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Model Paper 10 — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsModel Paper 102 Marks
Question
Find the matrix $X$ for which: $\left[\begin{array}{ll}3 & 2 \\ 7 & 5\end{array}\right] X\left[\begin{array}{ll}-1 & 1 \\ -2 & 1\end{array}\right]=\left[\begin{array}{ll}2 & -1 \\ 0 & 4\end{array}\right]$
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Answer

$
\text { Let } A=\left[\begin{array}{ll}
3 & 2 \\
7 & 5
\end{array}\right] B=\left[\begin{array}{cc}
-1 & 1 \\
-2 & 1
\end{array}\right] C=\left[\begin{array}{cc}
2 & -1 \\
0 & 4
\end{array}\right]
$
Then The given equation becomes as
$
\begin{array}{l}
AXB=C \\
\Rightarrow X=A^{-1} CB^{-1} \\
\text { how }|A|=35-14=21 \\
\text { and }|B|=-1+2=1
\end{array}
$
$
\begin{array}{l}
A^{-1}=\frac{\operatorname{adj}(A)}{|A|}=\frac{1}{21}\left[\begin{array}{cc}
5 & -2 \\
-7 & 3
\end{array}\right] \\
\text { and } B^{-1}=\frac{a d(B)}{|B|}=\frac{1}{1}\left[\begin{array}{cc}
1 & -1 \\
2 & -1
\end{array}\right]
\end{array}
$
$
\begin{array}{l}
\Rightarrow X=A^{-1} CB^{-1}=\frac{1}{21}\left[\begin{array}{cc}
5 & -2 \\
-7 & 3
\end{array}\right]\left[\begin{array}{cc}
2 & -1 \\
0 & 4
\end{array}\right]\left[\begin{array}{ll}
1 & -1 \\
2 & -1
\end{array}\right] \\
=\frac{1}{21}\left[\begin{array}{cc}
10+0 & -5-8 \\
-14+0 & 7+12
\end{array}\right]\left[\begin{array}{ll}
1 & -1 \\
2 & -1
\end{array}\right] \\
=\frac{1}{21}\left[\begin{array}{cc}
10 & -13 \\
-14 & 19
\end{array}\right]\left[\begin{array}{cc}
1 & -1 \\
2 & -1
\end{array}\right] \\
=\frac{1}{21}\left[\begin{array}{cc}
10-26 & -10+13 \\
-14+38 & 14-19
\end{array}\right]
\end{array}
$
Hence, $x=\frac{1}{21}\left[\begin{array}{cc}-16 & 3 \\ 24 & -5\end{array}\right]$

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