Notes On Symmetric and Skew Symmetric Matrices - CBSE Class 12 Maths
Transpose of a matrix : If A = [aij] be an m x n matrix, then the matrix obtained by interchanging the rows and columns of A is called the transpose of A. Ex : A = $\left[\begin{array}{ccc}\text{2}& \text{1}& \text{-4}\\ \text{3}& \text{2}& \text{5}\end{array}\right]$  B = $\left[\begin{array}{cc}\text{2}& \text{3}\\ \text{1}& \text{2}\\ \text{-4}& \text{5}\end{array}\right]$                     The transpose of a matrix, A, is denoted as A' or AT. (i) (X' )' = X, (ii) (kX' )' = kX (where kis any constant) (iii) (X + Y)' = X' + Y' (iv) (X Y)' = Y' X' Let X =     Y =  are  two matrices i) (A')' = A X =  X' = $\left[\begin{array}{cc}\text{1}& \text{3}\\ \text{1}& \text{1}\\ \text{2}& \text{5}\end{array}\right]$ (X')' =  = X          (ii) (kA')' = kA (where kis any constant) LHS: (5Y)' 5 x Y = 5 x  =   (5Y)' = $\left[\begin{array}{cc}\text{10}& \text{5}\\ \text{5}& \text{20}\\ \text{25}& \text{10}\end{array}\right]$     RHS: 5Y' Y' =  5 x Y' = 5 x     =   (5Y)' = $\left[\begin{array}{cc}\text{10}& \text{5}\\ \text{5}& \text{20}\\ \text{25}& \text{10}\end{array}\right]$    = 5Y'    LHS = RHS (iii) (A + B)' = A' + B' (X + Y)' X + Y =   +    =     (X+Y)' =  $\left[\begin{array}{cc}\text{3}& \text{4}\\ \text{2}& \text{5}\\ \text{7}& \text{7}\end{array}\right]$     X' + Y' =  $\left[\begin{array}{cc}\text{1}& \text{3}\\ \text{1}& \text{1}\\ \text{2}& \text{5}\end{array}\right]$  +  $\left[\begin{array}{cc}\text{2}& \text{1}\\ \text{1}& \text{4}\\ \text{5}& \text{2}\end{array}\right]$    = $\left[\begin{array}{cc}\text{3}& \text{4}\\ \text{2}& \text{5}\\ \text{7}& \text{7}\end{array}\right]$  (X+Y)' =  $\left[\begin{array}{cc}\text{3}& \text{4}\\ \text{2}& \text{5}\\ \text{7}& \text{7}\end{array}\right]$   = X' + Y' (iv) (AB)' = B' A' X = $\left[\begin{array}{c}\text{0}\\ \text{1}\\ \text{2}\end{array}\right]$ and Y = $\left[\begin{array}{ccc}\text{1}& \text{5}& \text{7}\end{array}\right]$ LHS: (XY)' XY =  =  $\left[\begin{array}{ccc}\text{0}& \text{0}& \text{0}\\ \text{1}& \text{5}& \text{7}\\ \text{2}& \text{10}& \text{14}\end{array}\right]$       (XY)' = $\left[\begin{array}{ccc}\text{0}& \text{1}& \text{2}\\ \text{0}& \text{5}& \text{10}\\ \text{0}& \text{7}& \text{14}\end{array}\right]$ RHS: Y' X' X =  $\left[\begin{array}{c}\text{0}\\ \text{1}\\ \text{2}\end{array}\right]$  X' =   Y =   Y' =    Y'X' =   = $\left[\begin{array}{ccc}\text{0}& \text{1}& \text{2}\\ \text{0}& \text{5}& \text{10}\\ \text{0}& \text{7}& \text{14}\end{array}\right]$           (XY)' =   $\left[\begin{array}{ccc}\text{0}& \text{1}& \text{2}\\ \text{0}& \text{5}& \text{10}\\ \text{0}& \text{7}& \text{14}\end{array}\right]$     = Y'X' Symmetric matrix : A square matrix A = [aij] is said to be symmetric if A' = A, that is, [aij] = [aij] for all possible values of i and j. Ex: X =  $\left[\begin{array}{ccc}\text{0}& \text{1}& \text{2}\\ \text{0}& \text{5}& \text{10}\\ \text{0}& \text{7}& \text{14}\end{array}\right]$      $\underset{\to }{\text{Transpose operation}}$          X' = X , X is a symmetric matrix. Skew symmetric matrix: A square matrix A = [aij] is said to be skew symmetric if A' = - A that is, [aij] = - [aij] for all possible values of i and j. Ex: Y =                  $\underset{\to }{\text{Transpose operation}}$                                                 Y' = -Y ⇒ Y is a skew symmetric matrix. Theorem 1: For any square matrix A with real number entries, A + A′ is a symmetric matrix and A - A′ is a skew symmetric matrix. Proof1): A is a square