Symmetric and Non-symmetric Matrices MCQ Quiz - Objective Question with Answer for Symmetric and Non-symmetric Matrices - Download Free PDF

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

(A + B)T = A + B (symmetric)

(A - B)T = - (A - B) (skew-symmetric)

Calculation

Given:

A + B is symmetric, A - B is skew-symmetric, D = CT

A = \(\begin{bmatrix} -1 & 2 & 3 \\ 4 & 3 & -2 \\ 3 & -4 & 5 \end{bmatrix}\), C = \(\begin{bmatrix} 0 & 1 & -2 \\ 2 & -1 & 0 \\ 0 & 2 & 1 \end{bmatrix}\)

⇒ D = C^T = \(\begin{bmatrix} 0 & 2 & 0 \\ 1 & -1 & 2 \\ -2 & 0 & 1 \end{bmatrix}\) 

A^T + B^T = A + B ...(1)

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

⇒ (1) + (2): 2A^T = 2B

⇒ B = A^T

⇒ B = \(\begin{bmatrix} -1 & 4 & 3 \\ 2 & 3 & -4 \\ 3 & -2 & 5 \end{bmatrix}\)

B + D = \(\begin{bmatrix} -1 & 4 & 3 \\ 2 & 3 & -4 \\ 3 & -2 & 5 \end{bmatrix}\) + \(\begin{bmatrix} 0 & 2 & 0 \\ 1 & -1 & 2 \\ -2 & 0 & 1 \end{bmatrix}\)

⇒ B + D = \(\begin{bmatrix} -1+0 & 4+2 & 3+0 \\ 2+1 & 3-1 & -4+2 \\ 3-2 & -2+0 & 5+1 \end{bmatrix}\)

⇒ B + D = \(\begin{bmatrix} -1 & 6 & 3 \\ 3 & 2 & -2 \\ 1 & -2 & 6 \end{bmatrix}\)

∴ B + D = \(\begin{bmatrix} -1 & 6 & 3 \\ 3 & 2 & -2 \\ 1 & -2 & 6 \end{bmatrix}\)

Hence option 2 is correct

Concept Used:

If A is a symmetric matrix, then A^T = A.

If B is a skew-symmetric matrix, then B^T = -B.

If C = A + B, then C^T = A^T + B^T.

Calculation:

Given:

A is a symmetric matrix.

B is a skew-symmetric matrix.

 A + B = \(\begin{pmatrix} 1 & 3 \\ -2 & 5 \end{pmatrix}\) ...(1)

⇒ (A + B)^T = \(\begin{pmatrix} 1 & 3 \\ -2 & 5 \end{pmatrix}^T\)

⇒ A^T + B^T = \(\begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix}\)

⇒ A - B = \(\begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix}\) ...(2)

Adding (1) and (2):

⇒ (A + B) + (A - B) = \(\begin{pmatrix} 1 & 3 \\ -2 & 5 \end{pmatrix}\) + \(\begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix}\)

⇒ 2A = \(\begin{pmatrix} 2 & 1 \\ 1 & 10 \end{pmatrix}\)

⇒ A = \(\begin{pmatrix} 1 & 1/2 \\ 1/2 & 5 \end{pmatrix}\)

Subtracting (2) from (1):

⇒ (A + B) - (A - B) = \(\begin{pmatrix} 1 & 3 \\ -2 & 5 \end{pmatrix}\) - \(\begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix}\)

⇒ 2B = \(\begin{pmatrix} 0 & 5 \\ -5 & 0 \end{pmatrix}\)

⇒ B = \(\begin{pmatrix} 0 & 5/2 \\ -5/2 & 0 \end{pmatrix}\)

A - B = \(\begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix}\)

∴ A - B = \(\begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix}\)

Hence option 4 is correct

Concept

Square matrix A is said to be symmetric if aij = aij for all i and j.

