10 True or False Problems about Basic Matrix Operations
Problem 104
Test your understanding of basic properties of matrix operations.
There are 10 True or False Quiz Problems.
These 10 problems are very common and essential.
So make sure to understand these and don’t lose a point if any of these is your exam problems.
(These are actual exam problems at the Ohio State University.)
You can take the quiz as many times as you like.
The solutions will be given after completing all the 10 problems.
Click the View question button to see the solutions.
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Notations: $I$ denotes an identity matrix and $O$ denotes a zero matrix. The sizes of these matrices should be determined from the context.
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Determine whether each of the following sentences are True or False.
Notations: $I$ denotes an identity matrix and $O$ denotes a zero matrix. The sizes of these matrices should be determined from the context.
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Question 1 of 10
1. Question
True or False. Suppose that $A$ is an $m \times n$ matrix and $B$ is an $n \times m$ matrix. Then $(AB)^{\trans}=A^{\trans} B^{\trans}$.
Correct
In general, we have
\[(AB)^{\trans}=B^{\trans}A^{\trans}.\]Note that the size of $AB$ is $m\times m$, and thus the size of $(AB)^{\trans}$ is also $m\times m$.
The size of $A^{\trans}$ is $n\times m$, and the size of $B^{\trans}$ is $m\times n$.
Hence the size of $A^{\trans}B^{\trans}$ is $n\times n$.It follows that in general if $m\neq n$, then the matrices $(AB)^{\trans}$ and $A^{\trans}B^{\trans}$ can be the same since the sizes are different.
Incorrect
In general, we have
\[(AB)^{\trans}=B^{\trans}A^{\trans}.\]Note that the size of $AB$ is $m\times m$, and thus the size of $(AB)^{\trans}$ is also $m\times m$.
The size of $A^{\trans}$ is $n\times m$, and the size of $B^{\trans}$ is $m\times n$.
Hence the size of $A^{\trans}B^{\trans}$ is $n\times n$.It follows that in general if $m\neq n$, then the matrices $(AB)^{\trans}$ and $A^{\trans}B^{\trans}$ can be the same since the sizes are different.

Question 2 of 10
2. Question
True or False. Suppose that $A$ and $B$ are $n \times n$ matrices. Then
\[(A+B)(A+B)=A^2+2AB+B^2.\]Correct
The matrix product is distributive but not commutative. With this in mind, we compute
\begin{align*}
(A+B)(A+B)&=(A+B)A+(A+B)B\\
&=A^2+BA+AB+B^2.
\end{align*}Thus, to have the equality in question, we must have
\[2AB=BA+AB,\]
and hence
\[AB=BA.\]However, in general the matrix product is not commutative: $AB\neq BA$ in general.
Hence the statement is false.
For example, matrices
\[A=\begin{bmatrix}
1 & 0\\
0& 0
\end{bmatrix}, B=\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix}\]
do not commute.
In fact we have
\[AB=\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix} \text{ and } BA=\begin{bmatrix}
0 & 0\\
0& 0
\end{bmatrix},\]
and hence $AB\neq BA$.
So with these matrices $A, B$, we do not have the equality in question.Incorrect
The matrix product is distributive but not commutative. With this in mind, we compute
\begin{align*}
(A+B)(A+B)&=(A+B)A+(A+B)B\\
&=A^2+BA+AB+B^2.
\end{align*}Thus, to have the equality in question, we must have
\[2AB=BA+AB,\]
and hence
\[AB=BA.\]However, in general the matrix product is not commutative: $AB\neq BA$ in general.
Hence the statement is false.
For example, matrices
\[A=\begin{bmatrix}
1 & 0\\
0& 0
\end{bmatrix}, B=\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix}\]
do not commute.
In fact we have
\[AB=\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix} \text{ and } BA=\begin{bmatrix}
0 & 0\\
0& 0
\end{bmatrix},\]
and hence $AB\neq BA$.
So with these matrices $A, B$, we do not have the equality in question. 
