10 True of False Problems about Nonsingular / Invertible Matrices
by Yu ·
Problem 500
10 questions about nonsingular matrices, invertible matrices, and linearly independent vectors.
The quiz is designed to test your understanding of the basic properties of these topics.
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The solutions will be given after completing all the 10 problems.
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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.
10 True or False Problems about Nonsingular Matrix Operations
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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 $A$ and $B$ are nonsingular $n\times n$ matrices. Then $A+B$ is nonsingular.
Correct
False. For example, let $A=I$ and $B=I$. Then both matrices are nonsingular but $A+B=O$ is singular.
Incorrect
False. For example, let $A=I$ and $B=I$. Then both matrices are nonsingular but $A+B=O$ is singular.

Question 2 of 10
2. Question
True or False. If a square matrix has no zero rows or columns, then it has an inverse matrix.
Correct
False. As a counterexample, consider
\[A=\begin{bmatrix}
1 & 1\\
1& 1
\end{bmatrix}.\]
Then $A$ has no zero rows or columns, yet is does not have the inverse matrix as the determinant of $A$ is zero.Incorrect
False. As a counterexample, consider
\[A=\begin{bmatrix}
1 & 1\\
1& 1
\end{bmatrix}.\]
Then $A$ has no zero rows or columns, yet is does not have the inverse matrix as the determinant of $A$ is zero. 
Question 3 of 10
3. Question
True or False. Let $A$ be an $m \times n$ matrix.
If the equation $A\mathbf{x}=\mathbf{0}$ has only the trivial solution $\mathbf{x}\in \R^n$, then the columns of $A$ are linearly independent.Correct
True. Let
\[B=\begin{bmatrix}
B_1 & B_2 &\dots & B_n \\
\end{bmatrix},\]
where $B_i$ is the $i$th column vector of $B$ for $i=1, \dots, n$.
Suppose that we have a linear combination
\[c_1B_1+c_2B_2+\cdots+c_n B_n=\mathbf{0}\]
for some scalars $c_1, c_2, \dots, c_n$.
Then we can write it as
\[\begin{bmatrix}
B_1 & B_2 &\dots & B_n \\
\end{bmatrix}\begin{bmatrix}
c_1 \\
c_2 \\
\vdots \\
c_n
\end{bmatrix}=\mathbf{0}.\]Since $B\mathbf{x}=\mathbf{0}$ has only the trivial solution, we must have
\[\begin{bmatrix}
c_1 \\
c_2 \\
\vdots \\
c_n
\end{bmatrix}=\mathbf{0}.\]
Hence $c_1=c_2=\cdots=c_n=0$, and the column vectors $B_1, B_2, \dots, B_n$ are linearly independent.Incorrect
True. Let
\[B=\begin{bmatrix}
B_1 & B_2 &\dots & B_n \\
\end{bmatrix},\]
where $B_i$ is the $i$th column vector of $B$ for $i=1, \dots, n$.
Suppose that we have a linear combination
\[c_1B_1+c_2B_2+\cdots+c_n B_n=\mathbf{0}\]
for some scalars $c_1, c_2, \dots, c_n$.
Then we can write it as
\[\begin{bmatrix}
B_1 & B_2 &\dots & B_n \\
\end{bmatrix}\begin{bmatrix}
c_1 \\
c_2 \\
\vdots \\
c_n
\end{bmatrix}=\mathbf{0}.\]Since $B\mathbf{x}=\mathbf{0}$ has only the trivial solution, we must have
\[\begin{bmatrix}
c_1 \\
c_2 \\
\vdots \\
c_n
\end{bmatrix}=\mathbf{0}.\]
Hence $c_1=c_2=\cdots=c_n=0$, and the column vectors $B_1, B_2, \dots, B_n$ are linearly independent. 
Question 4 of 10
4. Question
True or False. Let $A$ be an $m \times n$ matrix.
If the equation $A\mathbf{x}=\mathbf{0}$ has only the trivial solution $\mathbf{x}\in \R^n$, then the rows of $A$ are linearly independent.Correct
False. As a counterexample, consider the $3\times 2$ matrix
\[A=\begin{bmatrix}
1 & 0 \\
1 & 0 \\
0 &1
\end{bmatrix}.\]
Then the equation $A\mathbf{x}=\mathbf{0}$ has only the trivial solution $\mathbf{x}=\mathbf{0}$, and yet the first and the second row of $A$ are linearly dependent.Incorrect
False. As a counterexample, consider the $3\times 2$ matrix
\[A=\begin{bmatrix}
1 & 0 \\
1 & 0 \\
0 &1
\end{bmatrix}.\]
Then the equation $A\mathbf{x}=\mathbf{0}$ has only the trivial solution $\mathbf{x}=\mathbf{0}$, and yet the first and the second row of $A$ are linearly dependent. 
Question 5 of 10
5. Question
True or False. The row echelon form of an $3\times 3$ matrix is invertible.
Correct
True. The reduced row echelon form of an invertible matrix is the identity matrix, which is invertible.
Incorrect
True. The reduced row echelon form of an invertible matrix is the identity matrix, which is invertible.

Question 6 of 10
6. Question
True or False. There is a nonzero nonsingular matrix $A$ such that $A^2=O$.
Correct
False. Suppose $A$ is nonsingular such that $A^2=O$.
Since $A$ is nonsingular, it is invertible.
