Question

If $A=\left[\begin{array}{ccc}1& 3& 3\\ 1& 4& 3\\ 1& 3& 4\end{array}\right]$ then find $A^{-1}$

Answer 1

We can write, $A.A^{-1}=I$

By performing elementary row operations, we can find the inverse of A.

$\left[\begin{array}{ccc}1& 3& 3\\ 1& 4& 3\\ 1& 3& 4\end{array}\right].A^{-1}=$ $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

Performing $R_2\to R_2-R_{1}and{R}_3\to R_3-R_1$

$\left[\begin{array}{ccc}1& 3& 3\\ 0& 1& 0\\ 0& 0& 1\end{array}\right].A^{-1}=$ $\left[\begin{array}{ccc}1& 0& 0\\ -1& 1& 0\\ -1& 0& 1\end{array}\right]$

Performing $R_1\to R_1-3.R_2$

$\left[\begin{array}{ccc}1& 0& 3\\ 0& 1& 0\\ 0& 0& 1\end{array}\right].A^{-1}=$ $\left[\begin{array}{ccc}4& -3& 0\\ -1& 1& 0\\ -1& 0& 1\end{array}\right]$

Performing $R_1\to R_1-3.R_3$

$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right].A^{-1}=$ $\left[\begin{array}{ccc}7& -3& -3\\ -1& 1& 0\\ -1& 0& 1\end{array}\right]$

We have the product $I.A^{-1}$ on the L.H.S after performing elementary row operations. That implies on the R.H.S, we have $A^{-1}$.

$A^{-1}=\left[\begin{array}{ccc}7& -3& -3\\ -1& 1& 0\\ -1& 0& 1\end{array}\right]$