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# Using matrices,solve the system of linear equations,$2x-y+z=3$$-x+2y-z=-4$$x-y+2z=1$

Toolbox:
• If |A|$\neq$ 0,then it is a non-singular matrix.
• Hence it is invertible.
• $A^{-1}=\frac{1}{|A|}.adj\; A$
• X=$A^{-1}B.$
Given:

2x-y+z=3

-x+2y-z=-4

x-y+2z=1
The given system of equation is of the form

AX=B.

(i.e)$\begin{bmatrix}2 & -1 & 1\\-1 & 2 & -1\\1 & -1 & 2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}3\\-4\\1\end{bmatrix}$

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

To find $A^{-1}$,let us first see if [A] is singular or non-singular.

|A| can be determined by expanding along $R_1$

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

$\;\;=2(4-1)+1(-2+1)+1(1-2)$

$\;\;=6-1-1=4\neq 0$

Since $|A| \neq 0$ [A] is non-singular.

Now let us find the adj A.

To find adj A,let us find the minors and cofactors of the elements of [A].
$M_{11}=\begin{vmatrix}2 & -1\\-1 & 2\end{vmatrix}$=4-1=3.

$M_{12}=\begin{vmatrix}-1 & -1\\1 & 2\end{vmatrix}$=-2+1=-1.

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

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

$M_{22}=\begin{vmatrix}2 & 1\\1 & 2\end{vmatrix}$=4-1=3.

$M_{23}=\begin{vmatrix}2 & -1\\1 & -1\end{vmatrix}$=-2+1=-1.

$M_{31}=\begin{vmatrix}-1 & 1\\2 & -1\end{vmatrix}$=1-2=-1.

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

$M_{33}=\begin{vmatrix}2 & -1\\-1 & 2\end{vmatrix}$=4-1=3.

$A_{11}=(-1)^{1+1}$.3=3.

$A_{12}=(-1)^{1+2}$.-1=1.

$A_{13}=(-1)^{1+3}$.-1=-1.

$A_{21}=(-1)^{2+1}$.-1=1.

$A_{22}=(-1)^{2+2}$.3=3.

$A_{23}=(-1)^{2+3}$.-1=1.

$A_{31}=(-1)^{3+1}$.-1=-1.

$A_{32}=(-1)^{3+2}$.-1=1.

$A_{33}=(-1)^{3+3}$.3=3.

Now adj A=$\begin{bmatrix}A_{11} & A_{21} & A_{31}\\A_{12} & A_{22} & A_{32}\\A_{13} & A_{23} & A_{33}\end{bmatrix}$

$\qquad\qquad=\begin{bmatrix}3 & 1 & -1\\1 & 3 & 1\\-1 & 1 & 3\end{bmatrix}$

$A^{-1}=\frac{1}{|A|}Adj \;A$,where |A|=4.

$A^{-1}=\frac{1}{4}\begin{bmatrix}3 & 1 & -1\\1 & 3 & 1\\-1 & 1 & 3\end{bmatrix}$

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

$\begin{bmatrix}x\\y \\z\end{bmatrix}=1/4\begin{bmatrix}3 & 1 & -1\\1 & 3 & 1\\-1 & 1 & 3\end{bmatrix}\begin{bmatrix}3\\-4\\1\end{bmatrix}$

Matrix multiplication can be done by multiplying rows of matrix A by column of matrix B.

$\begin{bmatrix}x\\y\\z\end{bmatrix}=1/4\begin{bmatrix}9-4-1\\3-12+1\\-3-4+3\end{bmatrix}=1/4\begin{bmatrix}4\\-8\\-4\end{bmatrix}$

$\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}4/4\\-8/4\\-4/4\end{bmatrix}=\begin{bmatrix}1\\-2\\-1\end{bmatrix}$

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