Linear Algebra Quiz: Transpose And Symmetric Matrices

20 questions · exam conditions

0:00

Transpose And Symmetric MatricesQuestion 1 of 20

Let $A$ be a $3 \times 2$ matrix. Which of the following operations results in a $2 \times 2$ symmetric matrix?

A^2

A A^T

A+A^T

A^T A

← Back to quizzes

Linear Algebra Quiz

Linear Algebra Quiz: Transpose And Symmetric Matrices

Practice Transpose And Symmetric Matrices in Linear Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Transpose And Symmetric Matrices, giving you a quick way to practice the rules, question types, and explanations that matter most for Linear Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Let $A$ be a $3 \times 2$ matrix. Which of the following operations results in a $2 \times 2$ symmetric matrix?

$A^2$
$A A^T$
$A+A^T$
$A^T A$ (correct answer)

Explanation: Let's analyze the dimensions of the resulting matrix for each option. The matrix

A

has dimensions

3 \times 2

, so its transpose

A^T

has dimensions

2 \times 3

. A)

A^2 = A A

. The product is not defined because the inner dimensions do not match (

3 \times 2

and

3 \times 2

). B)

A A^T

. The product dimensions are

(3 \times 2)(2 \times 3)

, resulting in a

3 \times 3

matrix. This is not a

2 \times 2

matrix. C)

A + A^T

. Addition is not defined because the matrices have different dimensions (

3 \times 2

and

2 \times 3

). D)

A^T A

. The product dimensions are

(2 \times 3)(3 \times 2)

, resulting in a

2 \times 2

matrix. To check if it's symmetric, we take the transpose:

(A^T A)^T = A^T (A^T)^T = A^T A

. Since the matrix is equal to its own transpose, it is symmetric. This is the correct choice.

Question 2

Let $A$ be an invertible, symmetric matrix. Which of the following is equivalent to the expression $(A^{-1} A^T)^T$ ?

$A^2$
$I$ (correct answer)
$A^{-2}$
$A$

Explanation: We are given that

A

is invertible and symmetric, so

A^{-1}

exists and

A^T = A

. We can simplify the expression in two ways. Method 1: Substitute

A^T=A

first. The expression becomes

(A^{-1} A)^T

. Since

A^{-1}A = I

(the identity matrix), the expression is

I^T

. The transpose of the identity matrix is itself, so

I^T = I

. Method 2: Use transpose properties first.

(A^{-1} A^T)^T = (A^T)^T (A^{-1})^T

. We know

(A^T)^T = A

and

(A^{-1})^T = (A^T)^{-1}

. So the expression becomes

A(A^T)^{-1}

. Now use the symmetric property

A^T=A

. The expression is

A A^{-1}

, which equals

I

. Both methods yield the identity matrix

I

Question 3

Consider the block matrix $M = \begin{pmatrix} A & B \\ B^T & C \end{pmatrix}$ where $A$ and $C$ are symmetric matrices. If $M^2 = M$ , what must be true about the relationship between $A$ , $B$ , and $C$ ?

$A^2 = A$ , $B = 0$ , and $C^2 = C$
$A^2 + BB^T = A$ , $AB + BC^T = B$ , and $C^2 + B^TB = C$
$A^2 + BB^T = A$ , $AB + BC = B$ , and $C^2 + B^TB = C$ (correct answer)
$A + C = I$ and $B$ is skew-symmetric

Explanation: When you encounter a block matrix equation like

M^2 = M

, you're dealing with an idempotent matrix. The key is to compute the matrix multiplication carefully using block matrix rules and then equate corresponding blocks. To find

M^2

, multiply the block matrix by itself:

Since

M^2 = M

, each block must equal its corresponding block in the original matrix. This gives us:

Top-left: $A^2 + BB^T = A$
Top-right: $AB + BC = B$
Bottom-left: $B^TA + CB^T = B^T$ (which is equivalent to the top-right when transposed, since $A$ and $C$ are symmetric)
Bottom-right: $B^TB + C^2 = C$

This matches option C exactly. Option A assumes

B = 0

, which is unnecessarily restrictive. While this would satisfy the idempotent condition, it's not required—non-zero

B

can work if the other conditions hold. Option B incorrectly uses

BC^T

instead of

BC

. Since

C

is symmetric (

C = C^T

), this seems minor but represents a computational error in the block multiplication. Option D introduces unrelated conditions about the sum and skew-symmetry that don't follow from the idempotent property. Study tip: When working with block matrices, always multiply block-by-block systematically, and remember that symmetry conditions (like

A = A^T

) often simplify your final expressions.

