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DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS5 Marks
Question
Solve the matrix equation $\begin{bmatrix}5 & 4 \\1 & 1 \end{bmatrix}\text{X}=\begin{bmatrix}1 & -2 \\1 & 3 \end{bmatrix},$ where $X$ is a $2 \times 2$ matrix.
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Answer

$\begin{bmatrix}5 & 4 \\1 & 1 \end{bmatrix}\text{X}=\begin{bmatrix}1 & -2 \\1 & 3 \end{bmatrix}$
Let $\text{A}=\begin{bmatrix}5 & 4 \\1 & 1 \end{bmatrix}\text{ and B}=\begin{bmatrix}1 & -2 \\1 & 3 \end{bmatrix}$
So, AX = B
or $X = A^{-1}B$ .....(i)
$|\text{A}|=1\neq0$
Cofactors of A are:
$C_{11} = 1, C_{12} = -1$
$C_{21} = -4, C_{22} = 5$
$\therefore\ \text{adj A}=\begin{bmatrix}\text{C}_{11} & \text{C}_{12} \\ \text{C}_{21} & \text{C}_{22} \end{bmatrix}$
$=\begin{bmatrix}1 & -1 \\-4 & 5 \end{bmatrix}^\text{T}$
$=\begin{bmatrix}1 & -4 \\-1 & 5 \end{bmatrix}$
Now, $\text{A}^{-1}=\frac{1}{|\text{A}|}\text{ adj A}$
$\text{A}^{1}=\frac{1}{2}=\begin{bmatrix}1 & -4 \\-1 & 5 \end{bmatrix}$
So from (i)
$\text{X}=\begin{bmatrix}1 & -4 \\-1 & 5 \end{bmatrix}=\begin{bmatrix}1 & 2 \\ 1 & 3 \end{bmatrix}$
$\text{X}=\begin{bmatrix}-3 & -14 \\ 4 & 17 \end{bmatrix}$

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