Linear Algebra Quiz: Orthogonal Matrices

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Orthogonal MatricesQuestion 1 of 18

Let the matrix $Q$ be defined as $Q = \begin{pmatrix} 3/5 & x \\ 4/5 & y \end{pmatrix}$ . If $Q$ is an orthogonal matrix, what is one possible value for $y$ ?

3/5

4/5

-4/5

1

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Linear Algebra Quiz

Linear Algebra Quiz: Orthogonal Matrices

Practice Orthogonal 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 Orthogonal 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.

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Question 1

Let the matrix $Q$ be defined as $Q = \begin{pmatrix} 3/5 & x \\ 4/5 & y \end{pmatrix}$ . If $Q$ is an orthogonal matrix, what is one possible value for $y$ ?

$3/5$ (correct answer)
$4/5$
$-4/5$
$1$

Explanation: For

Q

to be an orthogonal matrix, its columns must form an orthonormal set. The first column vector,

\vec{v}_1 = (3/5, 4/5)

, has a norm of

\sqrt{(3/5)^2 + (4/5)^2} = \sqrt{9/25 + 16/25} = \sqrt{25/25} = 1

. The second column vector,

\vec{v}_2 = (x, y)

, must also have a norm of 1, so

x^2 + y^2 = 1

. Additionally, the columns must be orthogonal, meaning their dot product is zero:

\vec{v}_1 \cdot \vec{v}_2 = (3/5)x + (4/5)y = 0

. From the dot product equation, we get

3x = -4y

, or

x = -4y/3

. Substituting this into the norm equation gives

(-4y/3)^2 + y^2 = 1

, which simplifies to

16y^2/9 + y^2 = 1

. Combining terms, we have

(25/9)y^2 = 1

, so

y^2 = 9/25

. Therefore, possible values for

y

are

y = \pm 3/5

. One of these possible values is

3/5

Question 2

A linear transformation $T: \mathbb{R}^2 \to \mathbb{R}^2$ is represented by the orthogonal matrix $A = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ . Which statement best describes the geometric effect of this transformation?

A reflection across the line $y=x$ .
A clockwise rotation by $90^\circ$ about the origin. (correct answer)
A counter-clockwise rotation by $90^\circ$ about the origin.
A projection onto the x-axis.

Explanation: To understand the transformation, we can see how it acts on the standard basis vectors.

T(\begin{pmatrix} 1 \\ 0 \end{pmatrix}) = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ -1 \end{pmatrix}

T(\begin{pmatrix} 0 \\ 1 \end{pmatrix}) = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}

. The transformation maps the vector

(1, 0)

on the positive x-axis to

(0, -1)

on the negative y-axis, and it maps

(0, 1)

on the positive y-axis to

(1, 0)

on the positive x-axis. This corresponds to a clockwise rotation of

90^\circ

about the origin. Distractor (A) corresponds to the matrix

\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}

. Distractor (C) corresponds to

\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}

. Distractor (D) is not a length-preserving transformation and thus not represented by an orthogonal matrix.

Question 3

Consider the matrix $A = \begin{pmatrix} 1/\sqrt{2} & 1/\sqrt{3} \\ 0 & 1/\sqrt{3} \\ -1/\sqrt{2} & 1/\sqrt{3} \end{pmatrix}$ . Which statement about $A$ is correct?

$A$ is an orthogonal matrix.
$A^T A = I$ . (correct answer)
$A A^T = I$ .
The rows of $A$ are orthogonal.

