# Find a Nonsingular Matrix Satisfying Some Relation ## 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) $B=\begin{bmatrix} -1 & 1 & -1 \\ 0 &-1 &0 \\ 2 & 1 & -4 \end{bmatrix}.$ Add to solve later

## 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. Add to solve later

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