Types of Matrices MCQ Quiz - Objective Question with Answer for Types of Matrices - Download Free PDF

Calculation:

Given,

Matrix A is defined as:

\(A = \begin{bmatrix} x & y & z \\ y & z & x \\ z & x & y \end{bmatrix}\)

It is mentioned that A is an orthogonal matrix. Therefore,

\(A^T A = I\)

Now, we calculate A²:

\(A^2 = A \cdot A\)

\(A^2 = \begin{bmatrix} x & y & z \\ y & z & x \\ z & x & y \end{bmatrix} \cdot \begin{bmatrix} x & y & z \\ y & z & x \\ z & x & y \end{bmatrix}\)

Since A is orthogonal, \(A^T A = I\), and hence:

\(A^2 = I\)

∴ A² = Identity matrix (I).

Calculation: 

Given, A^T = A, B^T = –B, C^T = –C

Let M = A¹³B²⁶ – B²⁶A¹³

⇒ Then, M^T = (A¹³B²⁶ – B²⁶A¹³)^T

= (A¹³B²⁶)T – (B²⁶A¹³)^T

= (B^T)²⁶(A^T)¹³ – (A^T)¹³(B^T)²⁶

= B²⁶A¹³ – A¹³ B²⁶ = –M

Hence, M is skew symmetric

Let, N = A²⁶C¹³ – C¹³A²⁶

⇒ then, N^T = (A²⁶C¹³)^T – (C¹³A²⁶)^T

= –(C)¹³(A)²⁶+ A²⁶C¹³ = N

Hence, N is symmetric.

∴ Only S2 is true.

Hence, the correct answer is Option 1.

Concept:

1. Symmetric Matrix: Square matrix A is said to be symmetric if the transpose of matrix A is equal to matrix A itself ⇔ A^T = A or A’ = A

2. Skew-Symmetric Matrix or Anti-symmetric: Square matrix A is said to be skew-symmetric if transpose of matrix A is equal to negative of matrix A ⇔ A^T = −A

3. Hermitian Matrix:  A Hermitian matrix is a complex square matrix that is equal to its own conjugate transpose ⇔ \({{\bf{A}}^{\bf{T}}} = \;{\bf{\bar A}}\)

4. Skew-Hermitian: \({{\bf{A}}^{\bf{T}}} = - {\bf{\bar A}}\)

5. Let z = x + iy be a complex number.

• Conjugate of z =  = x – iy

Calculation:

Let A = \(\left[ {\begin{array}{*{20}{c}} 0&{ - 4 + i}\\ {4 + i}&0 \end{array}} \right]\)

Now, transpose of matrix A

 A^T = \(\left[ {\begin{array}{*{20}{c}} 0&{4 + i}\\ { - 4 + i}&0 \end{array}} \right]\)

⇒ A^T ≠ A

∴ given matrix is not symmetric

A^T = \(\left[ {\begin{array}{*{20}{c}} 0&{4 + i}\\ { - 4 + i}&0 \end{array}} \right] \ne - {\rm{A}}\)

∴ given matrix is not skew-symmetric.

Now, Conjugate of matrix A

\({\rm{\bar A}} = \;\left[ {\begin{array}{*{20}{c}} 0&{\overline { - 4 + i} }\\ {\overline {4 + i} }&0 \end{array}} \right] = \;\left[ {\begin{array}{*{20}{c}} 0&{ - 4 - i}\\ {4 - i}&0 \end{array}} \right] = \; - \left[ {\begin{array}{*{20}{c}} 0&{4 + i}\\ { - \;4 + i}&0 \end{array}} \right]\)
 

\( \Rightarrow {\rm{\bar A}} = - \;{{\rm{A}}^{\rm{T}}}\)

∴ given matrix is skew-hermitian matrix.

Explanation:

\(\rm A=\frac{1}{3}\begin{bmatrix}1&2&2\\\ 2&1&-2\\\ x&2&y\end{bmatrix}\) is orthogonal matrix.

So, AA^t = I

⇒ \(\frac{1}{3}\begin{bmatrix}1&2&2\\\ 2&1&-2\\\ x&2&y\end{bmatrix}\)\(\frac{1}{3}\begin{bmatrix}1&2&x\\\ 2&1&2\\\ 2&-2&y\end{bmatrix}\) = \(\begin{bmatrix}1&0&0\\\ 0&1&0\\\ 0&0&1\end{bmatrix}\)

⇒ \(\frac{1}{9}\begin{bmatrix}9&0&x+4+2y\\\ 0&9&2x+2-2y\\\ x+4+2y&2x+2-2y&x^2+4y+y^2\end{bmatrix}\) = \(\begin{bmatrix}1&0&0\\\ 0&1&0\\\ 0&0&1\end{bmatrix}\)

Comparing both sides we get

x + 4 + 2y = 0....(i)

2x + 2 - 2y = 0

⇒ x - y + 1 = 0...(ii)

Subtracting them 

3y = - 3 ⇒ y = -1

putting in (i) we get

x + 1 + 1 = 0

x = -2

Hence x + y = - 1 - 2 = -3

Option (4) is true.

