Question

If A is a square matrix, then $A-A'$ is a:
A. Diagonal matrix.
B. Skew-symmetric matrix.
C. Symmetric matrix.
D. None of these.

Answer 1

Hint: Remember that (A')' = A and (A + B)' = A' + B'.
A symmetric matrix is a square matrix that is equal to its transpose. Formally, A is symmetric if and only if A = A'.
A skew-symmetric matrix is a square matrix that is equal to its transpose. Formally, A is skew-symmetric if and only if A = - A'.
A diagonal matrix is a matrix in which all the entries except the diagonal are 0.

Complete step-by-step answer:
Let's say that B = A - A', where A is a square matrix.
Taking transpose of both the sides, we get:
⇒ B' = (A - A')'
We know that the transpose of a sum / difference is equal to the sum of the transposes. Therefore:
⇒ B' = A' - (A')'
Using the fact that transpose of a transpose is equal to the original matrix [(A')' = A], we get:
⇒ B' = A' - A
⇒ B' = - B
We know that if the transpose is equal to the negative of a matrix, then it is a skew-symmetric matrix.
Therefore, the correct answer is B. Skew-symmetric matrix.

Note: Because equal matrices have equal dimensions, only square matrices can be symmetric.
The entries of a symmetric matrix are symmetric with respect to the main diagonal. i.e. ${{a}_{ij}}={{a}_{ji}}$ . In case of a skew-symmetric matrix, ${{a}_{ij}}=-{{a}_{ji}}$ .
The determinant of a diagonal matrix is equal to the product of its diagonal elements.