Matrices — Class 12 Maths Solution

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We are given that, $Y = \left[ {\begin{array}{cccccccccccccccccccc}3&2\\1&4\end{array}} \right]$ …..(i)

and $3X + Y = \left[ {\begin{array}{cccccccccccccccccccc}1&0\\{ - 3}&2\end{array}} \right]$ ….(ii)

Substuting the value of Y in (ii), we have

$2X + \left[ {\begin{array}{cccccccccccccccccccc}3&2\\1&4\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}1&0\\{ - 3}&2\end{array}} \right]$

$\Rightarrow$ $2X = \left[ {\begin{array}{cccccccccccccccccccc}1&0\\{ - 3}&2\end{array}} \right] - \left[ {\begin{array}{cccccccccccccccccccc}3&2\\1&4\end{array}} \right]$

$\Rightarrow$ $2X = \left[ {\begin{array}{cccccccccccccccccccc}{ - 2}&{ - 2}\\{ - 4}&{ - 2}\end{array}} \right] \Rightarrow X = \cfrac{1}{2}\left[ {\begin{array}{cccccccccccccccccccc}{ - 2}&{ - 2}\\{ - 4}&{ - 2}\end{array}} \right]$

Hence,$X = \left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&{ - 1}\\{ - 2}&{ - 1}\end{array}} \right]$