# If $A= \begin{bmatrix} 0 & 1 & 2\\ 1& 2 &3 \\ 3& 1&1 \end{bmatrix},$ find A-1. Using A-1 , solve the system of equations $y+2z= 5$ $x+2y+3z= 10$ $3x+y+z= 9$

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

$y+2z= 5$

$x+2y+3z= 10$

$3x+y+z= 9$
$\therefore \left | A \right |= 0 \left ( 2-3 \right )-1\left ( 1-9 \right )+2\left ( 1-6 \right )$
$= -1\left ( -8 \right )+2\left ( -5 \right )$
$\Rightarrow 8-10$
$= -2\: \: \: \therefore A^{-1}exist$
Now,
$A_{11}= -1\; \; A_{12}= 8\; \; A_{13}= -5$
$A_{21}= 1\; \; A_{22}= -6\; \; A_{23}= 3$
$A_{31}= -1\; \; A_{32}= 2\; \; A_{33}= -1$
$\therefore A^{-1}= \frac{1}{\left | A \right |}\left ( adj\: A \right )= \frac{-1}{2}\begin{bmatrix} -1 & 1 &-1 \\ 8& -6 &2 \\ -5& 3 & -1 \end{bmatrix}---(1)$
Now the given system of the equation can be written in the form of AX = B where
$A= \begin{bmatrix} 0 & 1 &2 \\ 1 & 2 &3 \\ 3 & 1 &1 \end{bmatrix}x= \begin{bmatrix} x\\ y \\z \end{bmatrix}\S B= \begin{bmatrix} 5\\ 10 \\9 \end{bmatrix}$
The solution of the given of equation is given by
$X= A^{-1}B$
$\begin{bmatrix} x\\ y \\ z \end{bmatrix}= \frac{-1}{2}\begin{bmatrix} -1 & 1 &-1 \\ 8 & -6 &2 \\ -5& 3 &-1 \end{bmatrix}\begin{bmatrix} 5\\ 10 \\ 9 \end{bmatrix}\: Using (1)$
$= \frac{1}{2}\begin{bmatrix} 1 & -1 & 1\\ -8& 6 & -2\\ 5 & -3 & 1 \end{bmatrix}\begin{bmatrix} 5\\ 10 \\ 9 \end{bmatrix}$
$= \frac{1}{2}\begin{bmatrix} 1 -10 +9\\ -40 +60 -18\\ 25 -30 +9 \end{bmatrix}$
Value of x,y,z are 0,1,2 respectively.

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