matrix with real numbers as its elements. Given, A = [aij]mxn Let X = A + A' If X is a symmetric matrix, X = X' X' = (A + A')' = (A)' + (A')'  '.' [(A + B)' = A' + B'] = A' + A = A' + A = X X is a symmetric matrix or (A' + A) is a symmetric matrix Proof 2): A - A' is a skew symmetric matrix. Let Y = A - A' Skew symmetric matrix: Y' = - Y Y' = (A - A')' = (A)' - (A')' = A' - A = - (A - A') = -Y Y is a skew symmetric matrix or (A - A') is a skew symmetric matrix. Ex: A =        A' =  $\left[\begin{array}{ccc}\text{2}& \text{4}& \text{1}\\ \text{3}& \text{3}& \text{0}\\ \text{5}& \text{2}& \text{7}\end{array}\right]$ To verify that A + A' is symmetric. A + A' =          (A + A' ) ' =     $\underset{\to }{\text{Transpose operation}}$      = A + A'    To verify that A - A' is a skew symmetric matrix. A - A' = $\left[\begin{array}{ccc}\text{0}& \text{-1}& \text{4}\\ \text{1}& \text{0}& \text{2}\\ \text{-4}& \text{-2}& \text{0}\end{array}\right]$            $\left[\begin{array}{ccc}\text{0}& \text{-1}& \text{4}\\ \text{1}& \text{0}& \text{2}\\ \text{-4}& \text{-2}& \text{0}\end{array}\right]$        $\underset{\to }{\text{Transpose operation}}$       $\left[\begin{array}{ccc}\text{0}& \text{1}& \text{4}\\ \text{-1}& \text{0}& \text{-2}\\ \text{-4}& \text{2}& \text{0}\end{array}\right]$           $\left[\begin{array}{ccc}\text{0}& \text{1}& \text{4}\\ \text{-1}& \text{0}& \text{-2}\\ \text{-4}& \text{2}& \text{0}\end{array}\right]$   = (-1)  = -(A - A' )       Theorem 2: Any square matrix can be expressed as the sum of a symmetric and a skew symmetric matrix. Proof: Let A = [aij]mxm We know that A + A' : Symmetric matrix                       A - A' : Skew Symmetric matrix = ½(A + A' ) + ½ (A - A') = (½A + ½A') + (½A - ½A') Using kA = k(A) = ½A + ½A = A.

Summary

Transpose of a matrix : If A = [aij] be an m x n matrix, then the matrix obtained by interchanging the rows and columns of A is called the transpose of A. Ex : A = $\left[\begin{array}{ccc}\text{2}& \text{1}& \text{-4}\\ \text{3}& \text{2}& \text{5}\end{array}\right]$  B = $\left[\begin{array}{cc}\text{2}& \text{3}\\ \text{1}& \text{2}\\ \text{-4}& \text{5}\end{array}\right]$                     The transpose of a matrix, A, is denoted as A' or AT. (i) (X' )' = X, (ii) (kX' )' = kX (where kis any constant) (iii) (X + Y)' = X' + Y' (iv) (X Y)' = Y' X' Let X =     Y =  are  two matrices i) (A')' = A X =  X' = $\left[\begin{array}{cc}\text{1}& \text{3}\\ \text{1}& \text{1}\\ \text{2}& \text{5}\end{array}\right]$ (X')' =  = X          (ii) (kA')' = kA (where kis any constant) LHS: (5Y)' 5 x Y = 5 x  =   (5Y)' = $\left[\begin{array}{cc}\text{10}& \text{5}\\ \text{5}& \text{20}\\ \text{25}& \text{10}\end{array}\right]$     RHS: 5Y' Y' =  5 x Y' = 5 x     =   (5Y)' = $\left[\begin{array}{cc}\text{10}& \text{5}\\ \text{5}& \text{20}\\ \text{25}& \text{10}\end{array}\right]$    = 5Y'    LHS = RHS (iii) (A + B)' = A' + B' (X + Y)' X + Y =   +    =     (X+Y)' =  $\left[\begin{array}{cc}\text{3}& \text{4}\\ \text{2}& \text{5}\\ \text{7}& \text{7}\end{array}\right]$     X' + Y' =  $\left[\begin{array}{cc}\text{1}& \text{3}\\ \text{1}& \text{1}\\ \text{2}& \text{5}\end{array}\right]$  +  $\left[\begin{array}{cc}\text{2}& \text{1}\\ \text{1}& \text{4}\\ \text{5}& \text{2}\end{array}\right]$    = $\left[\begin{array}{cc}\text{3}& \text{4}\\ \text{2}& \text{5}\\ \text{7}& \text{7}\end{array}\right]$  (X+Y)' =  $\left[\begin{array}{cc}\text{3}& \text{4}\\ \text{2}& \text{5}\\ \text{7}& \text{7}\end{array}\right]$   = X' + Y' (iv) (AB)' = B' A' X = $\left[\begin{array}{c}\text{0}\\ \text{1}\\ \text{2}\end{array}\right]$ and Y = $\left[\begin{array}{ccc}\text{1}& \text{5}& \text{7}\end{array}\right]$ LHS: (XY)' XY =  =  $\left[\begin{array}{ccc}\text{0}& \text{0}& \text{0}\\ \text{1}& \text{5}& \text{7}\\ \text{2}& \text{10}& \text{14}\end{array}\right]$       (XY)' = $\left[\begin{array}{ccc}\text{0}& \text{1}& \text{2}\\ \text{0}& \text{5}& \text{10}\\ \text{0}& \text{7}& \text{14}\end{array}\right]$ RHS: Y' X' X =  $\left[\begin{array}{c}\text{0}\\ \text{1}\\ \text{2}\end{array}\right]$  X' =   Y =   Y' =    Y'X' =   = $\left[\begin{array}{ccc}\text{0}& \text{1}& \text{2}\\ \text{0}& \text{5}& \text{10}\\ \text{0}& \text{7}& \text{14}\end{array}\right]$           (XY)' =   $\left[\begin{array}{ccc}\text{0}& \text{1}& \text{2}\\ \text{0}& \text{5}& \text{10}\\ \text{0}& \text{7}& \text{14}\end{array}\right]$     = Y'X' Symmetric matrix : A square matrix A = [aij] is said to be symmetric if A' = A, that is, [aij] = [aij] for all possible values of i and j. Ex: X =  $\left[\begin{array}{ccc}\text{0}& \text{1}& \text{2}\\ \text{0}& \text{5}& \text{10}\\ \text{0}& \text{7}& \text{14}\end{array}\right]$      $\underset{\to }{\text{Transpose operation}}$          X' = X , X is a symmetric matrix. Skew symmetric matrix: A square matrix A = [aij] is said to be skew symmetric if A' = - A that is, [aij] = - [aij] for all possible values of i and j. Ex: Y =                  $\underset{\to }{\text{Transpose operation}}$                                                 Y' = -Y ⇒ Y is a skew symmetric matrix. Theorem 1: For any square matrix A with real number entries, A + A′ is a symmetric matrix and A - A′ is a skew symmetric matrix. Proof1): A is a square matrix with real numbers as its elements. Given, A = [aij]mxn Let X = A + A' If X is a symmetric matrix, X = X' X' = (A + A')' = (A)' + (A')'  '.' [(A + B)' = A' + B'] = A' + A = A' + A = X X is a symmetric matrix or (A' + A) is a symmetric matrix Proof 2): A - A' is a skew symmetric matrix. Let Y = A - A' Skew symmetric matrix: Y' = - Y Y' = (A - A')' = (A)' - (A')' = A' - A = - (A - A') = -Y Y is a skew symmetric matrix or (A - A') is a skew symmetric matrix. Ex: A =        A' =  $\left[\begin{array}{ccc}\text{2}& \text{4}& \text{1}\\ \text{3}& \text{3}& \text{0}\\ \text{5}& \text{2}& \text{7}\end{array}\right]$ To verify that A + A' is symmetric. A + A' =          (A + A' ) ' =     $\underset{\to }{\text{Transpose operation}}$      = A + A'    To verify that A - A' is a skew symmetric matrix. A - A' = $\left[\begin{array}{ccc}\text{0}& \text{-1}& \text{4}\\ \text{1}& \text{0}& \text{2}\\ \text{-4}& \text{-2}& \text{0}\end{array}\right]$            $\left[\begin{array}{ccc}\text{0}& \text{-1}& \text{4}\\ \text{1}& \text{0}& \text{2}\\ \text{-4}& \text{-2}& \text{0}\end{array}\right]$        $\underset{\to }{\text{Transpose operation}}$       $\left[\begin{array}{ccc}\text{0}& \text{1}& \text{4}\\ \text{-1}& \text{0}& \text{-2}\\ \text{-4}& \text{2}& \text{0}\end{array}\right]$           $\left[\begin{array}{ccc}\text{0}& \text{1}& \text{4}\\ \text{-1}& \text{0}& \text{-2}\\ \text{-4}& \text{2}& \text{0}\end{array}\right]$   = (-1)  = -(A - A' )       Theorem 2: Any square matrix can be expressed as the sum of a symmetric and a skew symmetric matrix. Proof: Let A = [aij]mxm We know that A + A' : Symmetric matrix                       A - A' : Skew Symmetric matrix = ½(A + A' ) + ½ (A - A') = (½A + ½A') + (½A - ½A') Using kA = k(A) = ½A + ½A = A.

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