Square matrix A is said to be symmetric if the transpose of matrix A is equal to matrix A ⇔ AT = A

Calculation

For a symmetric matrix, AT = A

\(\left[\begin{array}{ccc} 2 & 3 & 4 \\ 3 & -3 & 4 \\ 4 & 4 & 4 \end{array}\right]\)  is symmetric as all aij = aij

∴ Option 2 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:

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:

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: 

A matrix X can be written as a sum of the symmetric and skew-symmetric matrix which are 

P(symmetric) = \(\rm 1\over2\)(X + XT)

Q(skew-symmetric) = \(\rm 1\over2\)(X - XT)

Where XT is transpose of matrix X

Given matrix X = \(\begin{bmatrix} 7& 5 & 7\\ 3 & 4 & 6\\ 2 & 3 & 2 \end{bmatrix}\)

Transpose matrix X^T = \(\begin{bmatrix} 7& 3 & 2\\ 5& 4 & 3\\ 7 & 6 & 2 \end{bmatrix}\)

Symmetric matrix (P) =  \(\rm 1\over2\)(X + XT)

⇒ P =  \(\rm {1\over2}\left(\begin{bmatrix} 7& 5 & 7\\ 3 & 4 & 6\\ 2 & 3 & 2 \end{bmatrix}+\begin{bmatrix} 7& 3 & 2\\ 5& 4 & 3\\ 7 & 6 & 2 \end{bmatrix}\right)\)

⇒ P = \(\rm {1\over2}\begin{bmatrix} 14& 8 & 9\\ 8& 8 & 9\\ 9 & 9 & 4 \end{bmatrix}\) 

⇒ P = \(\begin{bmatrix} 7& 4 & 4.5\\ 4& 4 & 4.5\\ 4.5 & 4.5 & 2 \end{bmatrix}\)

Skew-symmetric matrix (Q) = \(\rm 1\over2\)(X - XT)

⇒ Q =  \(\rm {1\over2}\left(\begin{bmatrix} 7& 5 & 7\\ 3 & 4 & 6\\ 2 & 3 & 2 \end{bmatrix}-\begin{bmatrix} 7& 3 & 2\\ 5& 4 & 3\\ 7 & 6 & 2 \end{bmatrix}\right)\)

⇒ Q = \(\rm {1\over2}\begin{bmatrix} 0& 2 & 5\\ -2& 0 & 3\\ -5 & -3 & 0 \end{bmatrix}\) 

⇒ Q = \(\begin{bmatrix} 0& 1& 2.5\\ -1& 0& 1.5\\ -2.5& -1.5& 0\end{bmatrix}\)

CONCEPT:

Symmetric Matrix:

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

Calculation:

\(A = \left[ {\begin{array}{*{20}{c}} 1&{3 + x}&2\\ {1 - x}&2&{y + 1}\\ 2&{5 - y}&3 \end{array}} \right]\)

A = At

 \( A^t=\left[ {\begin{array}{*{20}{c}} 1&{1 - x}&2\\ {3 + x}&2&{5 - y}\\ 2&{y + 1}&3 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&{3 + x}&2\\ {1 - x}&2&{y + 1}\\ 2&{5 - y}&3 \end{array}} \right] = A\)

On comparing 

3 + x = 1 - x

⇒ x = - 1

And, y + 1 = 5 - y

⇒ y = 2

3x + y = 3(-1) + 2

∴ 3x + y = -1 

Explanation:

Given, the matrix A = \(\left[ {\begin{array}{*{20}{c}} 0&1&{ - 1} \\ 4&{ - 3}&4 \\ 3&{ - 3}&4 \end{array}} \right]\) 

A = B + C

Here matrix A is expressed as the sum of symmetric and skew-symmetric matrices.

Then, \(B=\frac{1}{2}(A+A^{T}) \ and\ C=\frac{1}{2}(A-A^{T})\)

Where B is symmetric and C is a skew-symmetric matrix.

