$Assertion\ (A):$ For any symmetric matrix $A$, $B'AB$ is a skew-symmetric matrix. $Reason\ (R):$ A square matrix $P$ is skew-symmetric if $P' = -P$. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.

Assertion(A): For any symmetric matrix A, B'AB is a skew-symmetric matrix. Reason(R): A square matrix P is ske

Question

Assertion\ (A):

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The assertion claims $B'AB$ is skew-symmetric for symmetric $A$ , but this is false — it's actually symmetric. The reason correctly defines skew-symmetry. Answer: (D)

The heart of this problem lies in understanding what happens when you sandwich a symmetric matrix between a matrix and its transpose. Let's first be clear about what we're working with.

A symmetric matrix $A$ satisfies $A' = A$ . A skew-symmetric matrix $P$ satisfies $P' = -P$ . The reason (R) gives the correct definition of skew-symmetry, so we know immediately that R is true.

Now for the assertion: does $B'AB$ turn out to be skew-symmetric when $A$ is symmetric?

The key insight is to examine the transpose of $B'AB$ and see what we get. The transpose of a product reverses the order and transposes each factor.

Testing the Assertion

Start with the expression $B'AB$ and take its transpose:

(B'AB)' = B'A'(B')'

using the reversal property $(XYZ)' = Z'Y'X'$ .

Simplify using the transpose properties:
- $(B')' = B$ (transpose of transpose returns the original)
- $A' = A$ (since $A$ is symmetric)
Therefore:

(B'AB)' = B'A'B = B'AB

What does this tell us? We've shown that $(B'AB)' = B'AB$ , which means $B'AB$ is symmetric, not skew-symmetric.

Watch out

A common mistake is confusing the conditions: for skew-symmetry we need $(B'AB)' = -B'AB$ , but we actually get $(B'AB)' = +B'AB$ .

Verify the logic:
- For $B'AB$ to be skew-symmetric, we would need $(B'AB)' = -B'AB$
- But we proved $(B'AB)' = B'AB$
- These are contradictory unless $B'AB = 0$ (the zero matrix) …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 40 on this concept.

CBSE 2026Set 65/1/11 markMCQ
$Q. If A and B are skew symmetric matrices of same order, then which of the following matrices is also skew symmetric ? 1 (A) AB (B) AB + BA (C) (A + B) 2 (D) A - B$
›Reveal solutionSolution
A skew-symmetric matrix satisfies $A^T = -A$ . For two skew-symmetric matrices $A$ and $B$ of the same order, the combination $AB + BA$ is symmetric, not skew-symmetric, while $A - B$ remains skew-symmetric. The correct option is (D).

The key to this problem lies in the definition of a skew-symmetric matrix: a square matrix $M$ is skew-symmetric if its transpose equals its negative, i.e., $M^T = -M$ . For any two skew-symmetric matrices $A$ and $B$ of the same order, we have $A^T = -A$ and $B^T = -B$ .

When we combine $A$ and $B$ through operations like addition, multiplication, or squaring, the transpose of the result will involve the transposes of $A$ and $B$ in a specific way. The property $(XY)^T = Y^T X^T$ is crucial here — it reverses the order of multiplication. So, to check if a given expression is skew-symmetric, we compute its transpose and see if it equals the negative of the original expression.

Let’s examine each option step by step.

Option (A): $AB$

Compute $(AB)^T = B^T A^T = (-B)(-A) = BA$ .

For $AB$ to be skew-symmetric, we would need $(AB)^T = -AB$ , i.e., $BA = -AB$ . But this is not generally true for arbitrary skew-symmetric matrices — it would require $A$ and $B$ to anticommute, which is not guaranteed. So $AB$ is not necessarily skew-symmetric.

Option (B): $AB + BA$

Compute $(AB + BA)^T = (AB)^T + (BA)^T = B^T A^T + A^T B^T = (-B)(-A) + (-A)(-B) = BA + AB = AB + BA$ .

The transpose equals the original expression itself, meaning $AB + BA$ is symmetric, not skew-symmetric. So this is not the answer.

Option (C): $(A + B)^2$

First, note that $(A + B)^2 = A^2 + AB + BA + B^2$ .

Compute its transpose: $[(A + B)^2]^T = [(A + B)(A + B)]^T = (A + B)^T (A + B)^T = (A^T + B^T)(A^T + B^T) = (-A - B)(-A - B) = (A + B)^2$ .

So $(A + B)^2$ is symmetric, not skew-symmetric. …
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CBSE 2026Set 65/2/11 markMCQ
$A = \begin{bmatrix} 1 & a & b \\ -1 & 2 & c \\ 0 & 5 & 3 \end{bmatrix}$
›Reveal solutionSolution
A symmetric matrix equals its transpose, so corresponding off-diagonal entries must match. Equating $A = A^T$ gives $a = -1$ , $b = 0$ , $c = 5$ , hence $3a + b + c = 2$ .

A matrix is symmetric when it mirrors itself across the main diagonal — in other words, when $A = A^T$ . This means the entry in row $i$ , column $j$ must equal the entry in row $j$ , column $i$ for all positions. The diagonal entries stay put, but every pair of off-diagonal entries must be equal.

