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Matrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMatrices2 Marks
Question
Solve the equation for $\mathrm{x}, \mathrm{y}, \mathrm{z}$ and $\mathrm{t}$ if $2\left[\begin{array}{ll}x & z \\ y & t\end{array}\right]+3\left[\begin{array}{cc}1 & -1 \\ 0 & 2\end{array}\right]=3\left[\begin{array}{cc}3 & 5 \\ 4 & 6\end{array}\right]$.
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Answer

Given: $2\left[ {\begin{array}{*{20}{c}} x&z \\ y&t \end{array}} \right] + 3\left[ {\begin{array}{*{20}{c}} 1&{ - 1} \\ 0&2 \end{array}} \right] = 3\left[ {\begin{array}{*{20}{c}} 3&5 \\ 4&6 \end{array}} \right]$
$ \Rightarrow \left[ {\begin{array}{*{20}{c}} {2x}&{2z} \\ {2y}&{2t} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 3&{ - 3} \\ 0&6 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 9&{15} \\ {12}&{18} \end{array}} \right]$
$ \Rightarrow \left[ {\begin{array}{*{20}{c}} {2x + 3}&{2z - 3} \\ {2y + 0}&{2t + 6} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 9&{15} \\ {12}&{18} \end{array}} \right]$
Equating corresponding entries, we have
2x + 3 = 9 $\Rightarrow$ 2x = 9 - 3 $\Rightarrow$ 2x = 6 $\Rightarrow$ x = 3
and 2z - 3 = 15 $\Rightarrow$ 2z = 15 + 3 $\Rightarrow$ 2z = 18 $\Rightarrow$ z = 9
and 2y = 12 $\Rightarrow$ y = 6
And 2t + 6 = 18 $\Rightarrow$ 2t = 18 - 6 $\Rightarrow$ 2t = 12 $\Rightarrow$ t = 6
$\therefore$ x = 3, y = 6, z = 9, t = 6

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