Symmetric and Non-symmetric Matrices MCQ Quiz in मल्याळम - Objective Question with Answer for Symmetric and Non-symmetric Matrices - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

Square matrix A is said to be skew-symmetric if a_ij = −a_ij for all i and j.

Square matrix A is said to be skew-symmetric if the transpose of matrix A is equal to the negative of matrix A ⇔ A^T = −A

All the main diagonal elements in the skew-symmetric matrix are zero.

Calculation:

For a skew-symmetric matrix, diagonal elements are zero and AT = −A

So, both $\left[\begin{array}{ccc}0& 4& 6\\ -4& 0& -2\\ -6& 2& 0\end{array}\right]$and $\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]$are Skew-symmetric matrix.

Concept:

Symmetric Matrix: Any square matrix A is said to be a symmetric matrix if A' = A. where A' is the transpose of A.

Skew symmetric matrix: Any square matrix A is said to be a symmetric matrix if A' = - A. where A' is the transpose of A.

Calculation:

We have, P =  $\left[\begin{array}{ccc}0& -5& 8\\ 5& 0& 12\\ -8& -12& 0\end{array}\right]$

Transpose of Matrix P is

⇒ P' = $\left[\begin{array}{ccc}0& 5& -8\\ -5& 0& -12\\ 8& 12& 0\end{array}\right]$

⇒ P' = - P

∴ P is a skew-symmetric matrix.

Concept:

• Symmetric Matrix: Any square matrix A is said to be symmetric matrix if A' = A. where A' is the transpose of A.
• Skew symmetric matrix: Any square matrix A is said to be symmetric matrix if A' = - A. where A' is the transpose of A.

Calculation:

Let P = (AB'- BA’)

∴ P' = (AB' - BA’)'

= (AB’)' — (BA’)'

= (B')'A' — (A')'B'   [∵  (AB)' = BA]

= BA' - AB'

= - (AB'- BA’)

= - P

∴ P' = - P

So it is a skew symmetric matrix.

Concept:- Number of different n × n symmetric matrices with each element being 0 or 1 is equal to $2^{\frac{n\left(n+1\right)}{2}}$

Eg. For n = 1 possible matrices [0], [1] only 2 matrices are possible

For n = 2

 $\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$, $\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]$,$\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$, $\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]$, $\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]$, $\left[\begin{array}{cc}1& 1\\ 0& 0\end{array}\right]$, $\left[\begin{array}{cc}1& 0\\ 1& 0\end{array}\right]$, $\left[\begin{array}{cc}0& 1\\ 0& 1\end{array}\right]$, $\left[\begin{array}{cc}0& 0\\ 1& 1\end{array}\right]$, $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$, $\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$, $\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right]$, $\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]$, $\left[\begin{array}{cc}0& 1\\ 1& 1\end{array}\right]$, $\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$, $\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]$

Maximum 16 matrices are possible out of which 8 are symmetric

Application:-    ∴ for n = 2     $2^{\frac{n\left(n+1\right)}{2}}$ = 2³ 

hence 8 different matrices are possible. so option (2) is correct.

Explanation:

The Transpose of a matrix A represented by A' is obtained by interchanging its rows into columns and vice-versa.

A Symmetric Matrix is a square matrix such that a_ij = a_ji for all i, j where a_ij is the element of the ith row and jth column of the matrix.

∵ aij = aji ⇒ There will no effect on transposing the symmetric matrix.

Hence, If A' = A, then A is a symmetric matrix.

(i) is true

A Skew-symmetric Matrix is a square matrix such that aij = -aji for all i, j where aij is the element of the ith row and jth column of the matrix and a_ii = 0 or the leading diagonal elements are all zero.

(ii) is true

So on transposing A :

A' = - A,

Hence if A = - A', then A is a skew-symmetric matrix.

(iii) is true

Any square matrix can be expressed as the sum not the product of a symmetric and a skew-symmetric matrix

(iv) is false

Concept:

If matrix P is symmetric matrix then 

Transpose of P = P

Consider A and B matrices, (AB)T = (B)T(A)T

If matrix P is skew symmetric matrix then 

Transpose of P = -P

Calculation:

P and Q Symmetric Matrices therefore

Transpose of P = P                  ..... (1)

Transpose of Q = Q                 ..... (2)

Now,

Transpose of (PQ - QP) = (PQ – QP)^T

Using the property of Transpose , (A - B)^T = (A)^T - (B)^T

(PQ - QP)^T= (PQ)^T - (QP)^T                            

Using again property of transpose, (AB)^T = (B)^T(A)^T

(PQ)^T - (QP)^T = (Q)^T (P)^T - (P)^T (Q)^T        ............(3)

Using Equations (1) and (2) in (3) we get,

(PQ)^T - (QP)^T = QP - PQ

(PQ)^T - (QP)^T = - (PQ - QP)                                  

So,  

Transpose of (PQ - QP) = (PQ - QP)^T = -  (PQ - QP)

Which show that  

(PQ - QP) is a Skew Symmetric Matrix.

Hence, option (4) is correct.

