# True or False Problems on Midterm Exam 1 at OSU Spring 2018

## Problem 702

The following problems are True or False.

Let $A$ and $B$ be $n\times n$ matrices.

**(a) **If $AB=B$, then $B$ is the identity matrix.

**(b)** If the coefficient matrix $A$ of the system $A\mathbf{x}=\mathbf{b}$ is invertible, then the system has infinitely many solutions.

**(c)** If $A$ is invertible, then $ABA^{-1}=B$.

**(d)** If $A$ is an idempotent nonsingular matrix, then $A$ must be the identity matrix.

**(e)** If $x_1=0, x_2=0, x_3=1$ is a solution to a homogeneous system of linear equation, then the system has infinitely many solutions.

Contents

- Problem 702
- Solution.
- (a) True or False: if $AB=B$, then $B$ is the identity matrix.
- (b) True or False: if the coefficient matrix $A$ of the system $A\mathbf{x}=\mathbf{b}$ is invertible, then the system has infinitely many solutions.
- (c) True or False: if $A$ is invertible, then $ABA^{-1}=B$.
- (d) True or False: if $A$ is an idempotent nonsingular matrix, then $A$ must be the identity matrix.
- (e) True or False: if $x_1=0, x_2=0, x_3=1$ is a solution to a homogeneous system of linear equation, then the system has infinitely many solutions.

## Solution.

### (a) True or False: if $AB=B$, then $B$ is the identity matrix.

False. For example, if $B$ is the zero matrix, then of course we have $AB=B$ as both sides are the zero matrix.

### (b) True or False: if the coefficient matrix $A$ of the system $A\mathbf{x}=\mathbf{b}$ is invertible, then the system has infinitely many solutions.

False. If the coefficient matrix $A$ is invertible, the system has a unique solution $\mathbf{x}=A^{-1}\mathbf{b}$.

### (c) True or False: if $A$ is invertible, then $ABA^{-1}=B$.

False. The given equality is equivalent to $AB=BA$. Even $A$ is invertible, matrix multiplication is not commutative. As a counterexample, consider

\[A=\begin{bmatrix}

1 & 1\\

0& 1

\end{bmatrix} \text{ and } B=\begin{bmatrix}

1 & 0\\

-1& 1

\end{bmatrix}.\]
Note that the determinant of $A$ is $\det(A)=1\neq 0$. Hence $A$ is invertible.

Yet, we have

\[AB=\begin{bmatrix}

1 & 1\\

0& 1

\end{bmatrix}\begin{bmatrix}

1 & 0\\

-1& 1

\end{bmatrix}=\begin{bmatrix}

0 & 1\\

-1& 1

\end{bmatrix}\]
and

\[BA=\begin{bmatrix}

1 & 0\\

-1& 1

\end{bmatrix}\begin{bmatrix}

1 & 1\\

0& 1

\end{bmatrix}=\begin{bmatrix}

1 & 1\\

-1& 0

\end{bmatrix},\]
and hence $AB\neq BA$.

### (d) True or False: if $A$ is an idempotent nonsingular matrix, then $A$ must be the identity matrix.

True. Since $A$ is idempotent, we have $A^2=A$. As $A$ is nonsingular, it is invertible. Thus, the inverse matrix $A^{-1}$ exists. Then we have

\[I=A^{-1}A=A^{-1}A^2=IA=A.\]
Hence, such a matrix $A$ must be the identity matrix $I$.

### (e) True or False: if $x_1=0, x_2=0, x_3=1$ is a solution to a homogeneous system of linear equation, then the system has infinitely many solutions.

Note that any homogeneous system has the zero solution. In addition to the zero solution, this system has a solution $x_1=0, x_2=0, x_3=1$. So, it has at least two solutions. The only possibilities for the number of solutions of a system of linear equations are zero, one, or infinitely many.

So, we conclude that the system must have infinitely many solutions.

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