Orthogonal Matrix MCQ Quiz - Objective Question with Answer for Orthogonal Matrix - Download Free PDF

Last updated on Jul 2, 2025

Latest Orthogonal Matrix MCQ Objective Questions

Orthogonal Matrix Question 1:

Let A be an orthogonal 3 × 3 matrix with real entries. Pick out the true statements:

  1. The determinant of A is a rational number.
  2. d(Ax, Ay) = d(x, y) for any two vectors x and y in ℝ3, where d(u, v) denotes the usual Euclidean distance between vectors u, v ∈ 3.
  3. All the entries of A are positive.
  4. All the eigenvalues of A are real.

Answer (Detailed Solution Below)

Option :

Orthogonal Matrix Question 1 Detailed Solution

Concept:

A square matrix A is said to be an orthogonal matrix if AA' = A'A = I where I is the identity matrix and A' is the transpose of A.

Explanation:

A is an orthogonal 3 × 3 matrix.

So, AA' = I

(1): ∴ |AA'| = |I|

⇒ |A||A'| = 1 (as |AB| = |A||B|)

⇒ |A||A| = 1 (as |A'| = |A|)

⇒ |A|2 = 1 ⇒ |A| = ± 1

Hence, the determinant of A is a rational number.

(1) is correct

(2): d(Ax, Ay) = = = = = d(x, y)

(2) is correct

(3): Counterexample: A = [0110] is an orthogonal matrix but all the entries of A are not positive,

(3) is false

(4): Counterexample: A = [0110] is an orthogonal matrix but all the eigenvalues of A are ± i which are not real.

(4) is false

Orthogonal Matrix Question 2:

If A and B are invertible matrices such that A2 = I and B2 = I, then (AB)−1 is:

  1. B−1 A−1
  2. BA
  3. AB
  4. A−1 B−1

Answer (Detailed Solution Below)

Option 2 : BA

Orthogonal Matrix Question 2 Detailed Solution

Concept Used:-

When there are two matrices, say X and Y such that,

XY = YX = I

Then,

X = Y-1

Explanation:-

Given that A and B are invertible matrices such that

A2 = I and B2 = I., A-1 = A and B-1 = B

Here, multiply AB with B−1A−1 and solve it further,

⇒ ABB−1A−1 = AB × B−1A−1

⇒ ABB−1A−1 = A(BB−1)A−1

⇒ ABB−1A−1 = AIA−1            [ ∵ X(X−1) = (X−1)X = I ]

⇒ ABB−1A−1 = AA−1

⇒ ABB−1A−1 = I        ........(1)

Now, multiply B−1 A−1 to AB 

⇒ B−1 A−1 AB = B−1A−1 × AB

⇒ B−1 A−1 AB = B−1(A−1 × A)B

⇒ B−1 A−1 AB = B−1IB

⇒ B−1 A−1 AB = B−1B

⇒ B−1 A−1 AB = I       ........(2)

From equations (1) and (2), 

⇒ AB(B−1A−1) = (B−1A−1)AB = I

⇒ (AB)−1 = B−1A−1 = BA

This can also be found out with the fundamental theorem of the invertible matrix which says when both A and B are invertible matrices having the same size, then AB will be invertible, also

⇒ (AB)−1 = B−1A−1 = BA

So, the value of (AB)−1 is equal to BA

Hence, the correct option is 2.

Orthogonal Matrix Question 3:

If A is a square matrix of order 3 such that |A| = 3, then adj (adj A) is:

  1. 27A
  2. 3A
  3. 9A
  4. A

Answer (Detailed Solution Below)

Option 2 : 3A

Orthogonal Matrix Question 3 Detailed Solution

Concept Used:
A1=adjA|A|adjA=|A|A1
|kA|=kn|A|,(kA)1=1kA1 and 
|A1|=1|A|
where k is any scaler and n is the order of A.
 
