Question

If $A=\left[\begin{array}{ccc}3& 1& 2\\ 3& 2& -3\\ 2& 0& -1\end{array}\right]$, find $A^{-1}.$
Hence, solve the system of equations : 3x+3y +2z=1, x+2y=4, 2x-3y-z=5.

Answer 1

For the given matrix A, $|A|=\left|\begin{array}{ccc}3& 1& 2\\ 3& 2& -3\\ 2& 0& -1\end{array}\right|=3(-2)-1\left(3\right)+2(-4)=-17\ne 0\therefore A^{-1}$ exists.
Consider $A_{ij}$ be the cofactor of the element $a_{ij}$ of matrix A.
$\begin{array}{cc}{A}_{11}=-2,A_{12}=-3,A_{13}=-4,& \\ A_{21}=1,A_{22}=-7,A_{23}=2& \\ A_{31}=-7,A_{32}=15,A_{33}=3& \therefore \text{adj.A}=\left[\begin{array}{ccc}-2& 1& -7\\ -3& -7& 15\\ -4& 2& 3\end{array}\right]\end{array}$
So, $A^{-1}=\frac{adj.A}{|A|}=\frac{1}{-17}\left[\begin{array}{ccc}-2& 1& -7\\ -3& -7& 15\\ -4& 2& 3\end{array}\right]=\frac{1}{17}\left[\begin{array}{ccc}-2& -1& 7\\ 3& 7& -15\\ 4& -2& -3\end{array}\right]$
Now consider the equations: 3x+3y +2z=1, x+2y =4, 2x-3y -z =5
Let $P=\left[\begin{array}{ccc}3& 3& 2\\ 1& 2& 0\\ 2& -3& -1\end{array}\right]=A^T,B=\left[\begin{array}{c}1\\ 4\\ 5\end{array}\right]\text{and, X}=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$
Since $PX=B\Rightarrow X=P^{-1}B=\left(A^{-1}{)}^{T}B\right[\because P^{-1}=(A^T{)}^{-1}=(A^{-1}{)}^T$
So, $X=\frac{1}{17}\left[\begin{array}{ccc}2& 3& 4\\ -1& 7& -2\\ 7& -15& -3\end{array}\right]\left[\begin{array}{c}1\\ 4\\ 5\end{array}\right]\Rightarrow X=\frac{1}{17}\left[\begin{array}{c}34\\ 17\\ -68\end{array}\right]\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}2\\ 1\\ -4\end{array}\right]$
By equality of matrices, we get: x=2, y=1, z=-4.