Concept:

Every square matrix can be expressed as the sum of a symmetric and a skew-symmetric matrix uniquely, which is given as,

 A = $\frac{1}{2}$(A + A^T) + $\frac{1}{2}$(A - AT)

where $\frac{1}{2}$(A + AT) is a symmetric matrix and $\frac{1}{2}$(A - A^T)

​​Calculation:

Given

A = $\left(\begin{array}{ccc}0& {\sin }^2\theta & {\cos }^2\theta \\ {\cos }^2\theta & 0& {\sin }^2\theta \\ {\sin }^2\theta & {\cos }^2\theta & 0\end{array}\right)$ and A = P + Q where P is a symmetric matrix and Q is a skew-symmetric matrix.

⇒ P = $\frac{1}{2}$(A + A^T) 

⇒ P = $\frac{1}{2}$[$\left(\begin{array}{ccc}0& {\sin }^2\theta & {\cos }^2\theta \\ {\cos }^2\theta & 0& {\sin }^2\theta \\ {\sin }^2\theta & {\cos }^2\theta & 0\end{array}\right)$ + $\left(\begin{array}{ccc}0& {\cos }^2\theta & {\sin }^2\theta \\ {\sin }^2\theta & 0& {\cos }^2\theta \\ {\cos }^2\theta & {\sin }^2\theta & 0\end{array}\right)$]

⇒ P = $\frac{1}{2}$$\left(\begin{array}{ccc}0& {\sin }^2\theta +{\cos }^2\theta & {\sin }^2\theta +{\cos }^2\theta \\ {\sin }^2\theta +{\cos }^2\theta & 0& {\sin }^2\theta +{\cos }^2\theta \\ {\sin }^2\theta +{\cos }^2\theta & {\sin }^2\theta +{\cos }^2\theta & 0\end{array}\right)$ 

⇒ P = $\frac{1}{2}$$\left(\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right)$

⇒ P = $\left(\begin{array}{ccc}0& 1/2& 1/2\\ 1/2& 0& 1/2\\ 1/2& 1/2& 0\end{array}\right)$

∴ The correct answer is option (1).

Formula used:

a³ + b³ = (a + b)(a² - ab + b²)

cos2θ + sin2θ = 1

sin 2θ = 2sin θ cos θ 

Calculation:

Given

A = $\left(\begin{array}{ccc}0& {\sin }^2\theta & {\cos }^2\theta \\ {\cos }^2\theta & 0& {\sin }^2\theta \\ {\sin }^2\theta & {\cos }^2\theta & 0\end{array}\right)$ 

Expanding along R₁,

⇒ |A| = 0 - sin²θ (0 - sin4θ) + cos2θ (cos4θ  - 0)

⇒ |A| = cos6θ + sin6θ

⇒ |A| = (cos2θ)³ + (sin2θ)³

⇒ |A| = (cos2θ + sin2θ)(cos4θ - cos2θsin2θ + sin4θ)

⇒ |A| = (cos4θ + sin4θ - cos2θsin2θ)

⇒ |A| = (cos4θ + sin4θ + 2cos2θsin2θ - 3cos2θsin2θ)

⇒ |A| = ((cos2θ + sin2θ)² - 3cos2θsin2θ)

⇒ |A| = (1 - 3cos2θsin2θ)

⇒ |A| = $(1-\frac{3}{4}(\sin 2\theta {)}^2$

|A| is minimum when sin 2θ is maximum and maximum value of sin 2θ = 1

⇒ Minimum value of |A| = $(1-\frac{3}{4}(1))$ = $\frac{1}{4}$

∴ The correct option is (1).

CONCEPT:

If A is a symmetric matrix and B is a skew-symmetric matrix then,

A = A^T, B = - B^T        ...(1)

By the properties of the transpose of the matrix,

(A + B)^T = AT + BT       ...(2)

CALCULATION

Given:

A + B = $\left[\begin{array}{cc}2& 3\\ 5& -1\end{array}\right]$        ...(3)

Using 2,

⇒ A' + B' = $\left[\begin{array}{cc}2& 5\\ 3& -1\end{array}\right]$

Using equation 1,

⇒ A - B = $\left[\begin{array}{cc}2& 5\\ 3& -1\end{array}\right]$          ...(4) 

After adding equations (i) and (ii)

A = $\left[\begin{array}{cc}2& 4\\ 4& -1\end{array}\right]$, B = $\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$

⇒ AB = $\left[\begin{array}{cc}4& -2\\ -1& -4\end{array}\right]$

• So, the correct answer is option 3.

Concept:

(i) If Δ ≠ 0 and atleast one of Δx, Δy, Δz ≠ 0, then the given system of equations is consistent and has unique non trivial solution.

(ii) If Δ ≠ 0 & Δx = Δy = Δz = 0, then the given system of equations is consistent and has trivial solution only.

(iii) If Δ = Δx = Δy = Δz = 0, then the given system of equations is consistent and has infinite solutions 

Calculation:

Given, A is symmetric matrix and B is skew-symmetric matrix

⇒ AT = A, BT = –B

Let A2B2 − B2A2= P

∴ PT = (A2B2 – B2A2)T

= (A2B2)T – (B2A2)T

= (B2)T (A2)T– (A2)T (B2)T

= B2A2 – A2B = –P

⇒ P is a skew-symmetric matrix.

∴$\left[\begin{array}{ccc}0& a& b\\ -a& 0& c\\ -b& -c& 0\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

∴ ay + bz = 0 …(1)

–ax + cz = 0 …(2)

–bx – cy =0 …(3)

From equation (1), (2), (3)

Δ = 0 and Δ1 = Δ2= Δ3 = 0

∴ System of equations has infinite number of solutions.

The correct answer is Option 3.