Respuesta :
(a)Symmetric matrix -
Let matrix A be of order 3×3 =
[tex]\left[\begin{array}{ccc}1&3&8\\3&8&-4\\8&-4&6\end{array}\right][/tex]
So, the transpose of A will be -
A' =
[tex]\left[\begin{array}{ccc}1&3&8\\3&8&-4\\8&-4&6\end{array}\right][/tex]
So, we can say that A=A'. So, it is symmetric.
(b) Skew - symmetric matrix -
Let us consider a matrix B =
[tex]\left[\begin{array}{ccc}0&-6&4\\-6&0&7\\4&7&0\end{array}\right][/tex]
matrix B is said to be skew-symmetric if aij =−aji.
(B' =−B)
B' =
[tex]\left[\begin{array}{ccc}0&6&-4\\6&0&-7\\-4&-7&0\end{array}\right][/tex]
(c) A + A′ is a symmetric matrix for any square matrix A with real number entries, whereas A - A′ is a skew-symmetric matrix.
Proof: If B = A + A', then B ′ = (A + A ) ′
(A + B)′ = A′ + B′, thus = A′ + (A′)′
(as (A′)′ =A) = A′ + A
= B (because B + A = B + A′)
B = A+A′ is a symmetric matrix as a result.
Let C now equal A-A′.
Why does C' equal (A-A′)′=(A′)′?
= A′ - A (Why?)
=- (A - A′) = - C
A - A′ is a skew-symmetric matrix as a result.
To learn more about matrices from given link
https://brainly.in/question/14760668
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