# Solve the following systems of equations using matrix method : $x-y=3;\: 2x+3y+4x=17; \: y+2z=7$

Toolbox:
• If A is a nonsingular matrix,then its inverse exists.
• $A^{-1}=\frac{1}{|A|}.adj\; A$
• X=$A^{-1}B.$
Given:
x-y=3
2x+3y+4z=17.
y+2z=7
The given system of equation is of the form
AX=B.
(i.e)$\begin{bmatrix}1 & -1 & 0\\2 & 3 & 4\\0 & 1& 2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}3\\17\\7\end{bmatrix}$

Therefore x=$A^{-1}B.$

Let us now find $A^{-1}$

$|A|=\begin{bmatrix}1 & -1 & 0\\2 & 3 & 4\\0 & 1 & 2\end{bmatrix}$

To find |A|,let us expand along $R_1$

|A|=$1(3\times 2-4\times 1)-(-1)(2\times 2-4\times 0)+0$

$\;\;\;=(6-4)+4=6\neq 0$

|A|$\neq 0$,hence A is invertible.

Now let us find the adj A.

To find adj A,let us find the minors and cofactors of the elements of [A].

$M_{11}=\begin{vmatrix}3 & 4\\1 & 2\end{vmatrix}$=6-4=2.

$M_{12}=\begin{vmatrix}2 & 4\\0 & 2\end{vmatrix}$=4-0=4.

$M_{13}=\begin{vmatrix}2 & 3\\0 & 1\end{vmatrix}$=2-0=2.

$M_{21}=\begin{vmatrix}-1 & 0\\1 & 2\end{vmatrix}$=-2-0=-2.

$M_{22}=\begin{vmatrix}1 & 0\\0 & 2\end{vmatrix}$=2-0==2.

$M_{23}=\begin{vmatrix}1 & -1\\0 & 1\end{vmatrix}$=1-0=1.

$M_{31}=\begin{vmatrix}-1 & 0\\3 & 4\end{vmatrix}$=-4-0=-4.

$M_{32}=\begin{vmatrix}1 & 0\\2 & 4\end{vmatrix}$=4-0=4.

$M_{33}=\begin{vmatrix}1 &- 1\\2 & 3\end{vmatrix}$=3+2=5.

$A_{11}=(-1)^{1+1}$.2=2.
$A_{12}=(-1)^{1+2}$.4=-4.
$A_{13}=(-1)^{1+3}$.2=2.
$A_{21}=(-1)^{2+1}$.-2=2.
$A_{22}=(-1)^{2+2}$.2=2.
$A_{23}=(-1)^{2+3}$.1=-1.
$A_{31}=(-1)^{3+1}$.-4=-4.
$A_{32}=(-1)^{3+2}$.4=-4.
$A_{33}=(-1)^{3+3}$.5=5.
Now adj A=$\begin{bmatrix}A_{11} & A_{21} & A_{31}\\A_{12} & A_{22} & A_{32}\\A_{13} & A_{23} & A_{33}\end{bmatrix}$

$\qquad\qquad=\begin{bmatrix}2 & 2& -4\\-4 & 2& -4\\2 & -1 & 5\end{bmatrix}$

$A^{-1}=\frac{1}{|A|}Adj \;A$,where |A|=6.

$A^{-1}=\frac{1}{6}\begin{bmatrix}2 & 2 & -4\\-4 & 2 & -4\\2 & -1 & 5\end{bmatrix}$

X=$A^{-1}B$,substituting for x,$A^{-1}$ and B,

$\begin{bmatrix}x\\y \\z\end{bmatrix}=1/6\begin{bmatrix}2 & 2 & -4\\-4 & 2 & -4\\2 & -1 & 5\end{bmatrix}\begin{bmatrix}3\\17\\7\end{bmatrix}$

$\begin{bmatrix}x\\y\\z\end{bmatrix}=1/6\begin{bmatrix}6+34-28\\-12+34-28\\6-17+35\end{bmatrix}=1/6\begin{bmatrix}-12\\-6\\-24\end{bmatrix}$

$\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}12/6\\-6/6\\24/6\end{bmatrix}$

$\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}2\\-1\\4\end{bmatrix}$

Hence x=2,y=-1,z=4.