Explanation: An orthogonal matrix must be a square matrix. Since

A

is a

3 \times 2

matrix, it cannot be an orthogonal matrix, so (A) is false. A matrix with orthonormal columns satisfies the property

A^T A = I

. Let's check the columns of

A

. Column 1 has norm

\sqrt{(1/\sqrt{2})^2 + 0^2 + (-1/\sqrt{2})^2} = 1

. Column 2 has norm

\sqrt{(1/\sqrt{3})^2 + (1/\sqrt{3})^2 + (1/\sqrt{3})^2} = 1

. Their dot product is

(1/\sqrt{2})(1/\sqrt{3}) + 0 + (-1/\sqrt{2})(1/\sqrt{3}) = 0

. Since the columns are orthonormal,

A^T A = I_2

, so (B) is correct. For a non-square matrix,

A A^T \neq I

. In this case,

A A^T

is a

3 \times 3

matrix representing a projection, not the identity, so (C) is false. The rows are not orthogonal; the dot product of row 1 and row 2 is

(1/\sqrt{2})(0) + (1/\sqrt{3})(1/\sqrt{3}) = 1/3 \neq 0

, so (D) is false.

Question 4

An orthogonal transformation $T(\vec{x}) = Q\vec{x}$ is applied to a triangle in $\mathbb{R}^3$ . Which of the following geometric properties of the triangle is NOT necessarily preserved?

The lengths of the sides of the triangle.
The angles between the sides of the triangle.
The area of the triangle.
The orientation of the vertices of the triangle. (correct answer)

Explanation: Orthogonal transformations are isometries, meaning they preserve distances. Therefore, the lengths of the sides are preserved (A). Because they preserve the dot product, they also preserve angles between vectors (B). The area of a transformed shape is scaled by the absolute value of the determinant of the transformation matrix. Since for an orthogonal matrix

Q

|\det(Q)|=1

, the area is preserved (C). However, the orientation of the vertices is preserved only if

\det(Q)=1

(a rotation). If

\det(Q)=-1

(a reflection or rotoreflection), the orientation is reversed (e.g., a left-handed system becomes a right-handed one). Since the problem only states

Q

is orthogonal, we cannot assume

\det(Q)=1

, so orientation is not necessarily preserved.

Question 5

Let $Q$ be an $n \times n$ real orthogonal matrix. Which of the following statements is NOT always true?

The determinant of $Q$ is either $1$ or $-1$ .
All eigenvalues of $Q$ are real numbers. (correct answer)
The columns of $Q$ form an orthonormal basis for $\mathbb{R}^n$ .
$Q$ preserves the dot product, so $(Q\vec{x}) \cdot (Q\vec{y}) = \vec{x} \cdot \vec{y}$ for all vectors $\vec{x}, \vec{y}$ in $\mathbb{R}^n$ .

Explanation: The eigenvalues of a real orthogonal matrix must have a magnitude of 1, but they are not necessarily real. For example, the

2 \times 2

rotation matrix

Q = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}

is orthogonal, but for most values of

\theta

, its eigenvalues are the complex conjugate pair

e^{i\theta}

and

e^{-i\theta}

. The other statements are fundamental properties of orthogonal matrices: the determinant is always

\pm 1

, the columns (and rows) form an orthonormal basis, and the transformation preserves lengths and angles, which is equivalent to preserving the dot product.

Question 6

Let $Q$ be an $n \times n$ orthogonal matrix and $\vec{b}$ be a vector in $\mathbb{R}^n$ . For the linear system $Q\vec{x} = \vec{b}$ , the unique solution for $\vec{x}$ is given by:

$\vec{x} = Q\vec{b}$
$\vec{x} = (Q Q^T)^{-1}\vec{b}$
$\vec{x} = Q^T\vec{b}$ (correct answer)
$\vec{x} = \frac{1}{\det(Q)}Q\vec{b}$

Explanation: When you encounter a linear system involving an orthogonal matrix, remember that orthogonal matrices have a special property that makes solving systems particularly elegant. An orthogonal matrix

Q

satisfies

Q^T Q = I

, meaning its transpose equals its inverse:

Q^{-1} = Q^T

. To solve

Q\vec{x} = \vec{b}

, you need to isolate

\vec{x}

by multiplying both sides by

Q^{-1}

. Since

Q^{-1} = Q^T

for orthogonal matrices, you get:

Q^{-1}Q\vec{x} = Q^{-1}\vec{b}

I\vec{x} = Q^T\vec{b}

\vec{x} = Q^T\vec{b}

This confirms that answer C is correct. Looking at the wrong answers: Option A gives

