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$X$ तथा $Y$ ज्ञात कीजिए यदि $Y =\left[\begin{array}{ll}3 & 2 \\ 1 & 4\end{array}\right]$ तथा $2 X + Y =\left[\begin{array}{rl}1 & 0 \\ -3 & 2\end{array}\right]$
$\left[\begin{array}{cc}-1 & -1 \\ -2 & -1\end{array}\right]$
$\left[\begin{array}{cc}-1 & -1 \\ -2 & -1\end{array}\right]$
$\left[\begin{array}{cc}-1 & -1 \\ -2 & -1\end{array}\right]$
$\left[\begin{array}{cc}-1 & -1 \\ -2 & -1\end{array}\right]$
Solution
$2 X+Y=\left[\begin{array}{cc}1 & 0 \\ -3 & 2\end{array}\right]$
$\Rightarrow 2 X+\left[\begin{array}{ll}3 & 2 \\ 1 & 4\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ -3 & 2\end{array}\right]$
$ \Rightarrow 2X = $ $\left[ {\begin{array}{*{20}{c}}
1&0 \\
{ – 3}&2
\end{array}} \right] – \left[ {\begin{array}{*{20}{l}}
3&2 \\
1&4
\end{array}} \right]$ $ = \left[ {\begin{array}{*{20}{c}}
{1 – 3}&{0 – 2} \\
{ – 3 – 1}&{2 – 4}
\end{array}} \right]$
$\Rightarrow 2 X=\left[\begin{array}{ll}-2 & -2 \\ -4 & -2\end{array}\right]$
$\therefore \quad X=\frac{1}{2}\left[\begin{array}{ll}-2 & -2 \\ -4 & -2\end{array}\right]=\left[\begin{array}{ll}-1 & -1 \\ -2 & -1\end{array}\right]$