If <img alt="A= \begin{bmatrix} 2 & -3 & 5\\ 3& 2 &-4 \\ 1&2 &-2 \end{bmatrix}," src="https://entrancecorner.codecogs.com/gif.latex?A%3D%20%5Cbegin%7Bbmatrix%7D%202%20%26%20-3%20%26%205%5C%5C%203%26%202%20%26-4%20%5C%5C%201%262%

If $A= \begin{bmatrix} 2 & -3 & 5\\ 3& 2 &-4 \\ 1&2 &-2 \end{bmatrix},$ find $A^{-1}$ . Use it to solve the system of equations
$2x-3y+5z= 11$
$3x+2y-4z= -5$
$x+y-2z= -3\cdot$

Answers (1)

$\left | A \right |= \begin{vmatrix} 2 & -3 &5 \\ 3 &2 & -4\\ 1& 1& -2 \end{vmatrix}$
$= 2\left ( -4+4 \right )+3\left ( -6+4 \right )+5\left ( 3-2 \right )$
$\left | A \right |= 0-6+5= -1\neq 0$
$\therefore A^{-1}$ exists
Let Aij be the cofactor of the matrix   $A= \left [ aij \right ]$ ,
Then
       $A_{11}= \left ( -1 \right )^{1+1}\begin{vmatrix} 2 &-4 \\ 1&-2 \end{vmatrix}= -4+4= 0$
       $A_{12}= \left ( -1 \right )^{1+2}\begin{vmatrix} 3 &-4 \\ 1&-2 \end{vmatrix}= -\left ( -6+4 \right )= 2$
        $A_{13}= \left ( -1 \right )^{1+3}\begin{vmatrix} 3 &2 \\ 1&1 \end{vmatrix}= 3-2= 1$
      $A_{21}= \left ( -1 \right )^{2+1}\begin{vmatrix} -3 &5 \\ 1&-2 \end{vmatrix}= -\left ( 6-5 \right )= -1$
      $A_{22}= \left ( -1 \right )^{2+2}\begin{vmatrix} 2 &5 \\ 1&-2 \end{vmatrix}= -4-5= -9$
      $A_{23}= \left ( -1 \right )^{2+3}\begin{vmatrix} 2 &-3 \\ 1&1 \end{vmatrix}= -\left ( 2+3 \right )= -5$      $A_{31}= \left ( -1 \right )^{3+1}\begin{vmatrix} -3 &5 \\ 2&-4 \end{vmatrix}=\left ( 12-10 \right )= 2\; \; A_{32}= \left ( -1 \right )^{3+2}\begin{vmatrix} 2 &5 \\ 3&-4 \end{vmatrix}=-\left ( -8+15 \right )= 23$
$A_{33}= \left ( -1 \right )^{3+3}\begin{vmatrix} 2 &-3 \\ 3&2 \end{vmatrix}=4+9= 13$
$\therefore Aij= \begin{bmatrix} 0 & 2& 1\\ -1&-9 & -5\\ 2&23 & 13 \end{bmatrix}$
$adj\left ( A \right )= \left [ Aij \right ]^{T}= \begin{bmatrix} 0 & -1& 2\\ 2&-9 & 23\\ 1&-5 & 13 \end{bmatrix}$
$A^{-1}= \frac{adj\left ( A \right )}{\left | A \right |}= \frac{1}{\left ( -1 \right )}\begin{bmatrix} 0 & -1& 2\\ 2& -9&23 \\ 1 &-15 & 13 \end{bmatrix}$
       $= \begin{bmatrix} 0 & 1&- 2\\ -2& 9&-23 \\ -1 &15 & -13 \end{bmatrix}$
Now, given system of equations can be written as
$\begin{bmatrix} 2 & -3&- 5\\ 3& 2&-4 \\ 1 &1 & -2 \end{bmatrix}\begin{bmatrix} x\\ y \\ z \end{bmatrix}=\begin{bmatrix} 11\\ -5 \\ -3 \end{bmatrix}$
$\Rightarrow Ax= B$
$\therefore x= A^{-1}B$
$x= \begin{bmatrix} 0 & 1 &-2 \\ -2& 9 &-23 \\ -1& 5 & -13 \end{bmatrix}\begin{bmatrix} 11\\ -5 \\ -3 \end{bmatrix}$
$\Rightarrow \begin{bmatrix} x\\ y \\ z \end{bmatrix}= \begin{bmatrix} 0 -5+6 \\ -22 -45 +69 \\ -11 -25 +39 \end{bmatrix}= \begin{bmatrix} 1\\ 2 \\ 3 \end{bmatrix}$
$\therefore x= 1,y= 2,z= 3$