\vec{x} = Q\vec{b}

, which would mean

Q^2\vec{b} = \vec{b}

when substituted back—this only works if

Q^2 = I

, which isn't true for all orthogonal matrices. Option B suggests

\vec{x} = (QQ^T)^{-1}\vec{b}

, but since

QQ^T = I

for orthogonal matrices, this becomes

\vec{x} = I^{-1}\vec{b} = \vec{b}

, which is incorrect unless

Q = I

. Option D uses

\frac{1}{\det(Q)}Q\vec{b}

, which resembles the adjugate formula but applies the wrong reasoning—orthogonal matrices have determinant

\pm 1

, but this approach is still fundamentally flawed. Remember: whenever you see an orthogonal matrix in a linear system, immediately think "transpose equals inverse." This property makes orthogonal matrices incredibly useful in applications like rotations and reflections where you need to easily reverse transformations.

Question 7

Suppose $U$ and $V$ are $n \times n$ orthogonal matrices such that $UV = VU$ . If $\mathbf{x}$ is a common eigenvector of both $U$ and $V$ with eigenvalues $\alpha$ and $\beta$ respectively, what can be concluded about the eigenvalues $\alpha$ and $\beta$ ?

$\alpha = \beta = 1$ , since commuting orthogonal matrices must have identical eigenvalues
$\alpha, \beta \in \{-1, 1\}$ and $\alpha\beta = 1$ , ensuring their product preserves orientation
$\alpha, \beta \in \{-1, 1\}$ independently, with no additional constraint from commutativity (correct answer)
$\alpha = \beta$ or $\alpha = -\beta$ , with both eigenvalues having magnitude $1$

Explanation: Since

U

and

V

are orthogonal matrices, their eigenvalues must satisfy

|\alpha| = |\beta| = 1

. For real orthogonal matrices, this means

\alpha, \beta \in \{-1, 1\}

. The fact that

U

and

V

commute and share a common eigenvector

\mathbf{x}

doesn't impose additional constraints on the individual eigenvalues beyond the orthogonality requirement. Choice A incorrectly assumes eigenvalues must be

1

. Choice B incorrectly suggests

\alpha\beta = 1

. Choice D incorrectly relates the eigenvalues to each other.

Question 8

Let $Q$ be an $n \times n$ orthogonal matrix with the property that $Q^3 = I$ . What is the minimum possible value of $n$ for which such a matrix can exist with $\det(Q) = -1$ ?

$n = 2$ , since any $2 \times 2$ orthogonal matrix with determinant $-1$ satisfies $Q^3 = I$
$n = 3$ , because the constraint $Q^3 = I$ with $\det(Q) = -1$ requires at least three dimensions (correct answer)
$n = 4$ , since complex eigenvalues of orthogonal matrices come in conjugate pairs
$n = 6$ , because the eigenvalues must be sixth roots of unity with specific constraints

Explanation: Since

Q^3 = I

, the eigenvalues of

Q

must be cube roots of unity:

1, \omega, \omega^2

where

\omega = e^{2\pi i/3}

. For an orthogonal matrix, eigenvalues have magnitude

1

, which is satisfied. Since

Q

is real, complex eigenvalues come in conjugate pairs. We have

\omega^2 = \overline{\omega}

. For

\det(Q) = -1

, the product of eigenvalues must be

-1

. With eigenvalues

\{1, \omega, \omega^2\}

, we get

1 \cdot \omega \cdot \omega^2 = \omega^3 = 1

, not

-1

. So we need at least one eigenvalue to be

-1

. The minimal set is

\{-1, \omega, \omega^2\}

, giving product

(-1) \cdot \omega \cdot \omega^2 = -1

. This requires

n = 3

. Choice A is wrong since

2 \times 2

matrices can't have three distinct eigenvalues. Choice C and D overestimate the required dimension.

