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

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAdjoint and Inverse of a Matrix5 Marks
Question
Find the matrix X satisfying the matrix equation.$\text{X}\begin{bmatrix}5 & 3 \\-1 & -2 \end{bmatrix}\begin{bmatrix}14 & 7 \\7 & 7 \end{bmatrix}$
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Answer

$\text{X}\begin{bmatrix}5 & 3 \\-1 & -2 \end{bmatrix}\begin{bmatrix}14 & 7 \\7 & 7 \end{bmatrix}=-10+3=-7\neq0$Hence, A is invertible.
If $C_{ij}$ is a cofactor of $a_{ij}$ in A, then $C_{11}=-2, C_{12}=1, C_{21}=-3$ and $C_{22}=5$.
Now, $\text{adj A}\begin{bmatrix}-2 & 1 \\-3 & 5 \end{bmatrix}^\text{T}=\begin{bmatrix}-2 & -3 \\1 & 5 \end{bmatrix}$
$\Rightarrow\ \text{A}^{-1}=\frac{1}{|\text{A}|}\text{ adj A}=\frac{-1}{7}\begin{bmatrix}-2 & -3 \\1 & 5 \end{bmatrix}$
Let: $\text{B}=\begin{bmatrix}14 & 7 \\7 & 7 \end{bmatrix}$
$\Rightarrow\ \text{B}=\begin{bmatrix}14 & 7 \\7 & 7 \end{bmatrix}=98-49=49\neq0$
Hence, B is invertible.
The given matrix equation becomes XA = B.
$⇒ (XA)A^{-1}= BA^{-1}$
$\Rightarrow\ \text{X(AA}^{-1})=\text{B}=\begin{bmatrix}14 & 7 \\7 & 7 \end{bmatrix}\times\frac{-1}{7}\times\begin{bmatrix}-2 & -3 \\1 & 5 \end{bmatrix}$
$\Rightarrow\ \text{X}=\frac{-1}{7}\begin{bmatrix}-28+7 & -42+35 \\ -14+7 & -21+35 \end{bmatrix}$
$\Rightarrow\ \text{X}=\frac{-1}{7}\begin{bmatrix}-21 & -7 \\ -7 & 14 \end{bmatrix}$
$\Rightarrow\ \text{X}=\begin{bmatrix} 3 & 1 \\1 & -2 \end{bmatrix}$

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