Explanation:

A = (aij)2×2 and \(\rm a_{ij}=\left\{\begin{matrix}i^2+j^2;i\ne j\\\ i-j; i=j\end{matrix}\right\}\)

Then

Concept:

Properties of identity matrix:

If A is the square matrix of order n × n

• AI = IA = A
• Iⁿ = I           (Where n ∈ N)

Calculation:

Given

A² = I

Now, A3 + (A + I)2 - 9A - I² - A²

= A². A + A² + I² + 2AI - 9A - I² - A²

= I. A + I + I + 2AI - 9A - I - I           [∵ A2 = I and AI = IA = A]

= AI + 2AI - 9A 

= 3AI - 9A

= 3A - 9A

= - 6A

Concept:

• Elementary matrix: An elementary matrix is a matrix which differs from the identity matrix by one single elementary row operation
• An elementary matrix has each diagonal element 1.

Calculation:

Let's check option b,

Let E = \(\left[ {\begin{array}{*{20}{c}} 1&5&0\\ 0&1&0\\ 0&0&1 \end{array}} \right]\)

Apply R₁ → R₁ – 5R₂

\( \Rightarrow {\rm{E\;}} = \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right] = \;{{\rm{I}}_{3{\rm{\;}} \times 3}}\)

We can see that option B converted in to an identity matrix by one elementary operation.

Hence Option B is correct.

Shortcut Method:

We know that an elementary matrix has each diagonal element 1.

Only B has diagonal element is 1. (By Definition)

So, Option B is correct.

Concept:

A square matrix A = [a_ij]_{n × n} is said to be symmetric if A^T = A

A^T (Transpose) is obtained by changing rows to columns and columns to rows

Calculation:

Let A = \(\rm \begin{bmatrix} 3 & 0 & 0\\ 0 & 4 & 0\\ 0 & 0 & 0 \end{bmatrix}\)

 A^T = \(\rm \begin{bmatrix} 3 & 0 & 0\\ 0 & 4 & 0\\ 0 & 0 & 0 \end{bmatrix}\) = A

 A is  a symmetric matrix

Concept:

Identity matrix: A square matrix in which elements in the main diagonal are all '1' and the rest are all zero is called an identity matrix or unit matrix.

Thus, the square matrix A = [\(a_{ij}\)], if \(a_{ij}=\left\{\begin{matrix}1,if \hspace{3mm} i= j \\ 0,if \hspace{3mm} i\neq j \end{matrix}\right.\)

e.g., \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\), \(\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}\)are identity matrices of order 2 and 3 respectively.

For any natural number n, Iⁿ = I.ij]nneeeIⁿ_{\(I^n = I\)}

Calculation:

We have, A² =I

Also A and i are commutative, so we can expand (A + I)ⁿ using the expansion of (a+ b)ⁿ.

∴ (A – I)3 + (A + I)3 – 7A

= A³ - 3A² + 3A - I³ + A³ + 3A² + 3A + I² - 7A

= 2A³ + 6A - 7A

= 2A² ⋅A + 6A - 7A

= 2I ⋅A + 6A - 7A

= 2A + 6A - 7A

= 8A - 7A

= A

CONCEPT:

Symmetric Matrix:

Any real square matrix A = (a_ij) is said to be symmetric matrix if and only if a_ij = a_ji, ∀ i and j or in other words we can say that if A is a real square matrix such that A = A^t then A is said to be a symmetric matrix.

Skew-symmetric Matrix:

Any real square matrix A = (a_ij) is said to be skew-symmetric matrix if and only if a_ij = - a_ji, ∀ i and j or in other words we can say that if A is a real square matrix such that A = - A^t then A is said to be a skew-symmetric matrix.

Properties of Transpose of a Matrix:

• If A is a matrix of order m × n, then (A^t)^t = A
• If k ∈ R is a scalar and A is a matrix of order m × n, then (k × A)^t = k × A^t
• If A and B are matrices of same order m × n, then (A ± B)^t = A^t ± B^t.

CALCULATION:

Given: A is skew symmetric matrix

As we know that, if A is a skew symmetric matrix i.e A = - At 

⇒ (A²)^t = (A^t)²

∵ A is skew symmetric matrix

⇒ (A2)t = (- A)² = A²

So, A² is a symmetric matrix.

