If A, B are symmetric matrices of the same order then AB - BA is a

Question

If A, B are symmetric matrices of the same order then AB - BA is a (A) Skew symmetric matrix (B) Symmetric matrix (C) Zero matrix (D) Identity matrix.

Tardigrade Admin · Accepted Answer

A and B are symmetric matrices ⇒ A' = A ⇒ B' = B Now (AB - BA)' = (AB)' - (BA)' = B'A' - A'B' = BA - AB ∵ B' = B, A' = A = - (AB - BA) ⇒ (AB - BA) is a skew symmetric matrix.

Q. If A, B are symmetric matrices of the same order then AB - BA is a

Skew symmetric matrix

Symmetric matrix

Zero matrix

Identity matrix.

A and B are symmetric matrices $\Rightarrow A^{'} = A$
$\Rightarrow B^{'} = B$
Now $(A B - B A)^{'} = (A B)^{'} - (B A)^{'}$
$= B^{'} A^{'} - A^{'} B^{'} = B A - A B$
$∵ B^{'} = B, A^{'} = A = - (A B - B A)$
$\Rightarrow (A B - B A)$ is a skew symmetric matrix.

Q. If A, B are symmetric matrices of the same order then AB - BA is a

Skew symmetric matrix

Symmetric matrix

Zero matrix

Identity matrix.

A and B are symmetric matrices ⇒A′=A ⇒B′=B Now (AB−BA)′=(AB)′−(BA)′ =B′A′−A′B′=BA−AB ∵B′=B,A′=A=−(AB−BA) ⇒(AB−BA) is a skew symmetric matrix.

A and B are symmetric matrices $\Rightarrow A^{'} = A$
$\Rightarrow B^{'} = B$
Now $(A B - B A)^{'} = (A B)^{'} - (B A)^{'}$
$= B^{'} A^{'} - A^{'} B^{'} = B A - A B$
$∵ B^{'} = B, A^{'} = A = - (A B - B A)$
$\Rightarrow (A B - B A)$ is a skew symmetric matrix.