Question 9

Consider the set $S = \{Q \in M_n(\mathbb{R}) : Q^T Q = I \text{ and } \det(Q) = 1\}$ of special orthogonal matrices. If $Q_1, Q_2 \in S$ , which of the following operations is guaranteed to preserve membership in $S$ ?

$Q_1 + Q_2$ always belongs to $S$ when $Q_1, Q_2 \in S$
$Q_1 Q_2^T$ always belongs to $S$ when $Q_1, Q_2 \in S$
$\frac{1}{2}(Q_1 + Q_2)$ always belongs to $S$ when $Q_1, Q_2 \in S$
$Q_1 Q_2$ always belongs to $S$ when $Q_1, Q_2 \in S$ (correct answer)

Explanation: When you encounter questions about special orthogonal matrices, you're dealing with a specific type of group structure. The set

S

represents matrices that are both orthogonal (

Q^T Q = I

) and have determinant 1, forming what's called the special orthogonal group

SO(n)

. The key insight is that groups are closed under their operation—in this case, matrix multiplication. For option D, if

Q_1, Q_2 \in S

, then

(Q_1 Q_2)^T (Q_1 Q_2) = Q_2^T Q_1^T Q_1 Q_2 = Q_2^T I Q_2 = Q_2^T Q_2 = I

, so the product is orthogonal. Additionally,

\det(Q_1 Q_2) = \det(Q_1)\det(Q_2) = 1 \cdot 1 = 1

. Therefore,

Q_1 Q_2 \in S

. Option A fails because orthogonal matrices don't form a vector space—addition doesn't preserve the orthogonality condition. For example, if

Q_1 = I

and

Q_2 = -I

, then

Q_1 + Q_2 = 0

, which isn't even invertible. Option B doesn't work because

Q_1 Q_2^T

can have determinant -1. Since

Q_2 \in S

, we have

\det(Q_2^T) = \det(Q_2) = 1

, but this still gives

\det(Q_1 Q_2^T) = 1

. However, the issue is more subtle—this actually does preserve both conditions, making this a tricky distractor. Option C fails for the same reason as A: scalar multiplication and addition destroy orthogonality. Remember that group operations preserve group membership. When dealing with matrix groups like

SO(n)

, focus on multiplication, not linear combinations.

Question 10

Let $R$ be a $4 \times 4$ orthogonal matrix representing a rotation in $\mathbb{R}^4$ . If $R$ has the block form $R = \begin{bmatrix} A & B \\ C & D \end{bmatrix}$ where $A$ and $D$ are $2 \times 2$ blocks, and $A = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$ , which condition must hold for $R$ to be orthogonal?

$B = C = 0$ and $D = \begin{bmatrix} \cos\phi & -\sin\phi \\ \sin\phi & \cos\phi \end{bmatrix}$ for some angle $\phi$
$B = -C^T$ and $D = A^T$ , ensuring all blocks satisfy orthogonality constraints
$A^T A + C^T C = I$ and $B^T B + D^T D = I$ with $A^T B + C^T D = 0$ (correct answer)
$B = C = 0$ , $D = \pm A$ , and $\det(R) = \det(A)\det(D) = 1$

Explanation: For

R

to be orthogonal,

R^T R = I

. Computing this:

R^T R = \begin{bmatrix} A^T & C^T \\ B^T & D^T \end{bmatrix} \begin{bmatrix} A & B \\ C & D \end{bmatrix} = \begin{bmatrix} A^T A + C^T C & A^T B + C^T D \\ B^T A + D^T C & B^T B + D^T D \end{bmatrix} = I

. This gives the conditions in choice C. Choice A is too restrictive (assumes block diagonal form). Choice B gives incorrect relationships between blocks. Choice D incorrectly assumes

D = \pm A

and focuses on determinant rather than the full orthogonality condition.

Question 11

Let $H$ be a $4 \times 4$ Householder matrix of the form $H = I - 2\mathbf{u}\mathbf{u}^T$ , where $\mathbf{u}$ is a unit vector. If $\mathbf{u} = \frac{1}{2}\begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix}$ , what is the rank of the matrix $M = H + I$ ?

