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# Show that the matrix $B'AB$ is symmetric or skew symmetric according as A is symmetric or skew symmetric.

Toolbox:
• A square matrix A=[a$_{ij}$] is said to be symmetric if A'=A that is $[a_{ij}]=[a_{ji}]$ for all possible value of i and j.
• A square matrix A=[a$_{ij}$] is said to be skew symmetric if A'=-A that is $[a_{ij}]= -[a_{ji}]$ for all possible value of i and j.
Step 1: Let A be symmetric matrix
A'=A
Now consider B'AB and take transpose of it.
(B'AB)'=(B'(AB))'
$\qquad$=(AB)'(B')'
$\qquad$=(B'A')B (From the property of transpose of a matrix we have (AB)'=B'A')
Replace A'=A in the above equation, we get
$\qquad$=B'AB
A matrix is said to be symmetric if A=A'
Thus (B'AB)= B'AB
Hence B'AB is a symmetric matrix
Step 2: Let A be skew symmetric matrix
A'=-A
Now consider B'AB and take transpose of it.
(B'(AB))'=(AB)'B'
$\qquad$=(B'A')B(From the property of transpose of a matrix we have (AB)'=B'A')
Replace A=-A in the above equation we get
$\qquad$=B'(-A)B
$\qquad$=-B'AB
A matrix is said to be skew symmetric if A'= - A
(B'AB) = -B'AB
B'AB is skew symmetric matrix
edited Mar 19, 2013