MCQ
If $\left[ {\begin{array}{*{20}{c}}x&0\\1&y\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 2}&1\\3&4\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3&5\\6&3\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}2&4\\2&1\end{array}} \right]$, then
  • A
    $x = - 3,y = - 2$
  • $x = 3,y = - 2$
  • C
    $x = 3,y = 2$
  • D
    $x = - 3,y = 2$

Answer

Correct option: B.
$x = 3,y = - 2$
b
(b) Since$x - 2 = 3 - 2 \Rightarrow x = 3$

and $y + 4 = 3 - 1 \Rightarrow y = - 2$.

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