1 markMCQSection A

If A and B are skew symmetric matrices of same order, then which of the following matrices is also skew symmetric?

Q: If A and B are skew symmetric matrices of same order, then which of the following matrices is also skew symmetric?

**Correct option: (D) A - B** Since $A' = -A \text{ and } B' = -B, \text{ we get } (A - B)' = A' - B' = -A - (-B) = -(A - B)$. $\text{This means } A - B \text{ satisfies the skew-symmetric condition } M' = -M$.

Matrices

Skew-Symmetric Matrices

Options

(A)

AB

(B)

AB + BA

(C)

(A + B)^2

(D)

A - B

Official Answer

Correct option: (D) A - B

Since $A' = -A \text{ and } B' = -B, \text{ we get } (A - B)' = A' - B' = -A - (-B) = -(A - B)$ . $\text{This means } A - B \text{ satisfies the skew-symmetric condition } M' = -M$ .

skew symmetric matrixA'=-Asymmetric matrixtranspose reversal lawmatrix combinationsAB and BA

Marking Scheme

11 mark: correct option (D) selected — requires verifying the skew-symmetric condition using transpose reversal rules.

Hint

Use $A' = -A,\ B' = -B$ , and the reversal rule $(XY)' = Y'X'$ to test each option against the skew-symmetric condition $M' = -M$ .

Quick Oral Answer

Using $A' = -A \text{ and } B' = -B$ , $(A-B)' \text{ works out to } -(A-B)$ , satisfying the skew-symmetric condition, so A - B is the answer, option (D).

Analysis & Explanation

This tests the ability to apply the definitions A' = -A, B' = -B and the transpose reversal law (XY)' = Y'X' to combinations of skew-symmetric matrices.

Setup

Given: $A' = -A \text{ and } B' = -B$ (definition of skew-symmetric).

Checking (A - B)

$(A - B)' = A' - B' = (-A) - (-B) = -A + B = -(A - B)$ . This equals -(A-B), so A - B is skew-symmetric. Option (D) is correct.

Why the other options fail

(AB)': $(AB)' = B'A' = (-B)(-A) = BA$ . For AB to be skew-symmetric we would need $BA = -AB$ , which is not true in general (AB and BA are usually different matrices), so AB is not necessarily skew-symmetric.
(AB + BA)': $(AB+BA)' = (AB)' + (BA)' = B'A' + A'B' = BA + AB = AB + BA$ . This equals the original matrix itself, so AB + BA is actually symmetric, not skew-symmetric.
((A+B)²)': Let $C = A + B$ , so $C' = A' + B' = -A - B = -C$ , meaning C is itself skew-symmetric. Then $(C^2)' = (C')^2 = (-C)^2 = C^2$ , so C² is symmetric, not skew-symmetric.

Common Mistakes

1Assuming AB is skew-symmetric just because both A and B individually are, without checking $(AB)' = B'A'$ properly.
2Mistaking $AB + BA$ for skew-symmetric when it is actually symmetric, since $(AB+BA)' \text{ simplifies back to } AB+BA$ itself.
3Forgetting to apply the reversal law $(XY)' = Y'X'$ when computing transposes of products, and instead writing $(AB)' = A'B'$ .

Interesting Facts

Every square matrix can be uniquely written as the sum of a symmetric matrix and a skew-symmetric matrix: $A = \frac{1}{2}(A+A') + \frac{1}{2}(A-A')$ , a decomposition used throughout physics (e.g., splitting stress tensors) and engineering.

All diagonal entries of a skew-symmetric matrix must be zero, since $a_{ii} = -a_{ii} \text{ forces } a_{ii} = 0$ — a quick way to instantly disqualify many matrices from being skew-symmetric.

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Frequently Asked Questions

What is the definition of a skew-symmetric matrix?

A square matrix M is skew-symmetric if its transpose equals its negative, i.e., $M' = -M$ . This forces every diagonal entry to be zero.

Is the product of two skew-symmetric matrices always skew-symmetric?

No. If A and B are skew-symmetric, $(AB)' = B'A' = (-B)(-A) = BA$ , which generally is not equal to -AB unless A and B commute in a special way. So the product AB is not skew-symmetric in general.