To Find: The matrix B

A = \(\left[ {\begin{array}{*{20}{c}} 0&1&{ - 1} \\ 4&{ - 3}&4 \\ 3&{ - 3}&4 \end{array}} \right]\) 

\(A^{T}=\begin{bmatrix} 0 & 4& 3 \\ 1 & -3 & -3 \\ -1& 4& 4 \\ \end{bmatrix}\)

\(A+ A^{T}=\left[ {\begin{array}{*{20}{c}} 0&1&{ - 1} \\ 4&{ - 3}&4 \\ 3&{ - 3}&4 \end{array}} \right] +\begin{bmatrix} 0 & 4& 3 \\ 1 & -3 & -3 \\ -1& 4& 4 \\ \end{bmatrix}\)

\(A+ A^{T}=\begin{bmatrix} 0 &5 & 2 \\ 5& -6 & 1 \\ 2& 1 & 8 \\ \end{bmatrix}\)

\(B=\frac{1}{2}(A+ A^{T})=\frac{1}{2}\begin{bmatrix} 0 &5 & 2 \\ 5& -6 & 1 \\ 2& 1 & 8 \\ \end{bmatrix}\)

Concept:

A matrix X can be written as a sum of symmetric and skew-symmetric matrix which are 

P(symmetric) = \(\rm 1\over2\)(X + X^T)

Q(skew-symmetric) = \(\rm 1\over2\)(X - XT)

Calculation:

A = \(\rm \begin{bmatrix} 2 & -4 & 3\\ 3 & 1 & -2\\ 1& -3 & 5 \end{bmatrix}\)

A^T = \(\rm \begin{bmatrix} 2 & 3 & 1\\ -4& 1 & -3\\ 3& -2 & 5 \end{bmatrix}\)

P(symmetric matrix) = \(\rm 1\over2\)(A + AT)

⇒ P = \(\rm {1\over2}\left(\begin{bmatrix} 2 & -4 & 3\\ 3 & 1 & -2\\ 1& -3 & 5 \end{bmatrix}+ \begin{bmatrix} 2 & 3 & 1\\ -4& 1 & -3\\ 3& -2 & 5 \end{bmatrix}\right)\)

⇒ P = \(\rm \begin{bmatrix} 2 & -0.5 & 2\\ -0.5& 1 & -2.5\\ 2& -2.5 & 5 \end{bmatrix}\)

Q(skew-symmetric) = \(\rm 1\over2\)(A - AT)

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’ then A is said to be a symmetric matrix.
• (A ± B)' = A' ± B'
• (A ⋅ B)' = B' ⋅ A'

Calculation:

Given: A and B are two symmetric matrices of order n

Statement 1: A + B is also a symmetric matrix.

Let's find out transpose of A + B

⇒ (A + B)' = A' + B'

∵ A and B are two symmetric matrices of order n i.e A' = A and B' = B

⇒ (A + B)' = A + B

Hence, statement 1 is true.

Statement 2: A ⋅ B is a symmetric matrix

Let's find out the transpose of A ⋅ B

⇒ (A ⋅ B)' = B' ⋅ A'

∵ A and B are two symmetric matrices of order n i.e A' = A and B' = B

⇒ (A ⋅ B)' = B ⋅ A

But we also know that matrix multiplication is not commutative in general

So, we cannot say that (A ⋅ B)' = A ⋅ B in general

Hence, statement 2 is false.

Concept:

• Symmetric Matrix: Any real square matrix A = (aij) is said to be a symmetric matrix if and only if aij = aji, ∀ i, and j in other words we can say that if A is a real square matrix such that A = A’ then A is said to be a symmetric matrix.
• Skew-symmetric Matrix: Any real square matrix A = (aij) is said to be a skew-symmetric matrix if and only if aij = - aji, ∀ I, and j or in other words we can say that if A is a real square matrix such that A =- A’ then A is said to be a skew-symmetric matrix.

• Any real square matrix says A, can be expressed as the sum of the symmetric and skew-symmetric matrix.

  i.e \(A = \frac{1}{2}\;\left( {A + A'} \right) + \frac{1}{2}\;\left( {A - A'} \right)\) where A + A’ is symmetric and A – A’ is a skew-symmetric matrix.

