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

Inverse Matrices Problems and Solutions

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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2 Responses

  1. 08/31/2017

    […] The solutions are given in the post↴ Find the Inverse Matrices if Matrices are Invertible by Elementary Row Operations […]

  2. 08/31/2017

    […] 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 […]

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