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 system of equations $x-y+z=4,x-2y-2z=9,2x+y+3z=1$

Answer 1

Let A=$\left[\begin{array}{ccc}-4& 4& 4\\ -7& 1& 3\\ 5& -3& -1\end{array}\right]$, B=$\left[\begin{array}{ccc}1& -1& 1\\ 1& -2& 2\\ 2& 1& 3\end{array}\right]$ $\therefore AB=\left[\begin{array}{ccc}-4& 4& 4\\ -7& 1& 3\\ 5& -3& -1\end{array}\right]\left[\begin{array}{ccc}1& -1& 1\\ 1& -2& 2\\ 2& 1& 3\end{array}\right]=\left[\begin{array}{ccc}8& 0& 0\\ 0& 8& 0\\ 0& 0& 8\end{array}\right]$

$\Rightarrow AB=8I....\left(i\right)$

Consider the given systems of eqautions: $x-y+z=4,x-2y-2z=9,2x+y+3z=1$

These equations can be expressed as :BX=D where B=$\left[\begin{array}{ccc}1& -1& 1\\ 1& -2& -2\\ 2& 1& 3\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right],C=\left[\begin{array}{c}4\\ 9\\ 1\end{array}\right]$

Therefore,$X=B^{-1}C=\frac{1}{2}AC$ $[\text{By (i),AB}=8I\Rightarrow \left(\frac{1}{8}A\right)B=I\therefore B^{-1}=\frac{1}{8}A]$

$X=\frac{1}{2}\left[\begin{array}{ccc}-4& 4& 4\\ -7& 1& 3\\ 5& -3& -1\end{array}\right]\left[\begin{array}{c}4\\ 9\\ 1\end{array}\right]\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}3\\ -2\\ 1\end{array}\right]\therefore \text{by equality of matrices:}:x=3,y=-2,z=-1$