Question 3 of 10
3. Question
True or False. Let $A$ and $B$ be $n\times n$ matrices. If $A$ and $B$ commute, then matrices $A^2$ and $B$ must commute.
Correct
Since matrices $A$ and $B$ commutes, we have $AB=BA$.
Using this and associativity of the matrix product, we have
\begin{align*}
A^2B=(AA)B=A(AB)=A(BA)=(AB)A=(BA)A=B(AA)=BA^2.
\end{align*}
Hence matrices $A^2$ and $B$ must commute.Incorrect
Since matrices $A$ and $B$ commutes, we have $AB=BA$.
Using this and associativity of the matrix product, we have
\begin{align*}
A^2B=(AA)B=A(AB)=A(BA)=(AB)A=(BA)A=B(AA)=BA^2.
\end{align*}
Hence matrices $A^2$ and $B$ must commute. 
Question 4 of 10
4. Question
True or False. Suppose that $A$ is an $n \times n$ matrix and assume $A^2=O$, where $O$ is the zero matrix. Then $A=O$.
Correct
As a counterexample, consider the $2\times 2$ matrix
\[A=\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix}.\]
Then the direct computation shows that
\[A^2=\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix}
\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix}
=\begin{bmatrix}
0 & 0\\
0& 0
\end{bmatrix}=O.\]
Hence $A$ is a nonzero matrix yet $A^2=O$.Incorrect
As a counterexample, consider the $2\times 2$ matrix
\[A=\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix}.\]
Then the direct computation shows that
\[A^2=\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix}
\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix}
=\begin{bmatrix}
0 & 0\\
0& 0
\end{bmatrix}=O.\]
Hence $A$ is a nonzero matrix yet $A^2=O$. 
Question 5 of 10
5. Question
Suppose that $A$ is an $n \times n$ matrix and assume $A^2=I$, where $I$ is the identity matrix. Then $A=I$.
Correct
For example, consider the matrix
\[A=\begin{bmatrix}
1 & 0\\
0& 1
\end{bmatrix}=I.\]
Then it follows that $A^2=I$.
Hence $A$ is not necessarily the identity matrix.Incorrect
For example, consider the matrix
\[A=\begin{bmatrix}
1 & 0\\
0& 1
\end{bmatrix}=I.\]
Then it follows that $A^2=I$.
Hence $A$ is not necessarily the identity matrix. 
Question 6 of 10
6. Question
True or False. There are $n\times n$ matrices $A$ and $B$ such that $AB=O$ but $BA \neq O$.
Correct
\item True. For example, consider
\[A=\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix} \text{ and } B=\begin{bmatrix}
1 & 0\\
0& 0
\end{bmatrix}.\]
Then we have
\begin{align*}
AB&=\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix}
\begin{bmatrix}
1 & 0\\
0& 0
\end{bmatrix}
=\begin{bmatrix}
0 & 0\\
0& 0
\end{bmatrix}=O\\[6pt]
BA&=\begin{bmatrix}
1 & 0\\
0& 0
\end{bmatrix}
\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix}
=\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix}\neq O.
\end{align*}Incorrect
\item True. For example, consider
\[A=\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix} \text{ and } B=\begin{bmatrix}
1 & 0\\
0& 0
\end{bmatrix}.\]
Then we have
\begin{align*}
AB&=\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix}
\begin{bmatrix}
1 & 0\\
0& 0
\end{bmatrix}
=\begin{bmatrix}
0 & 0\\
0& 0
\end{bmatrix}=O\\[6pt]
BA&=\begin{bmatrix}
1 & 0\\
0& 0
\end{bmatrix}
\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix}
=\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix}\neq O.
