Mathematics Transpose of a Matrix, symmetric and skew symmetric matrix For CBSE-NCERT
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### Topic Covered

color {red} ** Transpose of a Matrix
color {red} ** Properties of transpose of the matrices
color {red} ** Symmetric and Skew Symmetric Matrices

### Transpose of a Matrix

\color{green} ✍️ If A = [a_(ij)] be an m × n matrix, then the matrix obtained by interchanging the rows and columns of A is called the transpose of A.

\color{green} ✍️ Transpose of the matrix A is denoted by A′ or (A^T). In other words, if A = [a_(ij)]_(m × n), then A′ = A^T= [a_(ji)]_(n × m).

● For example, If A = [(3,5),(sqrt3,1),(0,{-1}/5) ] _(3xx2), then A' = [ (3, sqrt3, 0 ),(5,1 ,{-1}/5) ]_(2xx3)

### Properties of transpose of the matrices

● For any matrices A and B of suitable orders, and transpose are A' & B', then we have

(i) (A′)′ = A,
(ii) (kA)′ = kA′ (where k is any constant)
(iii) (A + B)′ = A′ + B′
(iv) (A B)′ = B′ A′

### Symmetric and Skew Symmetric Matrices

color{blue}{"Symmetric Matrix :"}

\color{green} ✍️ A square matrix A = [a_(ij)] is said to be symmetric

color{red}{"If "A′ = A, \ \ "that is,"\ \ [a_(ij)] = [a_(ji)]} for all possible values of i and j.

For example A = [(sqrt3, 2, 3),(2, -1.5, -1),(3, -1,1)] is a symmetric matrix as A′ = A

color{blue}{"Skew Symmetric Matrix :"}

\color{green} ✍️ A square matrix A = [aij] is said to be skew symmetric matrix

color{red}{"if" A′ = – A,\ \ "that is" \ \ a_(ji) = – a_(ij) for all possible values of i and j.

Now, if we put i = j, we have a_(ii) = – a_(ii). Therefore 2a_(ii) = 0 or a_(ii )= 0 for all i’s.

=> color{green}{"This means that all the diagonal elements of a skew symmetric matrix are zero."}

For example, the matrix B = [ (0,e,f),(-e ,0 ,g),(-f , -g , 0) ] is a skew symmetric matrix as B′= –B

Now, we are going to prove some results of symmetric and skew-symmetric matrices.

### Theorem 1 :

color{blue}{" For any square matrix A with real number entries, A + A′ is a symmetric matrix"}

color{blue}{" and A – A′ is a skew symmetric matrix."}

● Let B = A + A′, then

B′ = (A + A′)′

= A′ + (A′)′ (as (A + B)′ = A′ + B′)

= A′ + A (as (A′)′ = A)

= A + A′ (as A + B = B + A)

= B

● Therefore B = A + A′ is a symmetric matrix

Now let C = A – A′

C′ = (A – A′)′ = A′ – (A′)′

= A′ – A

= – (A – A′) = – C

Therefore C = A – A′ is a skew symmetric matrix.

### Theorem 2 :

color{blue}{"Any square matrix can be expressed as the sum of a symmetric and a skew symmetric matrix."}

● Let A be a square matrix, then we can write

A = 1/2(A +A') + 1/2 (A - A')

● From the Theorem 1, we know that (A + A′) is a symmetric matrix and (A – A′) is a skew symmetric matrix.

● Since for any matrix A, (kA)′ = kA′, it follows that 1/2(A +A') is symmetric matrix and 1/2 (A -A') is skew symmetric matrix.

● Thus, any square matrix can be expressed as the sum of a symmetric and a skew symmetric matrix.
Q 3114680550

Express the matrix B= [ (2,-2,-4), ( -1,3,4), ( 1,-2,-3) ]  as the sum of a symmetric and a skew symmetric matrix.
Class 12 Chapter 3 Example 22
Solution:

Here

B' = [ (2,-1,1), (-2,3,-2), ( -4,4,-3) ]

Let  P= 1/2 (B+B') =1/2 [ ( 4, -3,-3), ( -3,6,2), ( -3,2, -6) ] = [ (2, -3/2, -3/2), ( -3/2 ,3,1), (-3/2,1,-3) ]

Now P' = [ ( 2, -3/2 , -3/2 ), ( -3/2 ,3,1), ( -3/2 ,1, -3) ] =P

Thus P= 1/2 (B+B')  is a symmetric matrix

Also, let Q =1/2 (b -B') =1/2 [ (0,-1, -5), ( 1,0, 6), ( 5, -6 , 0 ) ] = [ (0 , -1/2, -5/2 ), ( 1/2, 0 ,3 ), ( 5/2 , -3, 0 ) ]

Then Q' = [ (0,1/2 , 5/3 ), (-1/2 , 0, -3 ), ( -5/2 ,3, 0 ) ] = -Q

Thus Q = 1/2 (B-B') is a skew symmetric matrix.

Now P+Q = [ (2, -3/2, -3/2 ), ( -3/2 ,3 ,1 ), ( -3/2 , 1, -3 ) ] + [ ( 0 , -1/2 , -5/2 ), ( 1/2 , 0 ,3 ), (5/2 , -3 , 0) ] = [ ( 2,-2, -4 ), (-1,3,4), ( 1,-2,-3) ] = B

Thus, B is represented as the sum of a symmetric and a skew symmetric matrix.