Question

If A and B are symmetric matrices, prove that AB + BA is symmetric.

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Answer 1

Answer:

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Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

↝ A and B are symmetric matrices.

We know,

A square matrix A is said to be symmetric iff A' = A.

So, using this definition,

$\rm :\longmapsto\:\boxed{ \tt{ \: A' = A \: }}$

and

$\rm :\longmapsto\:\boxed{ \tt{ \: B' = B \: }}$

Now, We have to prove that AB + BA is symmetric.

So, we have to prove that,

$\purple{\boxed{ \tt{ \: {(AB + BA)}' = AB + BA \: }}}$

So, Consider

$\rm :\longmapsto\:(AB + BA)'$

$\rm \: = \: (AB)' + (BA)'$

$\rm \: = \: B'A' + A'B'$

$\rm \: = \: BA + AB$

$\rm \: = \: AB + BA$

Hence,

$\rm \implies\:\boxed{ \bf{ \: (AB + BA)' = AB + BA \: }}$

$\bf\implies \:AB + BA \: is \: symmetric$

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Explore more :-

Skew Symmetric Matrix :- A square matrix A is said to be skew symmetric iff A' = - A.

The main diagonal elements of skew symmetric matrix are 0.

The determinant of skew symmetric matrix of odd order is 0.

Properties of transpose :-

$\boxed{ \tt{ \: (A')' = A \: }}$

$\boxed{ \tt{ \: (A + B)' = A' + B' \: }}$

$\boxed{ \tt{ \: (A - B)' = A' - B' \: }}$

$\boxed{ \tt{ \: (AB)' = B'A' \: }}$

$\boxed{ \tt{ \: (kA)' = kA' \: }}$