Hence, option D is the correct answer.

Concept:

1. Symmetric Matrix: Square matrix A is said to be symmetric if the transpose of matrix A is equal to matrix A itself ⇔ A^T = A or A’ = A

2. Skew-Symmetric Matrix or Anti-symmetric: Square matrix A is said to be skew-symmetric if transpose of matrix A is equal to negative of matrix A ⇔ A^T = −A

3. Hermitian Matrix:  A Hermitian matrix is a complex square matrix that is equal to its own conjugate transpose ⇔ \({{\bf{A}}^{\bf{T}}} = \;{\bf{\bar A}}\)

4. Skew-Hermitian: \({{\bf{A}}^{\bf{T}}} = - {\bf{\bar A}}\)

5. Let z = x + iy be a complex number.

• Conjugate of z =  = x – iy

Calculation:

Let A = \(\left[ {\begin{array}{*{20}{c}} 0&{ - 4 + i}\\ {4 + i}&0 \end{array}} \right]\)

Now, transpose of matrix A

 A^T = \(\left[ {\begin{array}{*{20}{c}} 0&{4 + i}\\ { - 4 + i}&0 \end{array}} \right]\)

⇒ A^T ≠ A

∴ given matrix is not symmetric

A^T = \(\left[ {\begin{array}{*{20}{c}} 0&{4 + i}\\ { - 4 + i}&0 \end{array}} \right] \ne - {\rm{A}}\)

∴ given matrix is not skew-symmetric.

Now, Conjugate of matrix A

\({\rm{\bar A}} = \;\left[ {\begin{array}{*{20}{c}} 0&{\overline { - 4 + i} }\\ {\overline {4 + i} }&0 \end{array}} \right] = \;\left[ {\begin{array}{*{20}{c}} 0&{ - 4 - i}\\ {4 - i}&0 \end{array}} \right] = \; - \left[ {\begin{array}{*{20}{c}} 0&{4 + i}\\ { - \;4 + i}&0 \end{array}} \right]\)
 

\( \Rightarrow {\rm{\bar A}} = - \;{{\rm{A}}^{\rm{T}}}\)

∴ given matrix is skew-hermitian matrix.

Concept:

Consider a matrix A is skew-symmetric, then A^T = −A
and A is symmetric, then A^T = A

Calculation:

Since, A is skew-symmetric.
A^T = −A
Since, A is symmetric.
A^T = A
⇒ −A = A
⇒2A = O
⇒A = O
Hence, A is a null matrix.

Hence, option (4) is correct.

Concept:

Singular Matrix: Any square matrix of order n is said to be singular if |A| = 0.

Involuntary Matrix: Any square matrix of order n is said to be an involuntary matrix if A² = I, where I is the identity matrix of order n.

Nilpotent Matrix: Any square matrix of order n is said to be nilpotent matrix if there exist least positive integer m such that A^m = O, where O is the null matrix of order n.

Idempotent Matrix: Any square matrix of order n is said to be an idempotent matrix if A² = A.

Calculation:

Given: \(A = \left[ {\begin{array}{*{20}{c}} 0&1\\ 1&0 \end{array}} \right],\)

\(\Rightarrow \;{A^2} = \;\left[ {\begin{array}{*{20}{c}} 0&1\\ 1&0 \end{array}} \right] \times \;\left[ {\begin{array}{*{20}{c}} 0&1\\ 1&0 \end{array}} \right] \\= \;\left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right] = I\)

Concept:

• For symmetric matrices, A = A' and B = B'
• For skew-symmetric matrices, A = - A'
• (A ± B)' = A' ± B'
• (AB)' = B'A'

Calculation:

Given: A and B are symmetric matrices

As we know that for symmetric matrices, we have A = A' and B = B'

(AB - BA)' = (AB)' - (BA)' --------(∵ (A ± B)' = A' ± B')

⇒ (AB - BA)' = B'A' - A'B' --------(∵ (AB)' = B'A')

⇒ (AB - BA)' = BA - AB ----------(∵ A = A' and B = B')

⇒ (AB - BA)' = - (AB - BA)

Hence Option 3 is correct.

Explanation:

Here we have to fill 4 entries it means we can form 2 × 2 matrices, 4 × 1 and 1 × 4 matrices

In 2 × 2 matrices, there are four places 

And each place has two ways to fill (1, 2) 

So, The total number of ways to fill the entries =  2 × 2 × 2 × 2 = 16.

In 4 × 1 matrices, there are four places

And each place has two ways to fill (1, 2) 

So, The total number of ways to fill the entries =  2 × 2 × 2 × 2 = 16.

Similarly in case of 1 × 4 matrices total number of ways = 16.

∴ Total number of matrices will be 48.