# Find the Inverse Matrices if Matrices are Invertible by Elementary Row Operations

## Problem 552

For each of the following $3\times 3$ matrices $A$, determine whether $A$ is invertible and find the inverse $A^{-1}$ if exists by computing the augmented matrix $[A|I]$, where $I$ is the $3\times 3$ identity matrix.

(a) $A=\begin{bmatrix} 1 & 3 & -2 \\ 2 &3 &0 \\ 0 & 1 & -1 \end{bmatrix}$

(b) $A=\begin{bmatrix} 1 & 0 & 2 \\ -1 &-3 &2 \\ 3 & 6 & -2 \end{bmatrix}$.

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## Elementary Row Operations and Inverse Matrices

Recall the following procedure of testing the invertibility of $A$ as well as finding the inverse matrix if exists.
If the augmented matrix $[A|I]$ is transformed into a matrix of the form $[I|B]$, then the matrix $A$ is invertible and the inverse matrix $A^{-1}$ is given by $B$.
If the reduced row echelon form matrix for $[A|I]$ is not of the form $[I|B]$, then the matrix $A$ is not invertible.

## Solution.

### (a) $A=\begin{bmatrix} 1 & 3 & -2 \\ 2 &3 &0 \\ 0 & 1 & -1 \end{bmatrix}$

We apply the elementary row operations as follows.
We have
\begin{align*}
&[A|I]= \left[\begin{array}{rrr|rrr}
1 & 3 & -2 & 1 &0 & 0 \\
2 & 3 & 0 & 0 & 1 & 0 \\
0 & 1 & -1 & 0 & 0 & 1 \\
\end{array} \right] \xrightarrow{R_2-2R_1}
\left[\begin{array}{rrr|rrr}
1 & 3 & -2 & 1 &0 & 0 \\
0 & -3 & 4 & -2 & 1 & 0 \\
0 & 1 & -1 & 0 & 0 & 1 \\
\end{array} \right]\6pt] &\xrightarrow{R_2\leftrightarrow R_3} \left[\begin{array}{rrr|rrr} 1 & 3 & -2 & 1 &0 & 0 \\ 0 & 1 & -1 & 0 & 0 & 1 \\ 0 & -3 & 4 & -2 & 1 & 0 \\ \end{array} \right] \xrightarrow{\substack{R_1-3R_2\\ R_3+3R_2}} \left[\begin{array}{rrr|rrr} 1 & 0& 1 & 1 &0 & -3 \\ 0 & 1 & -1 & 0 & 0 & 1 \\ 0 & 0 & 1 & -2 & 1 & 3 \\ \end{array} \right]\\[6pt] &\xrightarrow{\substack{R_1-R_3\\ R_2+R_3}} \left[\begin{array}{rrr|rrr} 1 & 0& 0 & 3 & -1 & -6 \\ 0 & 1 & 0 & -2 & 1 & 4 \\ 0 & 0 & 1 & -2 & 1 & 3 \\ \end{array} \right]. \end{align*} The left 3\times 3 part of the last matrix is the identity matrix. This implies that A is invertible and the inverse matrix is given by the right 3\times 3 matrix. Hence \[A^{-1}=\begin{bmatrix} 3 & -1 & -6 \\ -2 &1 &4 \\ -2 & 1 & 3 \end{bmatrix}.

### (b) $A=\begin{bmatrix} 1 & 0 & 2 \\ -1 &-3 &2 \\ 3 & 6 & -2 \end{bmatrix}$.

Now we consider the matrix $A=\begin{bmatrix} 1 & 0 & 2 \\ -1 &-3 &2 \\ 3 & 6 & -2 \end{bmatrix}$.

Applying elementary row operations, we obtain
\begin{align*}
&[A|I]= \left[\begin{array}{rrr|rrr}
1 & 0 & 2 & 1 &0 & 0 \\
-1 & -3 & 2 & 0 & 1 & 0 \\
3 & 6 & -2 & 0 & 0 & 1 \\
\end{array} \right] \xrightarrow{\substack{R_2+R_1\\ R_3-3R_1}}
\left[\begin{array}{rrr|rrr}
1 & 0 & 2 & 1 &0 & 0 \\
0 & -3 & 4 & 1 & 1 & 0 \\
0 & 6 & -8 & -3 & 0 & 1 \\
\end{array} \right]\6pt] &\xrightarrow{R_3-2R_1} \left[\begin{array}{rrr|rrr} 1 & 0 & 2 & 1 &0 & 0 \\ 0 & -3 & 4 & 1 & 1 & 0 \\ 0 & 0 & 0 & -5 & -2 & 1 \\ \end{array} \right] \xrightarrow{\frac{-1}{3}R_2} \left[\begin{array}{rrr|rrr} 1 & 0 & 2 & 1 &0 & 0 \\ 0 & 1 & -4/3 & -1/3 & -1/3 & 0 \\ 0 & 0 & 0 & -5 & -2 & 1 \\ \end{array} \right]. \end{align*} The last matrix is in reduced row echelon form but the left 3\times 3 part is not the identity matrix I. It follows that the matrix A is not invertible. ## Related Question. Problem. Find the inverse matrix of \[A=\begin{bmatrix} 1 & 1 & 2 \\ 0 &0 &1 \\ 1 & 0 & 1 \end{bmatrix} if it exists. If you think there is no inverse matrix of $A$, then give a reason.

This is a linear algebra exam problem at the Ohio State University.

The solution is given in the post↴
Find the Inverse Matrix of a $3\times 3$ Matrix if Exists

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