For the given matrix, the transpose swaps rows and columns:
$A^T = \begin{bmatrix} 1 & -1 & 0 \\ a & 2 & 5 \\ b & c & 3 \end{bmatrix}$
Now we impose the symmetry condition $A = A^T$ by equating corresponding entries.

Compare the $(1,2)$ and $(2,1)$ positions:

From $A$ : the $(1,2)$ entry is $a$ .

From $A^T$ : the $(1,2)$ entry is $-1$ .

Therefore $a = -1$ .

Compare the $(1,3)$ and $(3,1)$ positions:

From $A$ : the $(1,3)$ entry is $b$ .

From $A^T$ : the $(1,3)$ entry is $0$ .

Therefore $b = 0$ . …
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CBSE 2026Set 65/3/11 markMCQ
$A$
›Reveal solutionSolution
We classify the given expression by finding its transpose. Using the properties of transpose and the definitions of skew-symmetric matrices, we find that the transpose of $AB' + BA'$ is equal to the original expression itself. Thus, $AB' + BA'$ is a symmetric matrix.

To determine if a matrix expression is symmetric or skew-symmetric, the fundamental approach is to calculate its transpose. The classification depends on how the transpose relates to the original matrix.

A matrix $M$ is:

Symmetric if $M^T = M$ . This means the matrix is equal to its own transpose.

Skew-symmetric if $M^T = -M$ . This means the matrix is the negative of its own transpose.

We will use the following properties of matrix transpose:

$(P+Q)^T = P^T + Q^T$ (Transpose of a sum is the sum of transposes)

$(PQ)^T = Q^T P^T$ (Transpose of a product is the product of transposes in reverse order)

$(P^T)^T = P$ (Transpose of a transpose is the original matrix)

$(kP)^T = kP^T$ (Transpose of a scalar multiple is the scalar multiple of the transpose)

Let's apply these concepts to the given problem.

Understand the given information:

We are given that $A$ and $B$ are skew-symmetric matrices of the same order.

By definition, this means:

$A^T = -A$

$B^T = -B$

(Note: $A'$ and $B'$ are common notations for $A^T$ and $B^T$ respectively.)

Identify the expression to classify:

We need to classify the matrix $X = AB' + BA'$ .

Using the standard notation for transpose, this is $X = AB^T + BA^T$ .

Calculate the transpose of the expression:

To classify $X$ , we must find its transpose, $X^T$ .

$X^T = (AB^T + BA^T)^T$

Using the property $(P+Q)^T = P^T + Q^T$ :

$X^T = (AB^T)^T + (BA^T)^T$

Using the property $(PQ)^T = Q^T P^T$ :

$X^T = (B^T)^T A^T + (A^T)^T B^T$

Using the property $(P^T)^T = P$ :

$X^T = B A^T + A B^T$ …
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CBSE 2026Set A1 markMCQ
$A = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}$
(a) $\pi$
(b) $\frac{\pi}{3}$
(c) $\frac{3\pi}{2}$
(d) $\frac{\pi}{6}$
›Reveal solutionSolution

$A + A' = 2\cos\alpha\,I$ ; setting this equal to $I$ gives $\cos\alpha = \tfrac12$ , so $\alpha = \tfrac{\pi}{3}$ .

Here $A = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}$ , so its transpose is $A' = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$ .

$A + A' = \begin{bmatrix} 2\cos\alpha & 0 \\ 0 & 2\cos\alpha \end{bmatrix}.$ …

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CBSE 2026Set A1 markMCQ
$A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} \Rightarrow A + A' =$
(a) $\begin{bmatrix} 6 & -3 \\ -3 & -2 \end{bmatrix}$
(b) $\begin{bmatrix} 6 & 3 \\ 3 & 2 \end{bmatrix}$
(c) $\begin{bmatrix} 6 & 3 \\ -3 & -2 \end{bmatrix}$
(d) $\begin{bmatrix} 6 & 3 \\ -3 & 2 \end{bmatrix}$
›Reveal solutionSolution

Add $A$ and $A'$ term by term.

With $A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$ , the transpose is $A' = \begin{bmatrix} 3 & 1 \\ -4 & -1 \end{bmatrix}$ . …

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CBSE 2026Set ANNUAL1 markMCQ
$A = \begin{bmatrix} 4 & x+2 \\ 2x-3 & x+1 \end{bmatrix}$
(a) 3
(b) 4
(c) 5
(d) None of these
›Reveal solutionSolution

A matrix $A$ is symmetric when $A = A^T$ , so its $(1,2)$ and $(2,1)$ entries must be equal.

Given $A = \begin{bmatrix} 4 & x+2 \\ 2x-3 & x+1 \end{bmatrix}$ is symmetric.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. A matrix A is said to be symmetric matrix if$
(a) A' = A
(b) A' = −A
(c) det A = 0
(d) det A ≠ 0
›Reveal solutionSolution

Symmetry means the matrix is unchanged by transposing — entries mirror across the leading diagonal.