Calculation:

Given: \(A = \left[ {\begin{array}{*{20}{c}} 3&2&5\\ 4&1&3\\ 0&6&7 \end{array}} \right] = \frac{1}{2} \cdot (P + Q)\) 

where P is symmetric and Q is a skew-symmetric matrix

Here we have to find the matrix P and Q

As we know, any square matrix can be expressed as the sum of the symmetric and skew-symmetric matrices.

i.e If A is a square matrix then A can be expressed as

where A + A’ is symmetric and A – A’ is a skew-symmetric matrix.

By comparing \(A = \left[ {\begin{array}{*{20}{c}} 3&2&5\\ 4&1&3\\ 0&6&7 \end{array}} \right] = \frac{1}{2} \cdot (P + Q)\) with \(A = \frac{1}{2}\;\left( {A + A'} \right) + \frac{1}{2}\;\left( {A - A'} \right)\) we get,

⇒ P = A + A' and Q = A - A'

\(A = \left[ {\begin{array}{*{20}{c}} 3&2&5\\ 4&1&3\\ 0&6&7 \end{array}} \right] \Rightarrow A' = \left[ {\begin{array}{*{20}{c}} 3&4&0\\ 2&1&6\\ 5&3&7 \end{array}} \right]\)

\(\Rightarrow P = \;\;\left[ {\begin{array}{*{20}{c}} 3&2&5\\ 4&1&3\\ 0&6&7 \end{array}} \right] + \;\left[ {\begin{array}{*{20}{c}} 3&4&0\\ 2&1&6\\ 5&3&7 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 6&6&5\\ 6&2&9\\ 5&9&{14} \end{array}} \right]\)

Similarly,

 \(Q = \;\left[ {\begin{array}{*{20}{c}} 3&2&5\\ 4&1&3\\ 0&6&7 \end{array}} \right] - \;\left[ {\begin{array}{*{20}{c}} 3&4&0\\ 2&1&6\\ 5&3&7 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&{ - \;2}&5\\ 2&0&{ - \;3}\\ { - \;5}&3&0 \end{array}} \right]\)

Hence, \(P = \left[ {\begin{array}{*{20}{c}} 6&6&5\\ 6&2&9\\ 5&9&{14} \end{array}} \right]\;and\;Q = \;\left[ {\begin{array}{*{20}{c}} 0&{ - \;2}&5\\ 2&0&{ - \;3}\\ { - \;5}&3&0 \end{array}} \right]\)

Concept:  

A symmetric matrix 

• It is a square matrix A of size n × n when the square matrix is equal to the transposed form of that matrix, that is, A^T = A.
• If A = [a_ij]_n×n is the symmetric matrix, then a_ij = a_ji for 1 ≤ i ≤ n, and 1 ≤ j ≤ n.

A skew-symmetric matrix 

• It is a square matrix A of size n × n when the square matrix is equal to the transposed form of that matrix, that is, A^T = -A.
• Diagonal elements of the skew-symmetric is zero.

If A = [a_ij]_n×n is the skew-symmetric matrix, then a_ij = -a_ji for 1 ≤ i ≤ n, and 1 ≤ j ≤ n.

A square matrix is called a singular matrix when its determinant is equal to 0.

A square matrix is called a non-singular matrix when its determinant is not equal to 0.

Calculation:

Given:

The given matrix is A = \(\begin{vmatrix} 1& -2& -3 \\ 2& 1& -2 \\ 3& 2 & 1 \\ \end{vmatrix}\).

Transpose of the matrix is A^t = \(\begin{vmatrix} 1& 2& 3 \\ -2& 1& 2 \\ -3& -2 & 1 \\ \end{vmatrix}\).

Since A^t ≠ A, hence the matrix A is not symmetric.

Since diagonal elements of the matrix are not zero, then matrix A is not skew-symmetric.

The determinant ∆ of the matrix is given by,

∆ = 1(1(1) - 2(-2)) - (-2)(1(2) - 3(-2)) + (-3)(2(2) - 1(3))

\(\Rightarrow\) ∆ = 1(1 + 4) + 2(2 + 6) - 3(4 - 3)

\(\Rightarrow\) ∆ = 5 + 16 - 3

\(\Rightarrow\) ∆ = 18

The determinant of the matrix is not equal to zero, hence it is non-singular.

Hence, the correct answer is option 4.