Given A= [ ccc 2 & 2 & -4 -4 & 2 & -4 2 & -1 & 5 ], B= [ ccc 1 & -1 & 0 2 & 3 & 4 0 & 1 & 2 ], find B A and use this to solve the system of equations y + 2 z=7, x-y=3,2 x+3 y+4 z=17

We have, A = [ ccc 2 & 2 & -4 -4 & 2 & -4 2 & -1 & 5 ], B = [ ccc 1 & -1 & 0 2 & 3 & 4 0 & 1 & 2 ] BA = [ ccc 1 & -1 & 0 2 & 3 & 4 0 & 1 & 2 ] [ ccc 2 & 2 & -4 -4 & 2 & -4 2 & -1 & 5 ]= [ lll 6 & 0 & 0 0 & 6 & 0 0 & 0 & 6 ]=6 I B ^ -1 =A 6 =1 6 A=1 6 [ ccc 2 & 2 & -4 -4 & 2 & -4 2 & -1 & 5 ] Also, x-y=3,2 x+3 y+4 z=17 and y+2 z=7 [ ccc 1 & -1 & 0 2 & 3 & 4 0 & 1 & 2 ] [ l x y z ]= [ c 3 17 7 ] [ l x y z ]= [ ccc 1 & -1 & 0 2 & 3 & 4 0 & 1 & 2 ]^ -1 [ c 3 17 7 ] =1 6 [ ccc 2 & 2 & -4 -4 & 2 & -4 2 & -1 & 5 ] [ c 3 17 7 ] [using Eq. (i)] =1 6 [ c 6+34-28 -12+34-28 6-17+35 ] =1 6 [ c 12 -6 24 ]= [ c 2 -1 4 ] x=2, y=-1 and z=4

Given A= [ ccc 2 & 2 & -4 -4 & 2 & -4 2 & -1 & 5 ], B= [ ccc 1 & …

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Question

Given $A=\left[\begin{array}{ccc}2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5\end{array}\right], B=\left[\begin{array}{ccc}1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{array}\right]$, find $B A$ and use this to solve the system of equations $y + 2 z=7, x-y=3,2 x+3 y+4 z=17$

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Answer

We have,
$A =\left[\begin{array}{ccc}2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5\end{array}\right], B =\left[\begin{array}{ccc}1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{array}\right]$
$\therefore BA =\left[\begin{array}{ccc}1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{array}\right]\left[\begin{array}{ccc}2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5\end{array}\right]=\left[\begin{array}{lll}6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6\end{array}\right]=6 I$
$\therefore B ^{-1}=\frac{A}{6}=\frac{1}{6} A=\frac{1}{6}\left[\begin{array}{ccc}2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5\end{array}\right]$
Also, $x-y=3,2 x+3 y+4 z=17$ and $y+2 z=7$
$\Rightarrow\left[\begin{array}{ccc}1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}3 \\ 17 \\ 7\end{array}\right]$
$\therefore\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{ccc}1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{array}\right]^{-1}\left[\begin{array}{c}3 \\ 17 \\ 7\end{array}\right]$
$=\frac{1}{6}\left[\begin{array}{ccc}2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5\end{array}\right]\left[\begin{array}{c}3 \\ 17 \\ 7\end{array}\right]$
$[$using Eq.$ (i)] =\frac{1}{6}\left[\begin{array}{c}6+34-28 \\ -12+34-28 \\ 6-17+35\end{array}\right]$
$=\frac{1}{6}\left[\begin{array}{c}12 \\ -6 \\ 24\end{array}\right]=\left[\begin{array}{c}2 \\ -1 \\ 4\end{array}\right]$
$\therefore x=2, y=-1$ and $z=4$

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