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Question Bank [2022] — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsQuestion Bank [2022]4 Marks
Question
If $A=\left[\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right], B=\left[\begin{array}{ll}4 & 1 \\ 3 & 1\end{array}\right]$ and $C=\left[\begin{array}{ll}24 & 7 \\ 31 & 9\end{array}\right]$, then find the matrix $X$ such that $A X B=C$
✓

Answer

Since, $A X B=C$
$\therefore\left[\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}\right](X B)=\left[\begin{array}{ll} 24 & 7 \\ 31 & 9 \end{array}\right]$
First we perform the row transformations.
Applying $R_2 \rightarrow R_2-R_1$,
$\left[\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right](X B)=\left[\begin{array}{cc} 24 & 7 \\ 7 & 2 \end{array}\right]$
Applying $R_2 \rightarrow R_1-R_2$,
$\begin{aligned} & {\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right](X B)=\left[\begin{array}{cc} 17 & 5 \\ 7 & 2 \end{array}\right]} \end{aligned} $
$ \therefore X B=\left[\begin{array}{cc} 17 & 5 \\ 7 & 2 \end{array}\right]  $
$ \therefore X\left[\begin{array}{ll} 4 & 1 \\ 3 & 1 \end{array}\right]=\left[\begin{array}{cc} 17 & 5 \\ 7 & 2 \end{array}\right] $
Now, we perform the column transformations.
Applying $C_1 \leftrightarrow C_2$
$X\left[\begin{array}{ll} 1 & 4 \\ 1 & 3\ \end{array}\right]=\left[\begin{array}{cc} 5 & 17 \\ 2 & 7 \end{array}\right]$
Applying $C_2 \rightarrow C_2-4 C_1$,
$ X\left[\begin{array}{cc} 1 & 0 \\ 1 & -1 \end{array}\right]=\left[\begin{array}{cc} 5 & -3 \\ 2 & -1 \end{array}\right]$
$\begin{aligned} & \text { Applying } C_2 \rightarrow-C_2 \\ & X\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right]=\left[\begin{array}{ll}5 & 3 \\ 2 & 1\end{array}\right] \end{aligned}$
Applying $C_1 \rightarrow C_1-C_2$,
$X\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}2 & 3 \\ 1 & 1\end{array}\right]$
$ \therefore X=\left[\begin{array}{ll}2 & 3 \\ 1 & 1\end{array}\right]$

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