Hence we have
\begin{align*}
A=A^{1}A^2=A^{1}O=O,
\end{align*}
and the matrix $A$ must be the zero matrix.Incorrect
False. Suppose $A$ is nonsingular such that $A^2=O$.
Since $A$ is nonsingular, it is invertible.
Hence we have
\begin{align*}
A=A^{1}A^2=A^{1}O=O,
\end{align*}
and the matrix $A$ must be the zero matrix. 
Question 7 of 10
7. Question
True or False. If $A$ and $B$ are invertible $n\times n$ matrices, then $AB=BA$.
Correct
False. For example, consider
\[A=\begin{bmatrix}
1 & 1\\
0& 1
\end{bmatrix} \text{ and } B=\begin{bmatrix}
1 & 0\\
0& 2
\end{bmatrix}.\]
Then these matrices are invertible as their determinants are $\det(A)=1\neq 0$ and $\det(B)=2\neq 0$.
However they do not commute since
\begin{align*}
AB=\begin{bmatrix}
1 & 2\\
0& 2
\end{bmatrix} \text{ and } BA=\begin{bmatrix}
1 & 1\\
0& 2
\end{bmatrix}.
\end{align*}Incorrect
False. For example, consider
\[A=\begin{bmatrix}
1 & 1\\
0& 1
\end{bmatrix} \text{ and } B=\begin{bmatrix}
1 & 0\\
0& 2
\end{bmatrix}.\]
Then these matrices are invertible as their determinants are $\det(A)=1\neq 0$ and $\det(B)=2\neq 0$.
However they do not commute since
\begin{align*}
AB=\begin{bmatrix}
1 & 2\\
0& 2
\end{bmatrix} \text{ and } BA=\begin{bmatrix}
1 & 1\\
0& 2
\end{bmatrix}.
\end{align*} 
Question 8 of 10
8. Question
True or False. If $A$ and $B$ are $n\times n$ nonsingular matrices such that $A^2=I$ and $B^2=I$, then $(AB)^{1}=BA$.
Correct
True. Note that we have
\[A=A^{1}A^2=A^{1}I=A^{1}.\]
Similarly, we have $B^{1}=B$.It follows that
\begin{align*}
(AB)^{1}=B^{1}A^{1}=BA.
\end{align*}Incorrect
True. Note that we have
\[A=A^{1}A^2=A^{1}I=A^{1}.\]
Similarly, we have $B^{1}=B$.It follows that
\begin{align*}
(AB)^{1}=B^{1}A^{1}=BA.
\end{align*} 
Question 9 of 10
9. Question
True or False. If $A$ is an $m \times n$ matrix such that $A\mathbf{x}=\mathbf{0}$ for every vector $\mathbf{x}$ in $\R^n$, then $A$ is the $m\times n$ zero matrix.
Correct
True. For each $i=1, 2, \dots, n$, let $\mathbf{e}_i\in \R^n$ be the $n$dimensional vector whose $i$th entry is $1$ and $0$ elsewhere.
Then $A\mathbf{e}_i$ is the $i$th column vector of the matrix $A$.
Since by assumption $A\mathbf{e}_i=\mathbf{0}$, we see that the $i$th column of $A$ is the zero vector.
As this is true for any $i=1, \dots, n$, we conclude that $A$ is the zero matrix.Incorrect
True. For each $i=1, 2, \dots, n$, let $\mathbf{e}_i\in \R^n$ be the $n$dimensional vector whose $i$th entry is $1$ and $0$ elsewhere.
Then $A\mathbf{e}_i$ is the $i$th column vector of the matrix $A$.
Since by assumption $A\mathbf{e}_i=\mathbf{0}$, we see that the $i$th column of $A$ is the zero vector.
As this is true for any $i=1, \dots, n$, we conclude that $A$ is the zero matrix. 
Question 10 of 10
10. Question
True or False. Let $A$ be a $2 \times 2$ nonsingular matrix and let $\mathbf{v}_1$ and $\mathbf{v}_2$ be linearly independent vectors in $\R^2$.
Then the vectors $A\mathbf{v}_1, A\mathbf{v}_2$ are linearly independent vectors in $\R^2$.Correct
True. Consider a linear combination
\[c_1(A\mathbf{v}_1)+c_2(A\mathbf{v}_2)=\mathbf{0},\]
where $c_1, c_2$ are scalars.It yields that
\[A(c_1\mathbf{v}_1+c_2\mathbf{v}_2)=\mathbf{0}.\]Since $A$ is nonsingular, we have
\[c_1\mathbf{v}_1+c_2\mathbf{v}_2=\mathbf{0}.\]Because the vectors $\mathbf{v}_1, \mathbf{v}_2$ are linearly independent, we have
\[c_1=c_2=0.\]
Thus, the vectors $A\mathbf{v}_1, A\mathbf{v}_2$ are linearly independent vectors.Incorrect
True. Consider a linear combination
\[c_1(A\mathbf{v}_1)+c_2(A\mathbf{v}_2)=\mathbf{0},\]
where $c_1, c_2$ are scalars.It yields that
\[A(c_1\mathbf{v}_1+c_2\mathbf{v}_2)=\mathbf{0}.\]Since $A$ is nonsingular, we have
\[c_1\mathbf{v}_1+c_2\mathbf{v}_2=\mathbf{0}.\]Because the vectors $\mathbf{v}_1, \mathbf{v}_2$ are linearly independent, we have
\[c_1=c_2=0.\]
Thus, the vectors $A\mathbf{v}_1, A\mathbf{v}_2$ are linearly independent vectors.
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