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Adjoint and Inverse of a Matrix — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsAdjoint and Inverse of a Matrix5 Marks
Question
Find the matrix X for which:$\begin{bmatrix}3 & 2 \\ 7 & 5 \end{bmatrix}\text{X}\begin{bmatrix} -1 & 1 \\ -2 & 1 \end{bmatrix}=\begin{bmatrix} 2 & -1 \\ 0 & 4 \end{bmatrix}$
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Answer

Let $\text{A}=\begin{bmatrix}3 & 2 \\ 7 & 5 \end{bmatrix},\text{B}=\begin{bmatrix} -1 & 1 \\ -2 & 1 \end{bmatrix}\text{and C}=\begin{bmatrix} 2 & -1 \\ 0 & 4 \end{bmatrix}$ Now,$|\text{A}|=\begin{vmatrix}3 & 2 \\ 7 & 5 \end{vmatrix}=15-14=1$
$|\text{B}|=\begin{vmatrix} -1 & 1 \\ -2 & 1 \end{vmatrix}=-1+2=1$
Since, $|\text{A}|\neq0\text{ and }|\text{B}|\neq0$ Hence, A & B are invertible, so $A^{-1}$ and $B^{-1}$ exist. Cofactors of matrix A are $A_{11} = 5, A_{12} = -7, A_{21} = -2, A_{22} = 3$
Now, $\text{adj A}\begin{bmatrix}5 & -7 \\ -2 & 3 \end{bmatrix}^\text{T}=\begin{bmatrix}5 & -2 \\ -7 & 3 \end{bmatrix}$$\text{A}^{-1}=\frac{1}{|\text{A}|}\text{ adj A}=\begin{bmatrix}5 & -2 \\ -7 & 3 \end{bmatrix}$
Cofactors of matrix B are $B_{11} = 1, B_{12} = 2, B_{21} = -1, B_{22} = -1 $
​​​​​​​Now, $\text{adj B}=\begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix}^\text{T}=\begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix}$$\text{B}^{-1}=\frac{1}{|\text{B}|}\text{ adj B}=\begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix}$
The given equation Becomes $AXB = C$
$\Rightarrow (A^{-1} A) X (BB^{-1}) = A^{-1}CB^{-1}$
$\Rightarrow (I) X (I) = A^{-1}CB^{-1}​​​​​​​$​​​​​​​
$\Rightarrow\ \text{X}=\begin{bmatrix}5 & -2 \\ -7 & 3 \end{bmatrix}\begin{bmatrix}2 & -1 \\0 & 4 \end{bmatrix}\begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix}$
$\Rightarrow\ \text{X}=\begin{bmatrix}5 & -2 \\ -7 & 3 \end{bmatrix}\begin{bmatrix}2-2 & -2+1 \\0+8 & 0-4 \end{bmatrix}$
$\Rightarrow\ \text{X}=\begin{bmatrix}5 & -2 \\ -7 & 3 \end{bmatrix}\begin{bmatrix} 0 & -1 \\8 & -4 \end{bmatrix}$
$\Rightarrow\ \text{X}=\begin{bmatrix}-16 & 3 \\ 24 & -5 \end{bmatrix}$

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