Question 4

Given the matrix equation $X^T A X = B$ where $A$ and $B$ are $n \times n$ symmetric matrices and $X$ is $n \times n$ and invertible, which condition is necessary and sufficient for a solution to exist?

$A$ and $B$ must have the same determinant and trace
$A$ and $B$ must have the same rank and signature
$A$ and $B$ must have the same eigenvalues counting multiplicities
$A$ and $B$ must have the same number of positive and negative eigenvalues (correct answer)

Explanation: The transformation

X^TAX

is a congruence transformation. Two symmetric matrices are congruent if and only if they have the same inertia (same number of positive, negative, and zero eigenvalues) by Sylvester's law of inertia. This is equivalent to having the same signature. Choice A is incorrect since congruence doesn't preserve determinant or trace in general. Choice B mentions rank (preserved) but 'signature' usually refers to the difference between positive and negative eigenvalues, not their individual counts. Choice C is too strong - eigenvalues need not be identical, only their signs matter.

Question 5

Let $A$ be a $3 \times 3$ matrix. The matrix $B = A + A^T$ is given by $B = \begin{pmatrix} 4 & 6 & 0 \\ 6 & -2 & 5 \\ 0 & 5 & 8 \end{pmatrix}$ . Based on this information, what are the diagonal entries of matrix $A$ ?

The diagonal entries cannot be determined from the given information.
$4, -2, 8$
$2, -1, 4$ (correct answer)
$8, -4, 16$

Explanation: Let

A = (a_{ij})

and

B = (b_{ij})

. By definition,

B = A + A^T

, so the entries of

B

are given by

b_{ij} = a_{ij} + a_{ji}

. For the diagonal entries, where

i=j

, we have

b_{ii} = a_{ii} + a_{ii} = 2a_{ii}

. Therefore,

a_{ii} = b_{ii} / 2

. We can find the diagonal entries of

A

by halving the diagonal entries of

B

. The diagonal of

B

(4, -2, 8)

. Thus, the diagonal of

A

(4/2, -2/2, 8/2) = (2, -1, 4)

Question 6

Any square matrix $A$ can be written as the sum of a symmetric matrix $S$ and a skew-symmetric matrix $K$ . If $A = \begin{pmatrix} 3 & 5 \\ -1 & 7 \end{pmatrix}$ , what is the skew-symmetric component $K$ ?

$\begin{pmatrix} 0 & 3 \\ -3 & 0 \end{pmatrix}$ (correct answer)
$\begin{pmatrix} 3 & 2 \\ 2 & 7 \end{pmatrix}$
$\begin{pmatrix} 0 & 6 \\ -6 & 0 \end{pmatrix}$
$\begin{pmatrix} 0 & -3 \\ 3 & 0 \end{pmatrix}$

Explanation: The skew-symmetric component

K

of a matrix

A

is given by the formula

K = \frac{1}{2}(A - A^T)

. First, we find the transpose of

A

A^T = \begin{pmatrix} 3 & -1 \\ 5 & 7 \end{pmatrix}

. Next, we compute the difference

A - A^T

A - A^T = \begin{pmatrix} 3 & 5 \\ -1 & 7 \end{pmatrix} - \begin{pmatrix} 3 & -1 \\ 5 & 7 \end{pmatrix} = \begin{pmatrix} 3-3 & 5-(-1) \\ -1-5 & 7-7 \end{pmatrix} = \begin{pmatrix} 0 & 6 \\ -6 & 0 \end{pmatrix}

Finally, we multiply by

\frac{1}{2}

: $$K = \frac{1}{2}\begin{pmatrix} 0 & 6 \ -6 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 3 \ -3 & 0 \end{pmatrix}

Question 7

Consider the matrix equation $X + X^T = A$ where $A$ is a given $3 \times 3$ matrix. For this equation to have a solution, which condition must $A$ satisfy?

