Question

Determine the product
$\left[\begin{array}{ccc}-4& 4& 4\\ -7& 1& 3\\ 5& -3& -1\end{array}\right]\left[\begin{array}{ccc}1& -1& 1\\ 1& -2& -2\\ 2& 1& 3\end{array}\right]$
and use it to solve the system of equations
$x-y+z=4,x-2y-2z=9,2x+y+3z=1$.

Answer 1

Given product say
$AB=\left[\begin{array}{ccc}-4& 4& 4\\ -7& 1& 3\\ 5& -3& -1\end{array}\right]\left[\begin{array}{ccc}1& -1& 1\\ 1& -2& -2\\ 2& 1& 3\end{array}\right]=\left[\begin{array}{ccc}-4+4+8& 4-8+4& -4-8+12\\ -7+1+6& 7-2+3& -7-2+9\\ 5-3-2& -5+6-1& 5+6-3\end{array}\right]=\left[\begin{array}{ccc}8& 0& 0\\ 0& 8& 0\\ 0& 0& 8\end{array}\right]=8I_3$

Now, $AB$ = $8I_3\Rightarrow AB{B}^{-1}=8B^{-1}\Rightarrow \frac{1}{8}A=B^{-1}$

Given system of equations are :

$x-y+z=4,x-2y-2z=9$ and $2x+y+3z=1$

Their matrix form is : $BX=C$

$\Rightarrow X=B^{-1}C=\left(\frac{1}{8}A\right)C=\frac{1}{8}\left[\begin{array}{ccc}-4& 4& 4\\ -7& 1& 3\\ 5& -3& -1\end{array}\right]\left[\begin{array}{c}4\\ 9\\ 1\end{array}\right]=\frac{1}{8}\left[\begin{array}{c}-16+36+4\\ -28+9+3\\ 20-27-1\end{array}\right]=\frac{1}{8}\left[\begin{array}{c}24\\ -16\\ -8\end{array}\right]X=\left[\begin{array}{c}3\\ -2\\ -1\end{array}\right]$

Hence, $x=3,y=-2$ and $z=-1$