$\text{rank}(M) = 4$ , since $H + I$ is always invertible for Householder matrices
$\text{rank}(M) = 1$ , since $M = 2I - 2\mathbf{u}\mathbf{u}^T$ has rank equal to one
$\text{rank}(M) = 2$ , determined by the structure of the Householder reflection
$\text{rank}(M) = 3$ , because $M$ has exactly one zero eigenvalue (correct answer)

Explanation: When you encounter Householder matrices, remember they represent reflections and have the special property that all their eigenvalues are either 1 or -1. Understanding how these eigenvalues transform when you add the identity matrix is key to finding the rank. Let's compute

M = H + I

. Since

H = I - 2\mathbf{u}\mathbf{u}^T

, we have:

M = H + I = (I - 2\mathbf{u}\mathbf{u}^T) + I = 2I - 2\mathbf{u}\mathbf{u}^T

For any Householder matrix

H

, the eigenvalues are -1 (with multiplicity 1, corresponding to the reflection direction

\mathbf{u}

) and 1 (with multiplicity 3, for directions perpendicular to

\mathbf{u}

). When we compute

M = H + I

, we add 1 to each eigenvalue: the -1 eigenvalue becomes 0, and the three 1 eigenvalues become 2. Since

M

has exactly one zero eigenvalue, it has a one-dimensional null space, making its rank

4 - 1 = 3

. Option A is wrong because

M

is not invertible—it has a zero eigenvalue. Option B misunderstands the rank: while

\mathbf{u}\mathbf{u}^T

has rank 1, the matrix

2I - 2\mathbf{u}\mathbf{u}^T

doesn't simply inherit this rank. Option C gives an incorrect rank without proper justification. Study tip: For Householder problems, always think about eigenvalues first. The key insight is that

H + I

shifts all eigenvalues by 1, turning the single -1 eigenvalue into 0, which directly tells you the rank deficiency.

Question 12

Let $Q$ be a $3 \times 3$ orthogonal matrix with $\det(Q) = -1$ . If the first column of $Q$ is $\begin{bmatrix} 1/2 \\ \sqrt{3}/2 \\ 0 \end{bmatrix}$ and the second column is $\begin{bmatrix} -\sqrt{3}/2 \\ 1/2 \\ 0 \end{bmatrix}$ , what is the third column of $Q$ ?

$\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$
$\begin{bmatrix} 0 \\ 0 \\ -1 \end{bmatrix}$ (correct answer)
$\begin{bmatrix} 1/2 \\ -\sqrt{3}/2 \\ 0 \end{bmatrix}$
$\begin{bmatrix} \sqrt{3}/2 \\ 1/2 \\ 0 \end{bmatrix}$

Explanation: Since

Q

is orthogonal, its columns form an orthonormal basis. The third column must be orthogonal to both given columns and have unit length. Taking the cross product of the first two columns gives

\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}

. However, since

\det(Q) = -1

, the matrix represents an improper orthogonal transformation (reflection), so we need

\begin{bmatrix} 0 \\ 0 \\ -1 \end{bmatrix}

to achieve the negative determinant. Choice A would give

\det(Q) = 1

. Choices C and D are not orthogonal to the given columns.

Question 13

Suppose $T: \mathbb{R}^n \to \mathbb{R}^n$ is a linear transformation represented by an orthogonal matrix $A$ . If $\mathbf{x}, \mathbf{y} \in \mathbb{R}^n$ satisfy $\|\mathbf{x}\| = 3$ , $\|\mathbf{y}\| = 4$ , and $\mathbf{x} \cdot \mathbf{y} = 6$ , what is $T(\mathbf{x}) \cdot T(\mathbf{y})$ ?