$A$ must be invertible with positive determinant
$A$ must be symmetric with trace equal to zero
$A$ must be symmetric with all real eigenvalues
$A$ must be symmetric but may have any trace value (correct answer)

Explanation: Taking the transpose of both sides:

(X + X^T)^T = A^T

, which gives

X^T + X = A^T

. For consistency with the original equation

X + X^T = A

, we need

A = A^T

, so

A

must be symmetric. If

A

is symmetric, we can always construct a solution

X = \frac{1}{2}A + S

where

S

is any skew-symmetric matrix. The trace, determinant sign, and eigenvalue properties of

A

don't affect the existence of solutions.

Question 8

Let $S$ be the set of all $3 \times 3$ symmetric matrices and $T$ be the set of all $3 \times 3$ skew-symmetric matrices. If $A \in S$ , $B \in T$ , and $\det(A + B) \neq 0$ , what can be said about $(A + B)^{-T}$ ?

$(A + B)^{-T} = (A + B)^{-1}$ and belongs to $S$
$(A + B)^{-T} = (A^{-1} + B^{-1})$ when both inverses exist
$(A + B)^{-T}$ belongs to neither $S$ nor $T$ in general (correct answer)
$(A + B)^{-T} = -(A + B)^{-1}$ and belongs to $T$

Explanation: Since

A^T = A

and

B^T = -B

, we have

(A + B)^T = A^T + B^T = A - B

. Therefore

(A + B)^{-T} = ((A + B)^T)^{-1} = (A - B)^{-1}

. Taking transpose:

((A + B)^{-T})^T = ((A - B)^{-1})^T = ((A - B)^T)^{-1} = (A^T - B^T)^{-1} = (A - (-B))^{-1} = (A + B)^{-1}

. Since

(A + B)^{-T} = (A - B)^{-1}

and

((A + B)^{-T})^T = (A + B)^{-1}

, we have

(A - B)^{-1} \neq (A + B)^{-1}

in general, so

(A + B)^{-T}

is neither symmetric nor skew-symmetric.

Question 9

Given that $A$ is a $3 \times 3$ symmetric matrix with eigenvalues $\lambda_1, \lambda_2, \lambda_3$ , and $B = A^2 + 2A + 3I$ , what is the trace of $B^{-1}$ in terms of the eigenvalues of $A$ ?

$\frac{1}{\lambda_1^2 + 2\lambda_1 + 3} + \frac{1}{\lambda_2^2 + 2\lambda_2 + 3} + \frac{1}{\lambda_3^2 + 2\lambda_3 + 3}$ (correct answer)
$\frac{1}{(\lambda_1 + \lambda_2 + \lambda_3)^2 + 2(\lambda_1 + \lambda_2 + \lambda_3) + 9}$
$\frac{\lambda_1^2 + \lambda_2^2 + \lambda_3^2 + 2(\lambda_1 + \lambda_2 + \lambda_3) + 9}{(\lambda_1\lambda_2\lambda_3)^2}$
$\frac{3}{\lambda_1\lambda_2\lambda_3 + 2(\lambda_1 + \lambda_2 + \lambda_3) + 9}$

Explanation: Since

A

is symmetric, it has an orthonormal eigenbasis. If

A

has eigenvalues

\lambda_i

, then

B = A^2 + 2A + 3I

has eigenvalues

\lambda_i^2 + 2\lambda_i + 3

. The matrix

B^{-1}

has eigenvalues

\frac{1}{\lambda_i^2 + 2\lambda_i + 3}

. The trace of

B^{-1}

equals the sum of its eigenvalues. Choice B incorrectly treats the sum of eigenvalues as a single eigenvalue, C attempts to use determinant relationships incorrectly, and D uses an incorrect formula mixing trace and determinant.