\end{align*} 
Question 7 of 10
7. Question
True or False. There exists a $2\times 2$ matrix $A$ such that
\[A\begin{bmatrix}
x \\
y
\end{bmatrix}=\begin{bmatrix}
y \\
x
\end{bmatrix}\]
for any vector $\begin{bmatrix}
x \\
y
\end{bmatrix}\in \R^2$.Correct
True. In fact, if we put
\[A=\begin{bmatrix}
0 & 1\\
1& 0
\end{bmatrix},\]
then it is straightforward to see
\[A\begin{bmatrix}
x \\
y
\end{bmatrix}=\begin{bmatrix}
y \\
x
\end{bmatrix}\]
for any vector $\begin{bmatrix}
x \\
y
\end{bmatrix}\in \R^2$.Incorrect
True. In fact, if we put
\[A=\begin{bmatrix}
0 & 1\\
1& 0
\end{bmatrix},\]
then it is straightforward to see
\[A\begin{bmatrix}
x \\
y
\end{bmatrix}=\begin{bmatrix}
y \\
x
\end{bmatrix}\]
for any vector $\begin{bmatrix}
x \\
y
\end{bmatrix}\in \R^2$. 
Question 8 of 10
8. Question
True or False. There exists a $2\times 2$ matrix $A$ such that
\[A\begin{bmatrix}
x \\
y
\end{bmatrix}=\begin{bmatrix}
x^2 \\
y^2
\end{bmatrix}\]
for any vector $\begin{bmatrix}
x \\
y
\end{bmatrix}\in \R^2$.Correct
Seeking a contradiction, assume that there is a matrix such that
\[A\begin{bmatrix}
x \\
y
\end{bmatrix}=\begin{bmatrix}
x^2 \\
y^2
\end{bmatrix}\]
for any vector $\begin{bmatrix}
x \\
y
\end{bmatrix}\in \R^2$.Then consider the vector $\begin{bmatrix}
2 \\
2
\end{bmatrix} \in \R^2$.By assumption we have
\begin{align*}
A\begin{bmatrix}
2 \\
2
\end{bmatrix}=\begin{bmatrix}
2^2 \\
2^2
\end{bmatrix}=\begin{bmatrix}
4 \\
4
\end{bmatrix}.
\end{align*}On the other hand, we have
\begin{align*}
A\begin{bmatrix}
2 \\
2
\end{bmatrix}&=A\left(\, \begin{bmatrix}
1 \\
1
\end{bmatrix}+\begin{bmatrix}
1 \\
1
\end{bmatrix} \,\right)\\[6pt]
&=A\begin{bmatrix}
1 \\
1
\end{bmatrix}+A\begin{bmatrix}
1 \\
1
\end{bmatrix}\\[6pt]
&=\begin{bmatrix}
1^2 \\
1^2
\end{bmatrix}
+\begin{bmatrix}
1^2 \\
1^2
\end{bmatrix}\\[6pt]
&=\begin{bmatrix}
2 \\
2
\end{bmatrix}.
\end{align*}Since these computations do not agree, this is a contradiction.
Hence such a matrix $A$ does not exist.Incorrect
Seeking a contradiction, assume that there is a matrix such that
\[A\begin{bmatrix}
x \\
y
\end{bmatrix}=\begin{bmatrix}
x^2 \\
y^2
\end{bmatrix}\]
for any vector $\begin{bmatrix}
x \\
y
\end{bmatrix}\in \R^2$.Then consider the vector $\begin{bmatrix}
2 \\
2
\end{bmatrix} \in \R^2$.By assumption we have
\begin{align*}
A\begin{bmatrix}
2 \\
2
\end{bmatrix}=\begin{bmatrix}
2^2 \\
2^2
\end{bmatrix}=\begin{bmatrix}
4 \\
4
\end{bmatrix}.
\end{align*}On the other hand, we have
\begin{align*}
A\begin{bmatrix}
2 \\
2
\end{bmatrix}&=A\left(\, \begin{bmatrix}
1 \\
1
\end{bmatrix}+\begin{bmatrix}
1 \\
1
\end{bmatrix} \,\right)\\[6pt]
&=A\begin{bmatrix}
1 \\
1
\end{bmatrix}+A\begin{bmatrix}
1 \\
1
\end{bmatrix}\\[6pt]
&=\begin{bmatrix}
1^2 \\
1^2
\end{bmatrix}
+\begin{bmatrix}
1^2 \\
1^2
\end{bmatrix}\\[6pt]
&=\begin{bmatrix}
2 \\
2
\end{bmatrix}.
