Question

If $A=\left|\begin{array}{ccc}1& 2& 0\\ -2& -1& -2\\ 0& -1& 1\end{array}\right|$, then find the value of $A^{-1}$
Using $A^{-1}$, solve the system of linear equations x - 2y = 10, 2x- y - z = 8 and -2y + z = 7

Answer 1

We have, $A=\left|\begin{array}{ccc}1& 2& 0\\ -2& -1& -2\\ 0& -1& 1\end{array}\right|$ ..... (i)
$\therefore |A|=1(-3)-2(-2)+0=1\ne 0$
Now, $A_{11}=-3,A_{12}=2,A_{13}=2,A_{21}=-2,A_{22}=1,A_{23}=1,A_{31}=-4,A_{32}=2and{A}_{33}=3$
$\therefore adj\left(A\right)={\left|\begin{array}{ccc}-3& 2& 2\\ -2& 1& 1\\ -4& 2& 3\end{array}\right|}^T=\left|\begin{array}{ccc}-3& -2& -4\\ 2& 1& 2\\ 2& 1& 3\end{array}\right|\therefore A^{-1}=\frac{adjA}{|A|}=\frac{1}{1}\left|\begin{array}{ccc}-3& -2& -4\\ 2& 1& 2\\ 2& 1& 3\end{array}\right|$
$\Rightarrow A^{-1}=\left|\begin{array}{ccc}-3& -2& -4\\ 2& 1& 2\\ 2& 1& 3\end{array}\right|$ .... (ii)
Also, we have the system of linear equations as
x - 2y = 10,
2x - y -z = 8
and -2y + z=7
In the form of CX = D,
$\left[\begin{array}{ccc}1& 2& 0\\ 2& -1& -1\\ 0& -2& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}10\\ 8\\ 7\end{array}\right]$
We know that, $(A^T{)}^{-1}=(A^{-1}{)}^T$
$\therefore C^T=\left|\begin{array}{ccc}1& 2& 0\\ -2& -1& -2\\ 0& -1& 1\end{array}\right|=A$ [using Eq. (i)]
$\therefore X=C^{-1}D$
$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{ccc}-3& 2& 2\\ -2& 1& 1\\ -4& 2& 3\end{array}\right]\left[\begin{array}{c}10\\ 8\\ 7\end{array}\right]=\left[\begin{array}{c}-30+16+14\\ -20+8+7\\ -40+16+21\end{array}\right]=\left[\begin{array}{c}0\\ -5\\ -3\end{array}\right]$
$\therefore$ x = 0, y = -5 and z = -3