Find the matrix X for which : [ ll 3 & 2 7 & 5 ] X [ ll -1 & 1 -2 & 1 ]= [ ll 2 & -1 0 & 4 ]

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

Find the matrix X for which : [ ll 3 & 2 7 & 5 ] X [ ll -1 & 1 -2…

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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
$AXB=C$
$\Rightarrow X=A^{-1} CB^{-1}$
how $|A|=35-14=21$
and $|B|=-1+2=1$
$ \begin{array}{l} A^{-1}=\frac{\operatorname{adj}(A)}{|A|}=\frac{1}{21}\left[\begin{array}{cc} 5 & -2 \\ -7 & 3 \end{array}\right]\end{array} $
and $B^{-1}=\frac{a d(B)}{|B|}=\frac{1}{1}\left[\begin{array}{cc} 1 & -1 \\ 2 & -1 \end{array}\right] $
$ \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] \end{array} $
$ =\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] $
Hence, $x=\frac{1}{21}\left[\begin{array}{cc}-16 & 3 \\ 24 & -5\end{array}\right]$