Calculation:
adj(adjA)=|adjA|(adjA)1
 ||A|A1|(|A|A1)1=|A|3|A1|1|A|A
|A|311A=|A|A=3A

Orthogonal Matrix Question 4:

A and B are real non-zero 3 × 3 matrices and satisfy the equation (AB)T + B-1A = 0. If B is orthogonal, then A is

  1. symmetric
  2. anti-symmetric
  3. Hermitian
  4. anti-Hermitian

Answer (Detailed Solution Below)

Option 2 : anti-symmetric

Orthogonal Matrix Question 4 Detailed Solution

Concept:

Symmetric matrix:

A square matrix is said to be symmetric if AT= A. Where AT or  A1 is the transpose of matrix A. In the transposition of the matrix rows and columns are interchanged.

Anti-symmetric matrix: 

A square matrix is said to be anti-symmetric if AT= - A.

Hermitian matrix:

A square matrix is said to be Hermitian if AAθ. Where Aθ is the transpose of the conjugate of the matrix i.e. (A¯)T

Anti-Hermitian matrix: 

A square matrix is said to be Hermitian if A = - Aθ

Orthogonal Matrix: 

A square matrix is said to be orthogonal if A× AT= I. In other words, the transpose of an orthogonal matrix is equal to the inverse of the matrix i.e. AT=A-1. Where I is the identity matrix

Calculation:

(AB)T + BTA = 0

Property (AB)T = BTAT

BTAT + BTA = 0

B is orthogonal ie BT = B-1

B-1AT + B-1A = 0

if AT = - A then the equation will become zero.

 - B-1A + B-1A = 0

So A should be anti-symmertic matrix.

Orthogonal Matrix Question 5:

For an orthogonal matrix Q, the valid equality is

  1. QT = Q-1
  2. Q = Q-1
  3. QT = Q
  4. det(Q) = 0

Answer (Detailed Solution Below)

Option 1 : QT = Q-1

Orthogonal Matrix Question 5 Detailed Solution

Concept:

Orthogonal matrix: 

When the product of a matrix to its transpose gives identity matrix.

Suppose A is a square matrix with real elements and of n x n order and AT or A’ is the transpose of A.

AAT = I

Calculation:

Given: 

Q is an orthogonal matrix with real elements and of n x n order and QT or Q’ is the transpose of Q.

Then according to the definition;

QQT = I

multiply by Q- 1 on both sides

(Q- 1 Q)QT = Q- 1I

IQT = Q- 1

QT = Q- 1 or Q’ = Q- 1

Top Orthogonal Matrix MCQ Objective Questions

If A is [8576] then the value of |A121 - A120|

  1. 0
  2. 1
  3. 120
  4. 121

Answer (Detailed Solution Below)

Option 1 : 0

Orthogonal Matrix Question 6 Detailed Solution

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Concept:

Let B = |A121 - A120|

B = |A120 × (A – I)| 

B = |A120| × |A – I|

Calculation:

A = [8576]

Now, calculating matrix [A – I]

[A – I] = [8576][1001]

[A – I] = [7575]

Now determinant of |A – I|,

|A – I| = |7575|

|A – I| = 0   (Since two rows are repeated, therefore determinant = 0)

Hence, |A121 - A120|  A120|A – I|  = 0

For the given orthogonal matrix Q,

Q=[372767673727276737]

The inverse is __________

  1. [372767673727276737]

     

  2. [372767673727276737]
  3. [376727273767672737]
  4. [376727273767672737]

Answer (Detailed Solution Below)

Option 3 : [376727273767672737]

Orthogonal Matrix Question 7 Detailed Solution

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Concept:

For orthogonal matrix: A × AT = I ⇒ A-1 = AT

Calculation:

For given matrix (Q)=[372767673727276737]

∴ Q × QT = 1 ⇒ Q-1 = QT

Q1=QT=[376727273767672737]

Alternative Method:

Q1=1|Q|×Adj[Q]

AdjQ=[214942491449144921494249424914492142]

|Q|=37×(9491249)27×(1849449)+67×(3649649)=1

Q1=[376727273767672737]

Consider the matrix P=[1201201012012]

Which one of the following statements about P is INCORRECT?