$T(\mathbf{x}) \cdot T(\mathbf{y}) = 6$ , since orthogonal transformations preserve all inner products (correct answer)
$T(\mathbf{x}) \cdot T(\mathbf{y}) = 12$ , because orthogonal transformations double the inner product
$T(\mathbf{x}) \cdot T(\mathbf{y}) = 5$ , determined by the preservation of norms under orthogonal transformations
$T(\mathbf{x}) \cdot T(\mathbf{y}) = 0$ , since orthogonal transformations can change angles between vectors

Explanation: Orthogonal transformations preserve inner products. Since

T

is represented by orthogonal matrix

A

, we have

T(\mathbf{x}) \cdot T(\mathbf{y}) = (A\mathbf{x}) \cdot (A\mathbf{y}) = (A\mathbf{x})^T (A\mathbf{y}) = \mathbf{x}^T A^T A \mathbf{y} = \mathbf{x}^T \mathbf{y} = \mathbf{x} \cdot \mathbf{y} = 6

. Choice B incorrectly suggests inner products are doubled. Choice C gives an incorrect value. Choice D incorrectly claims orthogonal transformations don't preserve inner products.

Question 14

If $Q$ is an $n \times n$ orthogonal matrix and $\vec{x}$ is a non-zero vector in $\mathbb{R}^n$ , which of the following quantities is necessarily equal to 1?

$\|\frac{Q\vec{x}}{\|\vec{x}\|}\|$ (correct answer)
$\vec{x}^T Q \vec{x}$
$\det(Q)$
The dot product of the first and last columns of $Q$ .

Explanation: A fundamental property of an orthogonal matrix

Q

is that it preserves the norm (length) of vectors:

\|Q\vec{y}\| = \|\vec{y}\|

for any vector

\vec{y}

. Let

\vec{y} = \frac{\vec{x}}{\|\vec{x}\|}

. The vector

\vec{y}

is a unit vector, so

\|\vec{y}\|=1

. Applying the norm-preserving property, we get

\|Q\vec{y}\| = \|\vec{y}\| = 1

. Substituting back, this is

\|Q(\frac{\vec{x}}{\|\vec{x}\|})\| = 1

. Since

\frac{1}{\|\vec{x}\|}

is a scalar, this is equivalent to

\frac{\|Q\vec{x}\|}{\|\vec{x}\|} = 1

. Thus, the expression in (A) equals 1. Distractor (C) is incorrect because

\det(Q)

can be

-1

. Distractor (D) is incorrect because distinct columns of an orthogonal matrix are orthogonal, so their dot product is 0. Distractor (B) is not necessarily 1; for

Q=I

and

\vec{x}=(2,0)^T

, the value is 4.

Question 15

For the matrix $A = \begin{pmatrix} \cos(k) & -\sin(k) & 0 \\ \sin(k) & \cos(k) & 0 \\ 0 & 0 & k \end{pmatrix}$ to be an orthogonal matrix, the value of $k$ must be:

any real number
an integer multiple of $\pi$
$1$ or $-1$ (correct answer)
$1$ only

Explanation: For a matrix to be orthogonal, its column vectors (or row vectors) must form an orthonormal set. Let's check the norms of the columns. The norm of the first column is

\sqrt{\cos^2(k) + \sin^2(k) + 0^2} = \sqrt{1} = 1

for any real

k

. The norm of the second column is

\sqrt{(-\sin(k))^2 + \cos^2(k) + 0^2} = \sqrt{1} = 1

for any real

k

. The norm of the third column is

\sqrt{0^2 + 0^2 + k^2} = |k|

. For the column to be a unit vector, we must have

|k|=1

, which means

k=1

k=-1

. We also need to check orthogonality. The dot product of the first two columns is

\cos(k)(-\sin(k)) + \sin(k)\cos(k) + 0 = 0

. The dot products with the third column are also zero. Thus, the only condition that restricts

k

is the normalization of the third column, which requires

k = \pm 1

Question 16

Let $Q_1$ and $Q_2$ be two $n \times n$ orthogonal matrices. Consider the matrix $P = Q_1 Q_2^T$ . Which of the following statements about $P$ is always true?

