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Using matrices, solve the following system of linear equations:
$x+2y-3z= -4$
$2x+3y+2z= 2$
$3x-3y-4z= 11$

Answers (1)

we have $x+2y-3z= -4,\; 2x+3y+2z= 2,\; 3x-3y-4z= 11$
Let   $A= \begin{bmatrix} 1 & 2 & -3\\ 2&3 &2 \\ 3& -3 & -4 \end{bmatrix},\; B= \begin{bmatrix} -4\\ 2 \\ 11 \end{bmatrix},\; x= \begin{bmatrix} x\\ y \\ z \end{bmatrix}$
so that
   $Ax= B$ i.e $x= A^{-1}B---(i)$
   $\because \left | A \right |= \begin{vmatrix} 1 & 2 &-3 \\ 2& 3 &2 \\ 3&-3 & -4 \end{vmatrix}$
$= 1\left ( -6 \right )-2\left ( -14 \right )-3\left ( -15 \right )= -6+28+45= 67\not\equiv 0$
$\therefore A^{-1}$ exists
consider A_{ij be the cofactor of element aij of matrix A

by (i)}
By equality of matrices, we get
$\therefore x= 3,\; y= -2,z= 1$