  1. Determinant of P is equal to 1.
  2. P is orthogonal
  3. Inverse of P is equal to its transpose
  4. All eigenvalues of P are real numbers

Answer (Detailed Solution Below)

Option 4 : All eigenvalues of P are real numbers

Orthogonal Matrix Question 8 Detailed Solution

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Concept:

P=[1201201012012]

(i) |P|=12(12)0+12(12)=1

(ii) for Matrix to be orthogonal. PPT = PTP = I

PT=[1201201012012]

PPT=[1201201012012][1201201012012]

=[100010001]=I

As PPT = I, so P is orthogonal

(iii) P1=adjP|P|

=[1201201012012]=PT

As option A, B and C are correct. So option D is wrong.

If A is a square matrix then orthogonality property mandates

  1. AAT = A-1
  2. AAT = 0
  3. AAT = A2
  4. AAT = I

Answer (Detailed Solution Below)

Option 4 : AAT = I

Orthogonal Matrix Question 9 Detailed Solution

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Concept:

Orthogonal matrix: When the product of a matrix to its transpose gives identity matrix.

Suppose A is a square matrix with real elements and of n x n order and AT or A’ is the transpose of A.

AAT = I

Important Points

  • A square matrix such that A2 = I is called the involuntary matrix.
  • A square matrix such that A2 = A is called the Idempotent matrix.
  • Any square matrix A is said to be non-singular if |A| ≠ 0, and a square matrix A is said to be singular if |A| = 0.
  • A square matrix A = [aij] is said to be Hermitian matrix if aij = a̅ij

For an orthogonal matrix Q, the valid equality is

  1. QT = Q-1
  2. Q = Q-1
  3. QT = Q
  4. det(Q) = 0

Answer (Detailed Solution Below)

Option 1 : QT = Q-1

Orthogonal Matrix Question 10 Detailed Solution

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Concept:

Orthogonal matrix: 

When the product of a matrix to its transpose gives identity matrix.

Suppose A is a square matrix with real elements and of n x n order and AT or A’ is the transpose of A.

AAT = I

Calculation:

Given: 

Q is an orthogonal matrix with real elements and of n x n order and QT or Q’ is the transpose of Q.

Then according to the definition;

QQT = I

multiply by Q- 1 on both sides

(Q- 1 Q)QT = Q- 1I

IQT = Q- 1

QT = Q- 1 or Q’ = Q- 1

If A is a square matrix of order 3 such that |A| = 3, then adj (adj A) is:

  1. 27A
  2. 3A
  3. 9A
  4. A

Answer (Detailed Solution Below)

Option 2 : 3A

Orthogonal Matrix Question 11 Detailed Solution

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Concept Used:
A1=adjA|A|adjA=|A|A1
|kA|=kn|A|,(kA)1=1kA1 and 
|A1|=1|A|
where k is any scaler and n is the order of A.
 
Calculation:
adj(adjA)=|adjA|(adjA)1
 ||A|A1|(|A|A1)1=|A|3|A1|1|A|A
|A|311A=|A|A=3A

The eigenvectors of a real symmetric matrix corresponding to different eigenvalues are 

  1. orthogonal
  2. singular 
  3. non-singular 
  4. none of these 

Answer (Detailed Solution Below)

Option 1 : orthogonal

Orthogonal Matrix Question 12 Detailed Solution

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Explanation:

"λ" is called eigen value and "X" is called eigen vector of a square matrix "A", if

AX = λX

(i) Eigen values of a real symmetric matrix are always real.

(ii) Eigen vectors of a symmetric matrix corresponding to different eigenvalues are orthogonal to each other.

(iii) Sum of the eigen values of a matrix is always equal to the trace(sum of diagonal elements) of a matrix.

(iv) Eigen values of a matrix and its transpose are the same because the transpose matrix will also have the same characteristic equation.

Important Points

Characteristics of eigen values:

  • Tr (A) = Summation of eigen values
  • |A| = Product of eigen values
  • If A = Upper triangular matrix or lower triangular matrix or diagonal matrix, then its eigen values will be diagonal elements.
  • Eigen values of the hermitian matrix and real symmetric matrix are always real.
  • Eigen values of skew symmetric and skew hermitian matrix are either zero or purely imaginary.
  • Eigen values of the orthogonal matrix and unitary matrix have unit modulus.

