Question

If $\left[\begin{array}{ccc}X-Y& 2& -2\\ 4& X& 6\end{array}\right]+\left[\begin{array}{ccc}3& -2& 2\\ 1& 0& -1\end{array}\right]=\left[\begin{array}{ccc}6& 0& 0\\ 5& 2X+Y& 5\end{array}\right]$,
then

• A. $X=\frac{3}{2}$
• B. $X=\frac{5}{2}$
• C. $Y=-\frac{3}{2}$
• D. $Y=-\frac{5}{2}$

Answer 1

Answer: (A), (C)

$Y=-\frac{3}{2}$

$\left[\begin{array}{ccc}X-Y& 2& -2\\ 4& X& 6\end{array}\right]+\left[\begin{array}{ccc}3& -2& 2\\ 1& 0& -1\end{array}\right]=\left[\begin{array}{ccc}6& 0& 0\\ 5& 2X+Y& 5\end{array}\right]$
$=\left[\begin{array}{ccc}X-Y+3& 2-2& -2+2\\ 4+1& X+0& 6-1\end{array}\right]=\left[\begin{array}{ccc}6& 0& 0\\ 5& 2X+Y& 5\end{array}\right]$
$=\left[\begin{array}{ccc}X-Y+3& 0& 0\\ 5& X& 5\end{array}\right]=\left[\begin{array}{ccc}6& 0& 0\\ 5& 2X+Y& 5\end{array}\right]$
We know that corresponding entries of equal matrices are equal. Hence, we have:
$X-Y+3=6$
$\Rightarrow X-Y=3\cdots \left(1\right)$
$X=2X+Y$
$\Rightarrow X+Y=0\cdots \left(2\right)$
Hence, on solving, we have: $X=\frac{3}{2}$ and $Y=-\frac{3}{2}$