Question

Solve the following system of equations by matrix method:
 x − y + z = 2
2x − y = 0
2y − z = 1

Answer 1

Here, 
$$A=\left[\begin{array}{ccc}1& -1& 1\\ 2& -1& 0\\ 0& 2& -1\end{array}\right]$$
$$\left|A\right|=\left|\begin{array}{ccc}1& -1& 1\\ 2& -1& 0\\ 0& 2& -1\end{array}\right|$$
$$=1(1-0)+1(-2-0)+1(4-0)$$
$$=1-2+4$$
$$=3$$
$${\text{ Let }C}_{ij}{\text{ be the cofactors of the elements a }}_{ij}\text{ in }A\left[a_{ij}\right].\text{ Then,}$$
$$C_{11}={(-1)}^{1+1}\left|\begin{array}{cc}-1& 0\\ 2& -1\end{array}\right|=1,C_{12}={(-1)}^{1+2}\left|\begin{array}{cc}2& 0\\ 0& -1\end{array}\right|=2,C_{13}={(-1)}^{1+3}\left|\begin{array}{cc}2& -1\\ 0& 2\end{array}\right|=4$$
$$C_{21}={(-1)}^{2+1}\left|\begin{array}{cc}-1& 1\\ 2& -1\end{array}\right|=1,C_{22}={(-1)}^{2+2}\left|\begin{array}{cc}1& 1\\ 0& -1\end{array}\right|=-1,C_{23}={(-1)}^{2+3}\left|\begin{array}{cc}1& -1\\ 0& 2\end{array}\right|=-2$$
$$C_{31}={(-1)}^{3+1}\left|\begin{array}{cc}-1& 1\\ -1& 0\end{array}\right|=1,C_{32}={(-1)}^{3+2}\left|\begin{array}{cc}1& 1\\ 2& 0\end{array}\right|=2,C_{33}={(-1)}^{3+3}\left|\begin{array}{cc}1& -1\\ 2& -1\end{array}\right|=1$$
$$adjA={\left[\begin{array}{ccc}1& 2& 4\\ 1& -1& -2\\ 1& 2& 1\end{array}\right]}^T$$
$$=\left[\begin{array}{ccc}1& 1& 1\\ 2& -1& 2\\ 4& -2& 1\end{array}\right]$$
$$\Rightarrow A^{-1}=\frac{1}{\left|A\right|}adjA$$
$$=\frac{1}{1}\left[\begin{array}{ccc}1& 1& 1\\ 2& -1& 2\\ 4& -2& 1\end{array}\right]$$
$$X=A^{-1}B$$
$$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 2& -1& 2\\ 4& -2& 1\end{array}\right]\left[\begin{array}{c}2\\ 0\\ 1\end{array}\right]$$
$$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{3}\left[\begin{array}{c}2+1\\ 4+2\\ 8+1\end{array}\right]$$
$$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{1}\left[\begin{array}{c}3\\ 6\\ 9\end{array}\right]$$
$$\Rightarrow x=\frac{3}{3},y=\frac{6}{3}\text{ and }z=\frac{9}{3}$$
$$\therefore x=1,y=2\text{ and }z=3$$