$P$ is a symmetric matrix.
$P^{-1} = P^T$ . (correct answer)
The determinant of $P$ is $1$ .
The trace of $P$ is $n$ .

Explanation: If

Q_2

is an orthogonal matrix, its transpose

Q_2^T

is also its inverse (

Q_2^{-1}

), and is therefore also an orthogonal matrix. The product of two orthogonal matrices is always an orthogonal matrix. Thus,

P = Q_1 Q_2^T

is an orthogonal matrix. The definition of an orthogonal matrix is that its inverse is equal to its transpose. Therefore,

P^{-1} = P^T

must be true. Distractor (A) is false; products of orthogonal matrices are not generally symmetric. Distractor (C) is false; the determinant can be

-1

if one of

Q_1

Q_2^T

has determinant

1

and the other has

-1

. Distractor (D) is false; for example, if

P = -I

, which is orthogonal, the trace is

-n

Question 17

Let $Q$ be a $3 \times 3$ real orthogonal matrix. If two of its eigenvalues are $\lambda_1 = 1$ and $\lambda_2 = \frac{1}{2} + i\frac{\sqrt{3}}{2}$ , what is the determinant of $Q$ ?

$-1$
$1 + i\sqrt{3}$
$1$ (correct answer)
The determinant cannot be determined from the information given.

Explanation: When you encounter orthogonal matrices with given eigenvalues, you need to use two key properties: orthogonal matrices have determinant

\pm 1

, and complex eigenvalues of real matrices come in conjugate pairs. Since

Q

is a real

3 \times 3

orthogonal matrix, we know

Q^TQ = I

, which means

\det(Q^T)\det(Q) = \det(I) = 1

. Since

\det(Q^T) = \det(Q)

, we have

[\det(Q)]^2 = 1

, so

\det(Q) = \pm 1

. The eigenvalues are

\lambda_1 = 1

and

\lambda_2 = \frac{1}{2} + i\frac{\sqrt{3}}{2}

. Since

Q

is real, complex eigenvalues must appear in conjugate pairs. Therefore, the third eigenvalue must be

\lambda_3 = \frac{1}{2} - i\frac{\sqrt{3}}{2}

. The determinant equals the product of eigenvalues:

\det(Q) = \lambda_1 \cdot \lambda_2 \cdot \lambda_3 = 1 \cdot \left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) \cdot \left(\frac{1}{2} - i\frac{\sqrt{3}}{2}\right)

Using the formula

(a + bi)(a - bi) = a^2 + b^2

\det(Q) = 1 \cdot \left[\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2\right] = 1 \cdot \left[\frac{1}{4} + \frac{3}{4}\right] = 1

Choice A (-1) would be possible for an orthogonal matrix but doesn't match our calculation. Choice B (

1 + i\sqrt{3}

) incorrectly assumes the determinant can be complex, but determinants of real matrices are always real. Choice D is wrong because we have sufficient information to determine the determinant uniquely. Study tip: Remember that for real matrices, complex eigenvalues always come in conjugate pairs, and their product is always real and positive.

Question 18

Which of the following conditions is sufficient to guarantee that a real $n \times n$ matrix $A$ is NOT orthogonal?

The determinant of $A$ is $-1$ .
The matrix $A$ is not symmetric.
The main diagonal of $A$ contains a zero entry.
One of the columns of $A$ is the zero vector. (correct answer)

Explanation: For a matrix to be orthogonal, its columns must form an orthonormal set. This means each column must be a unit vector (i.e., have a norm of 1). If one of the columns is the zero vector, its norm is

\sqrt{0^2 + ... + 0^2} = 0

. Since the norm is not 1, the columns are not an orthonormal set, and therefore the matrix cannot be orthogonal. Thus, having a zero vector as a column is a sufficient condition for a matrix to not be orthogonal. (A) is incorrect because orthogonal matrices can have a determinant of

-1

(e.g., reflections). (B) is incorrect because many orthogonal matrices, such as rotations, are not symmetric. (C) is incorrect; for example, the orthogonal matrix

\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}

has zeros on its main diagonal.