# 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.

## 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.

### More from my site

#### You may also like...

This site uses Akismet to reduce spam. Learn how your comment data is processed.

##### Find the Vector Form Solution to the Matrix Equation $A\mathbf{x}=\mathbf{0}$

Find the vector form solution $\mathbf{x}$ of the equation $A\mathbf{x}=\mathbf{0}$, where \$A=\begin{bmatrix} 1 & 1 & 1 & 1 &2...

Close