 \end{array}\right]$
Here, a₁₁ = 2, M₁₁= 5
∴ A₁₁ = (-1)¹⁺¹(5) = 5
a₁₂ = -3, M₁₂ = 3
∴ A₁₂ = (-1)¹⁺²(3) = -3
a₂₁ = 3, M₂₁ = -3
∴ A A₂₁ = (-1)²⁺¹(-3) = 3
a₂₂ = 5, M₂₂ = 2
∴ A₂₂ = (-1)²⁺¹ = 2
∴ the co-factor matrix = $\left[\begin{array}{ll} A_{11} & A_{12} \\ A_{21} & A_{22} \end{array}\right]$
= $\left[\begin{array}{rr} 5 & -3 \\ 3 & 2 \end{array}\right]$
∴ adj A = $\left(\begin{array}{rr} 5 & 3 \\ -3 & 2 \end{array}\right)$

(ii) $\left[\begin{array}{ccc} 1 & -1 & 2 \\ -2 & 3 & 5 \\ -2 & 0 & -1 \end{array}\right]$
Solution:


A₁₁ = -3, A₁₂ = -12, A₁₃ = 6,
A₂₁ = -1, A₂₂ = 3, A₂₃ = 2,
A₃₁ = -11, A₃₂ = -9, A₃₃ = 1
∴ the co-factor matrix = $\left[\begin{array}{lll} \mathrm{A}_{11} & \mathrm{~A}_{12} & \mathrm{~A}_{15} \\ \mathrm{~A}_{21} & \mathrm{~A}_{22} & \mathrm{~A}_{23} \\ \mathrm{~A}_{31} & \mathrm{~A}_{32} & \mathrm{~A}_{33} \end{array}\right]$
= $\left[\begin{array}{rrr} -3 & -12 & 6 \\ -1 & 3 & 2 \\ -11 & -9 & 1 \end{array}\right]$
∴ adj A = $\left[\begin{array}{rrr} -3 & -1 & -11 \\ -12 & 3 & -9 \\ 6 & 2 & 1 \end{array}\right]$

Question 4.
If A = $\left[\begin{array}{ccc} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{array}\right]$, verify that A (adj A) = (adj A) A = | A | ∙ I
Solution:
A = $\left[\begin{array}{ccc} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{array}\right]$




From (1), (2) and (3), we get,
A(adj A) = (adj A)A = |A|∙I.
Note: This relation is valid for any non-singular matrix A.

Question 5.
Find the inverse of the following matrices by the adjoint method
(i) $\left[\begin{array}{ll} -1 & 5 \\ -3 & 2 \end{array}\right]$
Solution:
Let A = $\left[\begin{array}{ll} -1 & 5 \\ -3 & 2 \end{array}\right]$
∴ |A| = $\left|\begin{array}{ll} -1 & 5 \\ -3 & 2 \end{array}\right|$ = -2 + 15 = 13 ≠ 0
∴ A^-1 exists.
First we have to find the co-factor matrix
= [A_ij]_2×2, where A_ij = (-1)^i+jM_ij
Now, A₁₁ = (-1)¹⁺¹M₁₁ = 2
A₁₂ = (-1)¹⁺²M₁₂ = -(-3) = 3
A₂₁ = (-1)²⁺¹M₂₁ = -5
A₂₂ = (-1)²⁺²M₂₂ = -1
Hence, the co-factor matrix

(ii) $\left[\begin{array}{cc} 2 & -2 \\ 4 & 3 \end{array}\right]$
Solution:
Let A = $\left[\begin{array}{cc} 2 & -2 \\ 4 & 3 \end{array}\right]$
|A| = = 6 + 8 = 14 ≠ 0
∴ A^-1 exist
First we have to find the co-factor matrix
= [A_ij] _2×2 where A_ij = (-1)^i+jM_ij
Now, A₁₁ = (-1)¹⁺¹M₁₁ = 3
A₁₂ = (-1)¹⁺²M = -4
A₂₁ = (-2)²⁺¹M₂₁ = (-2) = 2
A₂₂ = (-1)²⁺²M₂₂ = 2
Hence the co-factor matrix
= $\left[\begin{array}{ll} A_{11} & A_{12} \\ A_{21} & A_{22} \end{array}\right]$ = $\left[\begin{array}{cc} 3 & -4 \\ 2 & 2 \end{array}\right]$
∴ adj A = $\left[\begin{array}{cc} 3 & 2 \\ -4 & 2 \end{array}\right]$
∴ A^-1 = $\frac{1}{|\mathrm{~A}|}$ (adj A) = $\frac{1}{14}\left(\begin{array}{cc} 3 & 2 \\ -4 & 2 \end{array}\right)$

(iii) $\left[\begin{array}{ccc} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{array}\right]$
Solution:
Let A = $\left[\begin{array}{ccc} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{array}\right]$



∴ A^-1 = $\frac{1}{3}\left[\begin{array}{rrr} 3 & 0 & 0 \\ -3 & 1 & 0 \\ 9 & 2 & -3 \end{array}\right]$

(iv) $\left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{array}\right]$
Solution:
Let A = $\left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{array}\right]$
∴ |A| = $\left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{array}\right]$
= 1(10 – 0) – 0 + 0
= 1(10) – 0 + 0
= 10 ≠ 0
∴ A^-1 exists.
First we have to find the co-factor matrix


∴ A^-1 = $\frac{1}{|\mathrm{~A}|}$ (adj A)
= $\frac{1}{10}\left(\begin{array}{rrr} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{array}\right)$
∴ A^-1 = $\frac{1}{10}\left(\begin{array}{rrr} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{array}\right)$

Question 6.
Find the inverse of the following matrices
(i) $\left[\begin{array}{cc} 1 & 2 \\ 2 & -1 \end{array}\right]$
Solution:
Let A = $\left[\begin{array}{cc} 1 & 2 \\ 2 & -1 \end{array}\right]$

(ii) $\left[\begin{array}{cc} 2 & -3 \\ -1 & 2 \end{array}\right]$
Solution:
Let A = $\left[\begin{array}{cc} 2 & -3 \\ -1 & 2 \end{array}\right]$

∴ A^-1 = $\left(\begin{array}{ll} 2 & 3 \\ 1 & 2 \end{array}\right)$

(iii) $\left[\begin{array}{lll} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}\right]$
Solution:
Let A = $\left[\begin{array}{lll} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}\right]$

(iv) $\left[\begin{array}{ccc} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{array}\right]$
Solution:
Let A = $\left[\begin{array}{ccc} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{array}\right]$