Find matrices A and B, if 2 A-B= [ ccc 6 & -6 & 0 -4 & 2 & 1 ] and A-2 B= [ ccc 3 & 2 & 8 -2 & 1 & -7 ]

Given equations are l 2 A-B= [ ccc 6 & -6 & 0 -4 & 2 & 1 ] (i) and A-2 B= [ ccc 3 & 2 & 8 -2 & 1 & -7 ] (ii) By (i) - (ii) 2, we get 3 B= [ ccc 6 & -6 & 0 -4 & 2 & 1 ]-2 [ ccc 3 & 2 & 8 -2 & 1 & -7 ] = [ ccc 6 & -6 & 0 -4 & 2 & 1 ]- [ ccc 6 & 4 & 16 -4 & 2 & -14 ] = [ ccc 6-6 & -6-4 & 0-16 -4+4 & 2-2 & 1+14 ] 3 B= [ ccc 0 & -10 & -16 0 & 0 & 15 ] B=1 3 [ ccc 0 & -10 & -16 0 & 0 & 15 ] B= [ ccc 0 & -10 3 & -16 3 0 & 0 & 5 ] By (i) 2- (ii), we get 3 A =2 [ ccc 6 & -6 & 0 -4 & 2 & 1 ]- [ ccc 3 & 2 & 8 -2 & 1 & -7 ] = [ ccc 12 & -12 & 0 -8 & 4 & 2 ]- [ ccc 3 & 2 & 8 -2 & 1 & -7 ] = [ ccc 12-3 & -12-2 & 0-8 -8+2 & 4-1 & 2+7 ] 3 A = [ ccc 9 & -14 & -8 -6 & 3 & 9 ] A =1 3 [ ccc 9 & -14 & -8 -6 & 3 & 9 ] A= [ ccc 3 & -14 3 & -8 3 -2 & 1 & 3 ]

Find matrices A and B, if 2 A-B= [ ccc 6 & -6 & 0 -4 & 2 & 1 ] an…

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Question

Find matrices $A$ and $B$, if $2 A-B=\left[\begin{array}{ccc}6 & -6 & 0 \\ -4 & 2 & 1\end{array}\right]$ and $A-2 B=\left[\begin{array}{ccc}3 & 2 & 8 \\ -2 & 1 & -7\end{array}\right]$

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Answer

Given equations are
$\begin{array}{l}2 A-B=\left[\begin{array}{ccc}6 & -6 & 0 \\-4 & 2 & 1\end{array}\right] \ldots \text { (i) } \\\text { and } A-2 B=\left[\begin{array}{ccc}3 & 2 & 8 \\-2 & 1 & -7\end{array}\right] \ldots \text{(ii)}\end{array}$
By $(i) - (ii) \times\ 2$, we get
$3 B=\left[\begin{array}{ccc}6 & -6 & 0 \\-4 & 2 & 1\end{array}\right]-2\left[\begin{array}{ccc}3 & 2 & 8 \\-2 & 1 & -7\end{array}\right] $
$=\left[\begin{array}{ccc}6 & -6 & 0 \\-4 & 2 & 1\end{array}\right]-\left[\begin{array}{ccc}6 & 4 & 16 \\-4 & 2 & -14\end{array}\right] $
$=\left[\begin{array}{ccc}6-6 & -6-4 & 0-16 \\-4+4 & 2-2 & 1+14\end{array}\right] $
$\therefore 3 B=\left[\begin{array}{ccc}0 & -10 & -16 \\0 & 0 & 15\end{array}\right]$
$\therefore B=\frac{1}{3}\left[\begin{array}{ccc}0 & -10 & -16 \\0 & 0 & 15\end{array}\right] $
$\therefore B=\left[\begin{array}{ccc}0 & \frac{-10}{3} & \frac{-16}{3} \\0 & 0 & 5\end{array}\right]$
By $(i) \times 2- (ii)$, we get
$3 A =2\left[\begin{array}{ccc}6 & -6 & 0 \\-4 & 2 & 1\end{array}\right]-\left[\begin{array}{ccc}3 & 2 & 8 \\-2 & 1 & -7\end{array}\right] $
$=\left[\begin{array}{ccc}12 & -12 & 0 \\-8 & 4 & 2\end{array}\right]-\left[\begin{array}{ccc}3 & 2 & 8 \\-2 & 1 & -7\end{array}\right] $
$=\left[\begin{array}{ccc}12-3 & -12-2 & 0-8 \\-8+2 & 4-1 & 2+7\end{array}\right] $
$\therefore 3 A =\left[\begin{array}{ccc}9 & -14 & -8 \\-6 & 3 & 9\end{array}\right] $
$\therefore A =\frac{1}{3}\left[\begin{array}{ccc}9 & -14 & -8 \\-6 & 3 & 9\end{array}\right]$
$\therefore A=\left[\begin{array}{ccc}3 & \frac{-14}{3} & -\frac{8}{3} \\ -2 & 1 & 3\end{array}\right]$