Question 10

Let $A$ and $B$ be $n \times n$ symmetric matrices. Which of the following statements is not always true?

$A+B$ is a symmetric matrix.
$cA$ is a symmetric matrix for any scalar $c$ .
$AB$ is a symmetric matrix. (correct answer)
$A^T$ is a symmetric matrix.

Explanation: Let's analyze the choices given that

A^T=A

and

B^T=B

. A)

(A+B)^T = A^T + B^T = A+B

. This is always true. B)

(cA)^T = cA^T = cA

. This is always true. C)

(AB)^T = B^T A^T = BA

. For

AB

to be symmetric, we need

(AB)^T = AB

, which means

BA=AB

. This is only true if

A

and

B

commute, which is not always the case for symmetric matrices. Thus, this statement is not always true. D) Since

A

is symmetric,

A^T = A

, which by definition is a symmetric matrix. This is always true.

Question 11

Let $A = \begin{pmatrix} k & 3 \\ 2k-1 & 4 \end{pmatrix}$ . The matrix $S = A + A^T$ is symmetric and has a trace of 10. What is the value of $k$ ?

$6$
$4$
$1$ (correct answer)
$3$

Explanation: For any square matrix

A

, the matrix

S = A + A^T

is always symmetric. The condition that

S

is symmetric is a distractor, as it holds for any

k

. The core of the problem is the trace. First, let's find

A+A^T

A^T = \begin{pmatrix} k & 2k-1 \\ 3 & 4 \end{pmatrix}

S = A + A^T = \begin{pmatrix} k+k & 3+(2k-1) \\ (2k-1)+3 & 4+4 \end{pmatrix} = \begin{pmatrix} 2k & 2k+2 \\ 2k+2 & 8 \end{pmatrix}

The trace of a matrix is the sum of its diagonal elements. So,

\mathrm{tr}(S) = 2k + 8

. We are given that the trace is 10.

2k + 8 = 10

2k = 2

k = 1

Question 12

Let $A$ be an $n \times n$ symmetric matrix and $B$ be an $n \times n$ skew-symmetric matrix. Which of the following expressions simplifies to the matrix $A$ ?

$(A+B)(A-B)$
$(A-B)^T + B$
The skew-symmetric part of $(A+B)$
The symmetric part of $(A+B)$ (correct answer)

Explanation: We are given

A^T = A

and

B^T = -B

. The symmetric part of any matrix

M

is given by

\frac{1}{2}(M + M^T)

. Let's find the symmetric part of the matrix

M = A+B

. First, find

M^T

M^T = (A+B)^T = A^T + B^T

. Using the given properties,

A^T=A

and

B^T=-B

, so:

M^T = A - B

. Now, substitute

M

and

M^T

into the formula for the symmetric part:

\frac{1}{2}(M + M^T) = \frac{1}{2}((A+B) + (A-B)) = \frac{1}{2}(2A) = A

. Therefore, the symmetric part of

(A+B)

is equal to

A

. Choice C, the skew-symmetric part, would be

\frac{1}{2}(M - M^T) = \frac{1}{2}((A+B) - (A-B)) = \frac{1}{2}(2B) = B

Question 13

Let $A$ and $B$ be arbitrary $n \times n$ matrices. Which of the following statements is always true?

If $A$ is symmetric, then $A$ is invertible.
The diagonal entries of $A - A^T$ are always zero. (correct answer)
$(AB)^T = A^T B^T$ .
If $A$ and $B$ are symmetric, then $AB$ is symmetric.

Explanation: Let's evaluate each statement. A) This is false. A counterexample is the zero matrix, which is symmetric (

O^T = O

) but not invertible. B) The diagonal entries of any matrix

M

are

m_{ii}

. For the matrix

M = A - A^T

, the diagonal entries are

m_{ii} = a_{ii} - (A^T)_{ii}

. By definition of transpose,

(A^T)_{ii} = a_{ii}

. So,

m_{ii} = a_{ii} - a_{ii} = 0

. This is always true. C) This is false. The correct property is

(AB)^T = B^T A^T

. Matrix multiplication is not generally commutative, so

A^T B^T \neq B^T A^T

. D) This is false. As shown in another question,

(AB)^T = B^T A^T

. If

A

and

B

are symmetric, this becomes

BA

AB

is only symmetric if

BA=AB

, i.e., if

A

and

B

commute.

