# 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

Sponsored Links

## 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

Sponsored Links