If A= [ ccc 2 & 3 & -1 4 & 7 & 5 ], B= [ ccc 1 & 3 & 2 4 & 6 & -1 ] and C= [ ccc 1 & -1 & 6 0 & 2 & -5 ], find the matrix X such that 3 ~A-2 ~B+4 X=5 C .

Since 3 ~A-2 ~B+4 X=5 C 4 X & =5 C-3 A+2 B 4 X & =5 [ ccc 1 & -1 & 6 0 & 2 & -5 ]-3 [ ccc 2 & 3 & -1 4 & 7 & 5 ] & +2 [ ccc 1 & 3 & 2 4 & 6 & -1 ] & = [ ccc 5 & -5 & 30 0 & 10 & -25 ]+ [ ccc -6 & -9 & 3 -12 & -21 & -15 ] & + [ ccc 2 & 6 & 4 8 & 12 & -2 ] & = [ ccc 5-6+2 & -5-9+6 & 30+3+4 0-12+8 & 10-21+12 & -25-15-2 ] & = [ ccc 1 & -8 & 37 -4 & 1 & -42 ] & X=1 4 [ ccc 1 & -8 & 37 -4 & 1 & -42 ]= [ ccc 1 4 & -2 & 37 4 -1 & 1 4 & -21 2 ] &

If A= [ ccc 2 & 3 & -1 4 & 7 & 5 ], B= [ ccc 1 & 3 & 2 4 & 6 & -1…

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Question

If $A=\left[\begin{array}{ccc}2 & 3 & -1 \\ 4 & 7 & 5\end{array}\right], B=\left[\begin{array}{ccc}1 & 3 & 2 \\ 4 & 6 & -1\end{array}\right]$ and
$C=\left[\begin{array}{ccc}1 & -1 & 6 \\ 0 & 2 & -5\end{array}\right]$, find the matrix $X$ such that
$
3 \mathrm{~A}-2 \mathrm{~B}+4 \mathrm{X}=5 \mathrm{C} \text {. }
$

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Answer

Since $3 \mathrm{~A}-2 \mathrm{~B}+4 \mathrm{X}=5 \mathrm{C}$
$
\begin{aligned}
\therefore 4 X & =5 C-3 A+2 B \\
\therefore 4 X & =5\left[\begin{array}{ccc}
1 & -1 & 6 \\
0 & 2 & -5
\end{array}\right]-3\left[\begin{array}{ccc}
2 & 3 & -1 \\
4 & 7 & 5
\end{array}\right] \\
& +2\left[\begin{array}{ccc}
1 & 3 & 2 \\
4 & 6 & -1
\end{array}\right] \\
& =\left[\begin{array}{ccc}
5 & -5 & 30 \\
0 & 10 & -25
\end{array}\right]+\left[\begin{array}{ccc}
-6 & -9 & 3 \\
-12 & -21 & -15
\end{array}\right] \\
& +\left[\begin{array}{ccc}
2 & 6 & 4 \\
8 & 12 & -2
\end{array}\right]
\end{aligned}
$

$\begin{aligned} & =\left[\begin{array}{ccc}5-6+2 & -5-9+6 & 30+3+4 \\ 0-12+8 & 10-21+12 & -25-15-2\end{array}\right] \\ & =\left[\begin{array}{ccc}1 & -8 & 37 \\ -4 & 1 & -42\end{array}\right] \\ & \therefore X=\frac{1}{4}\left[\begin{array}{ccc}1 & -8 & 37 \\ -4 & 1 & -42\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{4} & -2 & \frac{37}{4} \\ -1 & \frac{1}{4} & -\frac{21}{2}\end{array}\right] \\ & \end{aligned}$

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