A and B are real non-zero 3 × 3 matrices and satisfy the equation (AB)T + B-1A = 0. If B is orthogonal, then A is

  1. symmetric
  2. anti-symmetric
  3. Hermitian
  4. anti-Hermitian

Answer (Detailed Solution Below)

Option 2 : anti-symmetric

Orthogonal Matrix Question 13 Detailed Solution

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Concept:

Symmetric matrix:

A square matrix is said to be symmetric if AT= A. Where AT or  A1 is the transpose of matrix A. In the transposition of the matrix rows and columns are interchanged.

Anti-symmetric matrix: 

A square matrix is said to be anti-symmetric if AT= - A.

Hermitian matrix:

A square matrix is said to be Hermitian if AAθ. Where Aθ is the transpose of the conjugate of the matrix i.e. (A¯)T

Anti-Hermitian matrix: 

A square matrix is said to be Hermitian if A = - Aθ

Orthogonal Matrix: 

A square matrix is said to be orthogonal if A× AT= I. In other words, the transpose of an orthogonal matrix is equal to the inverse of the matrix i.e. AT=A-1. Where I is the identity matrix

Calculation:

(AB)T + BTA = 0

Property (AB)T = BTAT

BTAT + BTA = 0

B is orthogonal ie BT = B-1

B-1AT + B-1A = 0

if AT = - A then the equation will become zero.

 - B-1A + B-1A = 0

So A should be anti-symmertic matrix.

If A and B are invertible matrices such that A2 = I and B2 = I, then (AB)−1 is:

  1. B−1 A−1
  2. BA
  3. AB
  4. A−1 B−1

Answer (Detailed Solution Below)

Option 2 : BA

Orthogonal Matrix Question 14 Detailed Solution

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Concept Used:-

When there are two matrices, say X and Y such that,

XY = YX = I

Then,

X = Y-1

Explanation:-

Given that A and B are invertible matrices such that

A2 = I and B2 = I., A-1 = A and B-1 = B

Here, multiply AB with B−1A−1 and solve it further,

⇒ ABB−1A−1 = AB × B−1A−1

⇒ ABB−1A−1 = A(BB−1)A−1

⇒ ABB−1A−1 = AIA−1            [ ∵ X(X−1) = (X−1)X = I ]

⇒ ABB−1A−1 = AA−1

⇒ ABB−1A−1 = I        ........(1)

Now, multiply B−1 A−1 to AB 

⇒ B−1 A−1 AB = B−1A−1 × AB

⇒ B−1 A−1 AB = B−1(A−1 × A)B

⇒ B−1 A−1 AB = B−1IB

⇒ B−1 A−1 AB = B−1B

⇒ B−1 A−1 AB = I       ........(2)

From equations (1) and (2), 

⇒ AB(B−1A−1) = (B−1A−1)AB = I

⇒ (AB)−1 = B−1A−1 = BA

This can also be found out with the fundamental theorem of the invertible matrix which says when both A and B are invertible matrices having the same size, then AB will be invertible, also

⇒ (AB)−1 = B−1A−1 = BA

So, the value of (AB)−1 is equal to BA

Hence, the correct option is 2.

Orthogonal Matrix Question 15:

If A is [8576] then the value of |A121 - A120|

  1. 0
  2. 1
  3. 120
  4. 121

Answer (Detailed Solution Below)

Option 1 : 0

Orthogonal Matrix Question 15 Detailed Solution

Concept:

Let B = |A121 - A120|

B = |A120 × (A – I)| 

B = |A120| × |A – I|

Calculation:

A = [8576]

Now, calculating matrix [A – I]

[A – I] = [8576][1001]

[A – I] = [7575]

Now determinant of |A – I|,

|A – I| = |7575|

|A – I| = 0   (Since two rows are repeated, therefore determinant = 0)

Hence, |A121 - A120|  A120|A – I|  = 0

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