Using matrix method,Solve the following system of equation: <img alt="2x+y-z= 1" src="https://entrancecorner.codecogs.com/gif.latex?2x&

Using matrix method,Solve the following system of equation:
$3x-2y+3z= 8$
$2x+y-z= 1$
$4x-3y+2z= 4$

Answers (1)

$3x-2y+3z= 8$
$2x+y-z= 1$    $\left ( given \right )$
$4x-2y+2z= 4$
These equation can be written in the form $Ax= 13$
where
$A= \begin{bmatrix} 3 & -2& 3\\ 2 &1 &-1 \\ 4& -3 &2 \end{bmatrix}\: x= \begin{bmatrix} x\\ y \\ z \end{bmatrix}\: and\: B= \begin{bmatrix} 8\\ 1 \\4 \end{bmatrix}$
Now
$\left | A \right |= 3\left ( 2-3 \right )+2\left ( 4+4 \right )+3\left ( -6-4 \right )$
$= -17\neq 0$
Hence,A is non-singular and so its inverse exists cofactors of A are,
$A_{11}= -1 \: \:\; A_{12}= -8\: \:\; A_{13}= -10$
$A_{21}= -5 \: \:\; A_{22}= -6\: \:\; A_{23}= 1$
$A_{31}= -1 \: \:\; A_{32}= 9\: \:\; A_{33}= 7$
$\therefore adj\: A= \begin{bmatrix} -1 & -8& -10\\ -5&-6 & 1\\ -1& 9 & 7 \end{bmatrix}$
$= \begin{bmatrix} -1 & -5& -1\\ -8&-6 & 9\\ 10& 1 & 7 \end{bmatrix}$
Now $A^{-1}= \frac{1}{\left | A \right |}\: adj\: A$
$= \frac{1}{-17}\begin{bmatrix} -1 & -5& -1\\ -8& -6 &9 \\ -10& 1 &7 \end{bmatrix}$
$\Rightarrow x= A^{-1}B$
$=\frac{-1}{17}\begin{bmatrix} -1 & -5 &-1 \\ -8& -6&9 \\ -10& 1 & 7 \end{bmatrix}\begin{bmatrix} 8\\ 1 \\4 \end{bmatrix}$
$\begin{bmatrix} x\\ y \\z \end{bmatrix}= \frac{-1}{17}\begin{bmatrix} -17\\-34 \\ -51 \end{bmatrix}= \begin{bmatrix} 1\\ 2 \\ 3 \end{bmatrix}$
Hence, $x= 1,y= 2,z= 3$