# 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}$.

Contents

## 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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[…] 3 & -1 & -6 \ -2 &1 &4 \ -2 & 1 & 3 end{bmatrix}.] (See the post Find the Inverse Matrices if Matrices are Invertible by Elementary Row Operations for details of how to find the inverse matrix of this […]