Matrices — Class 12 Maths Solution

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$2\left[ {\begin{array}{cccccccccccccccccccc}x&z\\y&t\end{array}} \right] + 3\left[ {\begin{array}{cccccccccccccccccccc}1&{ - 1}\\0&2\end{array}} \right] = 3\left[ {\begin{array}{cccccccccccccccccccc}3&5\\4&6\end{array}} \right]$

$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}{2x}&{2z}\\{2y}&{2t}\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}3&{ - 3}\\0&6\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}9&{15}\\{12}&{18}\end{array}} \right]$

$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}{2x + 3}&{2z - 3}\\{2y + 0}&{2t + 6}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}9&{15}\\{12}&{18}\end{array}} \right]$

Now, 2x + 3 = 9 $\Rightarrow$ 2x = 6 $\Rightarrow$ x = 3

2z$-$ 3 = 15 $\Rightarrow$ 2z = 18 $\Rightarrow$ z = 9

2y = 12 $\Rightarrow$ y = 6

2t + 6 = 18 $\Rightarrow$ 2t = 12 $\Rightarrow$ t = 6

Hence, x = 3, y = 6, z = 9 and t = 6.