Question

By the method of matrix inversion, solve the system.
$\left[\begin{array}{ccc}1& 1& 1\\ 2& 5& 7\\ 2& 1& -1\end{array}\right]\left[\begin{array}{cc}{x}_1& y_1\\ x_2& y_2\\ x_3& y_3\end{array}\right]=\left[\begin{array}{cc}9& 2\\ 52& 15\\ 0& -1\end{array}\right]$. Find $x_1+y_3$

Answer 1

$\left[\begin{array}{ccc}1& 1& 1\\ 2& 5& 7\\ 2& 1& -1\end{array}\right]\left[\begin{array}{cc}{x}_1& y_1\\ x_2& y_2\\ x_3& y_3\end{array}\right]=\left[\begin{array}{cc}9& 2\\ 52& 15\\ 0& 1\end{array}\right]$$\Rightarrow AX=B$ (1)
Clearly $|A|=-4\ne 0$. Therefore,
$adjA={\left[\begin{array}{ccc}-12& 16& -8\\ 2& -3& 1\\ 2& -5& 3\end{array}\right]}^T=\left[\begin{array}{ccc}-12& 2& 2\\ 16& -3& -5\\ -8& 1& 3\end{array}\right]$
$\therefore A^{-1}=\frac{adj.A}{|A|}=\frac{-1}{4}\left[\begin{array}{ccc}-12& 2& 2\\ 16& -3& -5\\ -8& 1& 3\end{array}\right]$
Now, $A^{-1}B=\frac{-1}{4}\left[\begin{array}{ccc}-12& 2& 2\\ 16& -3& -5\\ -8& 1& 3\end{array}\right]\left[\begin{array}{cc}9& 2\\ 52& 15\\ 0& -1\end{array}\right]$
$=\frac{-1}{4}\left[\begin{array}{cc}-4& 4\\ -12& -8\\ -20& -4\end{array}\right]=\left[\begin{array}{cc}1& -1\\ 3& 2\\ 5& 1\end{array}\right]$
From Equation (1), we get
$X=A^{-1}B\Rightarrow \left[\begin{array}{cc}{x}_1& y_1\\ x_2& y_2\\ x_3& y_3\end{array}\right]=\left[\begin{array}{cc}1& -1\\ 3& 2\\ 5& 1\end{array}\right]$
$\Rightarrow x_1=1,x_2=3,x_3=5$ or $y_1=-1,y_2=2,y_3=1$

Thus, $x_1+y_3=2$.