Find a Nonsingular Matrix Satisfying Some Relation

Problems and solutions in Linear Algebra

Problem 280

Determine whether there exists a nonsingular matrix $A$ if
\[A^2=AB+2A,\] where $B$ is the following matrix.
If such a nonsingular matrix $A$ exists, find the inverse matrix $A^{-1}$.

(a) \[B=\begin{bmatrix}
-1 & 1 & -1 \\
0 &-1 &0 \\
1 & 2 & -2
\end{bmatrix}\]

(b) \[B=\begin{bmatrix}
-1 & 1 & -1 \\
0 &-1 &0 \\
2 & 1 & -4
\end{bmatrix}.\]

 
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Solution.

Suppose that a nonsingular matrix $A$ satisfying $A^2=AB+2A$ exists.
Then $A$ is invertible since $A$ is nonsingular, and thus the inverse $A^{-1}$ exists.

Multiplying by $A^{-1}$ on the left, we have
\begin{align*}
A=&A^{-1}A^2=A^{-1}(AB+2A)\\
&=A^{-1}AB+2A^{-1}A\\
&=B+2I,
\end{align*}
where $I$ is the $3\times 3$ identity matrix.
Therefore, if such a nonsingular matrix exists, it must be
\[A=B+2I. \tag{*}\]

(a) The first case

Let us consider the case
\[B\begin{bmatrix}
-1 & 1 & -1 \\
0 &-1 &0 \\
1 & 2 & -2
\end{bmatrix}.\] In this case, we have from (*)
\[A=B+2I=\begin{bmatrix}
1 & 1 & -1 \\
0 &1 &0 \\
1 & 2 & 0
\end{bmatrix}.\]

We still need to check that this matrix is in fact a nonsingular matrix.
To check the non-singularity and to find the inverse matrix at once, we consider the augmented matrix $[A\mid I]$ and apply elementary row operations.
We have
\begin{align*}
&[A\mid I] =
\left[\begin{array}{rrr|rrr}
1 & 1 & -1 & 1 &0 & 0 \\
0 & 1 & 0 & 0 & 1 & 0 \\
1 & 2 & 0 & 0 & 0 & 1 \\
\end{array} \right]\\[10pt] & \xrightarrow{R_3-R_1}
\left[\begin{array}{rrr|rrr}
1 & 1 & -1 & 1 &0 & 0 \\
0 & 1 & 0 & 0 & 1 & 0 \\
0 & 1 & 1 & -1 & 0 & 1 \\
\end{array} \right] \xrightarrow{\substack{R_1-R_2 \\ R_3-R_2}}
\left[\begin{array}{rrr|rrr}
1 & 0 & -1 & 1 & -1& 0 \\
0 & 1 & 0 & 0 & 1 & 0 \\
0 & 0 & 1 & -1 & -1 & 1 \\
\end{array} \right] \\[10pt] &\xrightarrow{R_1+R_3}
\left[\begin{array}{rrr|rrr}
1 & 0 & 0 & 0 & -2& 1 \\
0 & 1 & 0 & 0 & 1 & 0 \\
0 & 0 & 1 & -1 & -1 & 1 \\
\end{array} \right].
\end{align*}

The left part of the last matrix is the identity matrix, and thus the matrix $A$ is invertible and the inverse matrix is the right half:
\[A^{-1}=\begin{bmatrix}
0 & -2 & 1 \\
0 &1 &0 \\
-1 & -1 & 1
\end{bmatrix}.\]

(b) The second case

Next let us consider the case
\[B=\begin{bmatrix}
-1 & 1 & -1 \\
0 &-1 &0 \\
2 & 1 & -4
\end{bmatrix}.\] By (*), if a nonsingular matrix $A$ exists, it must be
\[A=B+2I=\begin{bmatrix}
1 & 1 & -1 \\
0 &1 &0 \\
2 & 1 & -2
\end{bmatrix}.\]

We need to determine whether this matrix is actually nonsingular.
In fact, we prove that this matrix is singular.
That is, we show that $A\mathbf{x}=\mathbf{0}$ has a nonzero solution.
Consider the augmented matrix $[A\mid \mathbf{0}]$. By Gauss-Jordan elimination, we have
\begin{align*}
&[A\mid \mathbf{0}] = \left[\begin{array}{rrr|r}
1 & 1 & -1 & 0 \\
0 &1 & 0 & 0 \\
2 & 1 & -2 & 0
\end{array} \right] \xrightarrow{R_3-2R_1}\\[10pt] & \left[\begin{array}{rrr|r}
1 & 1 & -1 & 0 \\
0 &1 & 0 & 0 \\
0 & -1 & 0 & 0
\end{array} \right] \xrightarrow{\substack{R_1-R_2\\ R_3+R_2}}
\left[\begin{array}{rrr|r}
1 & 0& -1 & 0 \\
0 &1 & 0 & 0 \\
0 & 0 & 0 & 0
\end{array} \right].
\end{align*}

The last matrix is in reduced row echelon form and it has a zero row. From this, we see that $x_3$ must be a free variable, and the matrix $A$ is singular.
(The general solution is $x_1=x_3, x_2=0$. Thus for example, $x_1=1, x_2=0, x_3=1$ is a nonzero solution of $A\mathbf{x}=\mathbf{0}$.)

Thus, we conclude that there is no nonsingular matrix $A$ satisfying $A^2=AB+2A$ in this case.


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