Question 8.4.4: Finding the Multiplicative Inverse of a 3 × 3 Matrix Find th...
Finding the Multiplicative Inverse of a 3 × 3 Matrix
Find the multiplicative inverse of
A=\left[\begin{array}{rrr}1 & -1 & 1 \\0 & -2 & 1 \\-2 & -3 & 0\end{array}\right].
Step 1 Form the augmented matrix \left[A \mid I_3\right].
Step 2 Perform row operations on \left[A \mid I_3\right] to obtain a matrix of the form \left[I_3 \mid B\right]. To the left of the vertical dividing line, we want 1 s down the diagonal from upper left to lower right and 0s elsewhere.
\left[\begin{array}{rrr|rrr}1 & -1 & 1 & 1 & 0 & 0 \\0 & -2 & 1 & 0 & 1 & 0 \\-2 & -3 & 0 & 0 & 0 & 1\end{array}\right]\overset{\text { Replace row } 3 \text { by } 2 R_1+R_3 \text {. }}{\longrightarrow } \left[\begin{array}{lll|lll}1 & -1 & 1 & 1 & 0 & 0 \\0 & -2 & 1 & 0 & 1 & 0 \\0 & -5 & 2 & 2 & 0 & 1\end{array}\right]\overset{-\frac{1}{2} R_2}{\longrightarrow } \left[\begin{array}{rrr|rrr}1 & -1 & 1 & 1 & 0 & 0 \\0 & 1 & -\frac{1}{2} & 0 & -\frac{1}{2} & 0 \\0 & -5 & 2 & 2 & 0 & 1\end{array}\right]\overset{\text{Replace row 1 by 1} R_2+R_1.\\\text{Replace row 3 by 5} R_2+R_3.}{\longrightarrow } \left[\begin{array}{rrr|rrr}1 & 0 & \frac{1}{2} & 1 & -\frac{1}{2} & 0 \\0 & 1 & -\frac{1}{2} & 0 & -\frac{1}{2} & 0 \\0 & 0 & -\frac{1}{2} & 2 & -\frac{5}{2} & 1\end{array}\right]\overset{-2 R_3}{\longrightarrow }
Step 3 Matrix B is A^{-1}. The last matrix shown at the bottom of the previous page is in the form \left[I_3 \mid B\right]. The multiplicative identity matrix is on the left of the vertical bar. Matrix B, the multiplicative inverse of A, is on the right. Thus, the multiplicative inverse of A is
A^{-1}=\begin{bmatrix} 3 & -3&1 \\ -2 & 2&-1 \\-4&5&-2\end{bmatrix}
Step 4 Verify the result by showing that AA^{-1} = I_3 and A^{-1}A = I_3. Try confirming the result by multiplying A and A^{-1} to obtain I_3. Do you obtain I_3 if you reverse the order of the multiplication?