Browse Questions

Using matrices, solve the following system of equation $x+y+z=3; 2x-y+z=2 ; x-2y+3z=2$

Toolbox:
• A matrix is said to be singular if |A|= 0.
• A matrix is said to be invertible if |A|$\neq 0$.
• If A is a non-singular matrix,AX=B,then $X=A^{-1}B.$
• Using this we can solve the system of equation which has unique solution.
Step 1:
$x+y+z=3$
$2x-y+z=2$
$x-2y+3z=2$
This system can be written as $AX=B$,
Where $A=\begin{bmatrix}1 & 1 & 1\\2 & -1 & 1\\1 & -2 & 3\end{bmatrix}$
$X=\begin{bmatrix}x\\y\\z\end{bmatrix}$
$B=\begin{bmatrix}3\\2\\2\end{bmatrix}$
$\mid A\mid=1(-3+2)-1(6-1)+1(-4+1)$
$\quad\;\;\;=-1-5-3$
$\quad\;\;\;=-9$
Step 2:
$A_{11}=1\begin{vmatrix}-1& 1\\-2& 3\end{vmatrix}=-1$
$A_{12}=-1\begin{vmatrix}2& 1\\1& 3\end{vmatrix}=-(5)$
$A_{13}=1\begin{vmatrix}2& -1\\1& -2\end{vmatrix}=-3$
$A_{21}=-1\begin{vmatrix}1& 1\\-2& 3\end{vmatrix}=-(5)$
$A_{22}=1\begin{vmatrix}1& 1\\1& 3\end{vmatrix}=2$
$A_{23}=-1\begin{vmatrix}1& 1\\1& -2\end{vmatrix}=3$
Hence $adj\; A=\begin{bmatrix}A_{11}&A_{21}&A_{31}\\A_{12} &A_{22}&A_{32}\\A_{13} &A_{23}&A_{33}\end{bmatrix}$
$\qquad\qquad\;\;\;=\begin{bmatrix}-1&-5&2\\-5 &2&1\\-3 &3&-3\end{bmatrix}$
$\therefore A^{-1}=\large\frac{1}{\mid A\mid}$$adj\;A \qquad=\large\frac{1}{-9}$$\begin{bmatrix}-1& -5&2\\-5&2&1\\-3&3&-3\end{bmatrix}$
Step 3:
$X=A^{-1}B$
$X=\large\frac{1}{-9}$$\begin{bmatrix}-1 & -5 & 2\\-5 & 2 &1\\-3 & 3 &-3\end{bmatrix}\begin{bmatrix}3\\2\\2\end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix}=-\large\frac{-1}{9}$$\begin{bmatrix}-3-10+4\\-15+4+2\\-9+6-6\end{bmatrix}$
$\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}\large\frac{-9}{-9}\\\large\frac{-9}{-9}\\\large\frac{-9}{-9}\end{bmatrix}=\begin{bmatrix}1\\1\\1\end{bmatrix}$
$\therefore x=1,y=1,z=1$