\end{align*}Since these computations do not agree, this is a contradiction.
Hence such a matrix $A$ does not exist. 
Question 9 of 10
9. Question
True or False. For any $n\times n$ matrices $A$ and $B$, we have
\[(AB)(A+B)=A^2B^2.\]Correct
Note that the matrix product is distributive but not commutative.
keep it in mind, we compute
\begin{align*}
(AB)(A+B)&=(AB)A+(AB)B\\
&=A^2BA+ABB^2.
\end{align*}Thus, if we have $(AB)(A+B)=A^2B^2$, then we must have
$AB=BA$.
But not every pair of matrices commute.
For example, see the matrices $A$ and $B$ in the solution of p[art (b).
Thus $(AB)(A+B) \neq A^2B^2$ in general.Incorrect
Note that the matrix product is distributive but not commutative.
keep it in mind, we compute
\begin{align*}
(AB)(A+B)&=(AB)A+(AB)B\\
&=A^2BA+ABB^2.
\end{align*}Thus, if we have $(AB)(A+B)=A^2B^2$, then we must have
$AB=BA$.
But not every pair of matrices commute.
For example, see the matrices $A$ and $B$ in the solution of p[art (b).
Thus $(AB)(A+B) \neq A^2B^2$ in general. 
Question 10 of 10
10. Question
True or False. If $A$ and $B$ are symmetric $n\times n$ matrices, then the product $AB$ is also symmetric.
Correct
Since $A$ and $B$ are symmetric, we have
\[A^{\trans}=A \text{ and } B^{\trans}=B.\]
By the property of the transpose, we have
\begin{align*}
(AB)^{\trans}&=B^{\trans}A^{\trans}\\
&=BA.
\end{align*}
Therefore, if $A$ and $B$ do not commute, then we have
\[(AB)^{\trans}\neq AB,\]
hence the matrix $AB$ is not symmetric.It remains to show that there exist symmetric matrices $A$ and $B$ such that they do not commute.
For example, let
\[A=\begin{bmatrix}
1 & 0\\
0& 0
\end{bmatrix} \text{ and } B=\begin{bmatrix}
0 & 1\\
1& 0
\end{bmatrix}.\]
These matrices are symmetric, and since we have
\begin{align*}
AB=\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix} \text{ and } BA=\begin{bmatrix}
0 & 0\\
1& 0
\end{bmatrix},
\end{align*}
they do not commute.Hence the product $AB$ is not symmetric.
Incorrect
Since $A$ and $B$ are symmetric, we have
\[A^{\trans}=A \text{ and } B^{\trans}=B.\]
By the property of the transpose, we have
\begin{align*}
(AB)^{\trans}&=B^{\trans}A^{\trans}\\
&=BA.
\end{align*}
Therefore, if $A$ and $B$ do not commute, then we have
\[(AB)^{\trans}\neq AB,\]
hence the matrix $AB$ is not symmetric.It remains to show that there exist symmetric matrices $A$ and $B$ such that they do not commute.
For example, let
\[A=\begin{bmatrix}
1 & 0\\
0& 0
\end{bmatrix} \text{ and } B=\begin{bmatrix}
0 & 1\\
1& 0
\end{bmatrix}.\]
These matrices are symmetric, and since we have
\begin{align*}
AB=\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix} \text{ and } BA=\begin{bmatrix}
0 & 0\\
1& 0
\end{bmatrix},
\end{align*}
they do not commute.Hence the product $AB$ is not symmetric.
(The Ohio State University, linear algebra exam)
Related Question.
Check out “10 True of False Problems about Nonsingular / Invertible Matrices” for True or False problems about nonsingular matrices, invertible matrices, and linearly independent vectors
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For question #8, Can’t A be equal to [x, 0; 0, y]?
Dear Mishy,
The matrix $A$ must be fixed. So it cannot contain variables like $x, y$.
Got it. Thanks!