If X and Y are 2 x 2 matrices, then solve the following matrix equations for X and Y
$2 \mathrm{X}+3 \mathrm{Y}=\left[\begin{array}{ll}2 & 3 \\ 4 & 0\end{array}\right], 3 \mathrm{X}+2 \mathrm{Y}=\left[\begin{array}{cc}-2 & 2 \\ 1 & -5\end{array}\right]$
Given, $\quad 2 X+3 Y=\left[\begin{array}{ll}2 & 3 \\ 4 & 0\end{array}\right]$ $\ldots(\mathrm{i})$
and $3 X+2 Y=\left[\begin{array}{cc}-2 & 2 \\ 1 & -5\end{array}\right]$ ...(ii)
On subractiong equations (i) and (ii), we get
$(3 X+2 Y)-(2 X+3 Y)=\left[\begin{array}{cc}-2-2 & 2-3 \\ 1-4 & -5-0\end{array}\right]$
Thus, $\quad X-Y=\left[\begin{array}{rr}-4 & -1 \\ -3 & -5\end{array}\right]$ ...(iii)
On adding equations (i) and (ii), we get
$5 X+5 Y=\left[\begin{array}{cc}0 & 5 \\ 5 & -5\end{array}\right]$
$X+Y=\frac{1}{5}\left[\begin{array}{cc}0 & 5 \\ 5 & -5\end{array}\right]=\left[\begin{array}{cc}0 & 1 \\ 1 & -1\end{array}\right]$$\ldots$ (iv)
On adding equations (iii) and (iv), we get
$(X-Y)+(X+Y)=\left[\begin{array}{cc}-4 & 0 \\ -2 & -6\end{array}\right]$
$2 X=\left[\begin{array}{cc}-4 & 0 \\ -2 & -6\end{array}\right]$
Thus, $X=\left[\begin{array}{cc}-2 & 0 \\ -1 & -3\end{array}\right]$
From equation (iv), we get
$\left[\begin{array}{cc}-2 & 0 \\ -1 & -3\end{array}\right]+Y=\left[\begin{array}{cc}0 & 1 \\ 1 & -1\end{array}\right]$
$Y=\left[\begin{array}{cc}0 & 1 \\ 1 & -1\end{array}\right]-\left[\begin{array}{cc}-2 & 0 \\ -1 & -3\end{array}\right]$
Thus, $Y=\left[\begin{array}{ll}2 & 1 \\ 2 & 2\end{array}\right]$
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