Question 14

A $3 \times 3$ matrix $A$ is constructed using the rule $a_{ij} = 2i + 2j$ for its entries. Which of the following properties does matrix $A$ have?

$A$ is skew-symmetric.
The determinant of $A$ is non-zero.
$A$ is symmetric. (correct answer)
The trace of $A$ is 0.

Explanation: To check if

A

is symmetric, we must verify if

a_{ij} = a_{ji}

for all

i, j

. According to the rule,

a_{ij} = 2i + 2j

. For the transposed entry, we have

a_{ji} = 2j + 2i

. Since addition is commutative,

2i + 2j = 2j + 2i

, which means

a_{ij} = a_{ji}

. Therefore, the matrix

A

is symmetric. Let's check the other options. (A) For

A

to be skew-symmetric,

a_{ij} = -a_{ji}

, which is not true here. (D) The diagonal entries are

a_{11}=4, a_{22}=8, a_{33}=12

, so the trace is

4+8+12=24 \neq 0

. (B) Let's write out the matrix:

A = \begin{pmatrix} 4 & 6 & 8 \\ 6 & 8 & 10 \\ 8 & 10 & 12 \end{pmatrix}

. The second row is the first row plus

(2, 2, 2)

, and the third row is the second row plus

(2, 2, 2)

. This means the rows are linearly dependent (e.g.,

R_1 + R_3 = 2R_2

), so the determinant must be zero. Thus,

A

is not invertible.

Question 15

For any $n \times n$ matrix $A$ , which of the following expressions is guaranteed to result in a skew-symmetric matrix?

$(A+A^T)^2$
$A^T A + A A^T$
$A^2 - (A^T)^2$ (correct answer)
$A^T A - A A^T$

Explanation: A matrix

X

is skew-symmetric if

X^T = -X

. Let's test each option. Let

X

be the expression in each choice. A) Let

S = A+A^T

, which is symmetric.

X = S^2

X^T = (S^2)^T = (S^T)^2 = S^2 = X

. So

(A+A^T)^2

is symmetric. B)

X^T = (A^T A + A A^T)^T = (A^T A)^T + (A A^T)^T = A^T(A^T)^T + (A^T)^T A^T = A^T A + A A^T = X

. So it is symmetric. C)

X^T = (A^2 - (A^T)^2)^T = (A^2)^T - ((A^T)^2)^T = (A^T)^2 - A^2 = -(A^2 - (A^T)^2) = -X

. This is skew-symmetric. D)

X^T = (A^T A - A A^T)^T = (A^T A)^T - (A A^T)^T = A^T A - A A^T = X

. So it is symmetric.

Question 16

Let $A$ and $P$ be $n \times n$ matrices. If $A$ is symmetric and $P$ is orthogonal (meaning $P^T P = I$ ), what property must the matrix $S = P A P^T$ have?

$S$ must be symmetric. (correct answer)
$S$ must be skew-symmetric.
$S$ must be equal to $A$ .
$S$ must be the identity matrix $I$ .

Explanation: To determine the property of

S = P A P^T

, we should examine its transpose,

S^T

. Using the property that

(XYZ)^T = Z^T Y^T X^T

, we have:

S^T = (P A P^T)^T = (P^T)^T A^T P^T

Since

(P^T)^T = P

, this simplifies to:

S^T = P A^T P^T

We are given that

A

is a symmetric matrix, which means

A^T = A

. Substituting this into our expression for

S^T

S^T = P A P^T

This is the original definition of

S

. Since

S^T = S

, the matrix

S

must be symmetric.

Question 17

Let $A$ be a $3 \times 3$ matrix such that $A^T = -A$ . If $\text{tr}(A^TA) = 18$ , what is $\text{tr}(A)$ ?

