Question

Let A and B be any two $$3\times 3$$ matrices. If A is symmetric and B is skew symmetric, then the matrix $$AB-BA$$ is:
A) Skew symmetric
B) Symmetric
C) Neither symmetric nor skew symmetric
D) I or –I, where I is the identity matrix

Answer 1

Hint: Here we will find the transpose of the given matrix and then use the concept of symmetric and skew symmetric matrix i.e.
If a matrix X is symmetric then $${\left(X\right)}^T=X$$
If a matrix Y is skew symmetric then $${\left(Y\right)}^T=-Y$$

Complete step-by-step answer:
The given matrix is:-
$$AB-BA$$
Taking transpose of the above matrix we get:-
$${\left(AB-BA\right)}^T={\left(AB\right)}^T-{\left(BA\right)}^T$$
Now we know that:
$${\left(XY\right)}^T=Y^T{X}^T$$
Hence, applying this property we get:-
$${\left(AB-BA\right)}^T=B^T{A}^T-A^T{B}^T$$……………………………………….(1)
Now since A is symmetric matrix
Therefore, $$A^T=A$$
Since B is skew symmetric matrix
Therefore,
$$B^T=-B$$
Hence substituting the values in equation 1 we get:-
$${\left(AB-BA\right)}^T=\left(-B\right)\left(A\right)-\left(A\right)\left(-B\right)$$
Simplifying it further we get:-
$$\begin{array}{c}{\left(AB-BA\right)}^T=-BA+AB\\ \Rightarrow {\left(AB-BA\right)}^T=AB-BA\end{array}$$
Hence, $$AB-BA$$ is a symmetric matrix.

Therefore, option A is the correct option.

Note: Students should note that only the square matrices can be symmetric or skew-symmetric form.
Also, matrix A is said to be symmetric if the transpose of matrix A is equal to matrix A and the upper triangular matrix is equal to the lower triangular matrix
$$\left[\begin{array}{ccc}a& b& c\\ b& d& f\\ c& f& e\end{array}\right]$$
Matrix A is said to be skew-symmetric if the transpose of matrix A is equal to negative of matrix A and the upper triangular matrix is negative to the lower triangular matrix or vice-versa.
$$\left[\begin{array}{ccc}a& b& c\\ -b& d& f\\ -c& -f& e\end{array}\right]$$