Question 16.7.2: Find the inverse of A =(1 3 3 1 3 4 1 4 3)....
Find the inverse of \mathbf{A}={\left(\begin{array}{l l}1&3&3 \\ 1& 3 & 4 \\ 1& 4 &3\end{array}\right)}.
First, write down the 3 ×6 matrix
(A:I)=\left ( \begin{array}{ccc|ccc} 1 & 3 & 3 & 1 & 0 & 0 \\ 1 & 3 & 4 &0 &1 & 0 \\ 1 & 4 & 3 & 0 & 0 & 1 \end{array} \right )three columns are the columns of A and whose next three columns are the columns of the 3 × 3 identity matrix.
The idea is now to use elementary operations on this matrix so that, in the end, the three first columns constitute an identity matrix. Then the last three columns will constitute the inverse of A.
To start, we multiply the first row by −1 and add the result to the second row. This gives a zero in the second row and the first column. You should be able then to understand the other operations used and why they are chosen.
We conclude that
A^{-1}=\left ( \begin{matrix} 7 & -3 & -3 \\ -1 & 0 & 1 \\ -1 & 1 & 0 \end{matrix} \right )This can be checked by using matrix multiplication to verify that AA^{−1} = I.