# using matrices, solve the following system of equations $x+2y-3z=-4;2x+3y+2z=2 ; 3x-3y-4z=11$

Toolbox:
• If the determinant value of a $3 \times 3$ square matrix is not equal to zero, then it is a non-singular matrix
• If it is a nonsingular matrix, then inverse exists
• $A^{-1}=\frac{1}{|A|} (adjoint A)$
• $A^{-1}B=X$
Step 1:
Given
$x+2y-3z=-4$
$2x+3y+2z=2$
$3x-3y-4z=11$
This system of the equation is of the form $AX=B$
$(ie)\begin{bmatrix} 1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} -4 \\ 2 \\ 11 \end{bmatrix}$
where $A=\begin{bmatrix} 1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4 \end{bmatrix}\qquad X=\begin{bmatrix} x \\ y \\ z \end{bmatrix}\;and \; B=\begin{bmatrix} -4 \\ 2 \\ 11 \end{bmatrix}$
Let us first find the determinant value of A
$|A|=1(3 \times -4 - 2 \times -3)-2(2 \times -4-2 \times 3)-3(2 \times -3-3\times 3)$
$=(-12+6)-2(-8-6)-3(-6-9)$
$=-6+28+45=67$
$|A| \neq 0$
Hence it is a non singular matrix
Next let us find the adjoint of A
Step 2:
$A_{11}=(-1)^{1+1} \begin{vmatrix} 3 & 2 \\ -3 & -4 \end{vmatrix}=-12+6=-6$
$A_{12}=(-1)^{1+2} \begin{vmatrix} 2 & 2 \\ 3 & -4 \end{vmatrix}=-(-8-6)=14$
$A_{13}=(-1)^{1+3} \begin{vmatrix} 2 & 3 \\ 3 & -3 \end{vmatrix}=-6-9=-15$
$A_{21}=(-1)^{2+1} \begin{vmatrix} 2 & -3 \\ -3 & -4 \end{vmatrix}=-(-8-9)=17$
$A_{22}=(-1)^{2+2} \begin{vmatrix} 1 & -3 \\ 3 & -4 \end{vmatrix}=-4+9=5$
$A_{23}=(-1)^{2+3} \begin{vmatrix} 1 & 2 \\ 3 & -3 \end{vmatrix}=-(-3-6)=9$
$A_{31}=(-1)^{3+1} \begin{vmatrix} 2 & -3 \\ 3 & 2 \end{vmatrix}=4+9=13$
$A_{32}=(-1)^{3+2} \begin{vmatrix} 1 & -3 \\ 2 & 2 \end{vmatrix}=-(2+6)=-8$
$A_{33}=(-1)^{3+3} \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix}=3-4=-1$
Adj $A=\begin{bmatrix} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{bmatrix}$
$=\begin{bmatrix} -6 & 17 & 13 \\ 14 & 8 & -8 \\ 15 & 9 & -1 \end{bmatrix}$
$A^{-1}=\frac{1}{|A|} adj (A),$
we know $|A|=67$
Hence $A^{-1}=\frac{1}{67}\begin{bmatrix} -6 & 17 & 13 \\ 14 & 8 & -8 \\ 15 & 9 & -1 \end{bmatrix}$
Step 3:
$AX=B \qquad => X=A^{-1}B$
Now Substituting for $A^{-1}$,B and X we get
$\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\frac{1}{67}\begin{bmatrix} -6 & 17 & 13 \\ 14 & 5 & -8 \\ -15 & 9 & -1 \end{bmatrix} \begin{bmatrix} -4 \\ 2 \\ 11 \end{bmatrix}$
$\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\frac{1}{67}\begin{bmatrix} 24+34+143 \\ -56+10-88 \\ 60+18-11 \end{bmatrix}=\begin{bmatrix} \frac{201}{67} \\ \frac{-134}{67} \\ \frac{67}{67} \end{bmatrix}$
$\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 3 \\ -2 \\ 1 \end{bmatrix}$
Therefore $x=3,y=-2,and \; z=1$

edited Apr 8, 2013 by meena.p