$0$ (correct answer)
$3$
$-3$
Cannot be determined from the given information

Explanation: Since

A^T = -A

, matrix

A

is skew-symmetric. For any skew-symmetric matrix, all diagonal entries must be zero (since

a_{ii} = -a_{ii}

implies

a_{ii} = 0

). Therefore,

\text{tr}(A) = 0

. The condition

\text{tr}(A^TA) = 18

provides information about the sum of squares of all entries but doesn't affect the trace of

A

Question 18

If $A$ is an $n \times n$ symmetric matrix and $B = A^3 - 2A^2 + A$ , which of the following must be true?

$B$ is symmetric and $B^T = B^3 - 2B^2 + B$
$B$ is symmetric and commutes with $A$ (correct answer)
$B$ is skew-symmetric and $B^T = -B$
$B$ may not be symmetric unless $A$ is diagonal

Explanation: Since

A

is symmetric,

A^T = A

. Taking the transpose:

B^T = (A^3)^T - 2(A^2)^T + A^T = (A^T)^3 - 2(A^T)^2 + A^T = A^3 - 2A^2 + A = B

. So

B

is symmetric. Also,

B = A(A^2 - 2A + I)

and since

A

commutes with any polynomial in

A

, we have

AB = BA

. Choice A is circular reasoning, C is incorrect since

B

is symmetric, and D is false since symmetry is preserved regardless of whether

A

is diagonal.

Question 19

Let matrices $A$ and $B$ be defined as $A = \begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}$ and $B^T = \begin{pmatrix} -1 & 0 \\ 4 & 2 \end{pmatrix}$ . What is the matrix $(AB)^T$ ?

$\begin{pmatrix} -1 & 0 \\ 8 & 6 \end{pmatrix}$ (correct answer)
$\begin{pmatrix} -1 & 8 \\ 0 & 6 \end{pmatrix}$
$\begin{pmatrix} -1 & 0 \\ 10 & 6 \end{pmatrix}$
$\begin{pmatrix} 7 & 12 \\ 4 & 6 \end{pmatrix}$

Explanation: The property of transposes states that

(AB)^T = B^T A^T

. We are given

B^T

and can find

A^T

. The transpose of

A

A^T = \begin{pmatrix} 1 & 0 \\ 2 & 3 \end{pmatrix}

. Now, we compute the product

B^T A^T

: $$B^T A^T = \begin{pmatrix} -1 & 0 \ 4 & 2 \end{pmatrix} \begin{pmatrix} 1 & 0 \ 2 & 3 \end{pmatrix} = \begin{pmatrix} (-1)(1)+(0)(2) & (-1)(0)+(0)(3) \ (4)(1)+(2)(2) & (4)(0)+(2)(3) \end{pmatrix} = \begin{pmatrix} -1 & 0 \ 8 & 6 \end{pmatrix}

Question 20

Let $P$ be a $4 \times 4$ matrix such that $P^T P = I$ . If $A$ is symmetric and $B = P^T A P$ , which statement about the relationship between $A$ and $B$ is most precise?

$B$ is symmetric and $\text{tr}(A) = \text{tr}(B)$ , but $\det(A)$ may not equal $\det(B)$
$B$ is symmetric, $\text{tr}(A) = \text{tr}(B)$ , and $\det(A) = \det(B)$ (correct answer)
$B$ is symmetric but $\text{tr}(A)$ and $\text{tr}(B)$ may differ
$B$ may not be symmetric unless $P$ is also symmetric

Explanation: Since

P^TP = I

, matrix

P

is orthogonal, so

P^{-1} = P^T

. We have

B^T = (P^TAP)^T = P^TA^T(P^T)^T = P^TAP = B

since

A^T = A

. So

B

is symmetric. For trace:

\text{tr}(B) = \text{tr}(P^TAP) = \text{tr}(APP^T) = \text{tr}(A)

since

PP^T = I

. For determinant:

\det(B) = \det(P^T)\det(A)\det(P) = \det(P)^2\det(A) = \det(A)

since

\det(P) = \pm 1

for orthogonal matrices.