A square matrix $A=[a_{ij}]$ is called symmetric if every entry equals its mirror-image entry across the main diagonal, i.e. $a_{ij}=a_{ji}$ for all $i,j$ . This condition is exactly captured by

$A' = A$

where $A'$ (or $A^T$ ) is the transpose of $A$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$A$
(a) Skew symmetric matrix
(b) Symmetric matrix
(c) Zero matrix
(d) Identity matrix
›Reveal solutionSolution

$AB-BA$ is skew-symmetric.

Given $A^T=A$ , $B^T=B$ . Then $(AB-BA)^T=(AB)^T-(BA)^T=B^TA^T-A^TB^T=BA-AB=-(AB-BA)$ .

…

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CBSE 2026Set ANNUAL1 mark
$B'AB$
›Reveal solutionSolution

Taking the transpose and using reversal law plus $A'=A$ returns $B'AB$ itself.

Given $A$ is symmetric, so $A'=A$ . Using $(XYZ)'=Z'Y'X'$ and $(B')'=B$ :

$(B'AB)'=B'\,A'\,(B')'=B'\,A'\,B=B'\,A\,B.$

…

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CBSE 2026Set SEM31 markMCQ
$a$
(a) $a = 3,\; b = 2,\; c = 4$
(b) $a = 2,\; b = 3,\; c = 1$
(c) $a = 1,\; b = 2,\; c = 3$
(d) $a = 0,\; b = 1,\; c = 3$
›Reveal solutionSolution
A symmetric matrix satisfies $a_{ij}=a_{ji}$ ; equate the three off-diagonal pairs and solve.

Recognising a symmetric matrix ( $A=A^T$ ) is a CBSE/NCERT Class 12 matrices concept.

For the matrix to be symmetric the $(i,j)$ and $(j,i)$ entries must match:

$(1,2)=(2,1)$ : $a+b-c = 1$ .

$(1,3)=(3,1)$ : $a+b+c = 9$ .

$(2,3)=(3,2)$ : $a-b+c = 5$ . …
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CBSE 2025Set 65/2/11 markMCQ
$A = \begin{bmatrix} 1 & 12 & 4y \\ 6x & 5 & 2x \\ 8x & 4 & 6 \end{bmatrix}$
›Reveal solutionSolution

A symmetric matrix satisfies $A = A^T$ , so corresponding off-diagonal entries are equal. Equating $a_{12} = a_{21}$ and $a_{13} = a_{31}$ gives $x = 2$ and $y = 4$ , so $(2x + y) = \boxed{8}$ .

A matrix is symmetric when it equals its own transpose. This means that if you flip the matrix across its main diagonal, you get the same matrix back. In practical terms, the entry in row $i$ , column $j$ must equal the entry in row $j$ , column $i$ for all positions.

The transpose of $A$ swaps rows and columns:
$A^T = \begin{bmatrix} 1 & 6x & 8x \\ 12 & 5 & 4 \\ 4y & 2x & 6 \end{bmatrix}$
For $A$ to be symmetric, we need $A = A^T$ . The diagonal entries $(1, 5, 6)$ already match themselves, so we focus on the off-diagonal pairs.

1. Compare the $(1,2)$ and $(2,1)$ entries:

From $A$ : the $(1,2)$ entry is $12$ .

From $A^T$ : the $(1,2)$ entry is $6x$ .

Setting them equal:
$6x = 12$ $x = 2$
2. Compare the $(1,3)$ and $(3,1)$ entries:

From $A$ : the $(1,3)$ entry is $4y$ .

From $A^T$ : the $(1,3)$ entry is $8x$ .

Setting them equal:
$4y = 8x$
Substituting $x = 2$ : …
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CBSE 2025Set 65/2/11 markMCQ
$Q. Which of the following can be both a symmetric and skew-symmetric matrix? (A) Unit Matrix (B) Diagonal Matrix (C) Null Matrix (D) Row Matrix$
›Reveal solutionSolution
A matrix that is both symmetric and skew-symmetric must satisfy $A = A^T$ and $A = -A^T$ , which forces every entry to be zero. The only such matrix is the Null Matrix, so option (C) is correct.

Why This Question Tests a Core Definition

Many students memorise the separate definitions of symmetric and skew-symmetric matrices but never pause to ask: Can a matrix satisfy both at once? That’s exactly what this problem does — it forces you to combine the two conditions algebraically and see what survives.

Let’s recall:

A matrix $A$ is symmetric if $A = A^T$ .

A matrix $A$ is skew-symmetric if $A = -A^T$ .

If a matrix is both, then both equalities hold simultaneously. That gives us a simple but powerful equation.

Step-by-Step Reasoning

1. Write down both conditions together.

If $A$ is symmetric:
$A = A^T$
If $A$ is also skew-symmetric:
$A = -A^T$
Since both are true, we can equate the right-hand sides:
$A^T = -A^T$
2. Solve the equation for $A^T$ .

Add $A^T$ to both sides:
$A^T + A^T = 0 \quad\Rightarrow\quad 2A^T = 0$
Dividing by 2:
$A^T = 0$
The zero matrix. And since $A = A^T$ , we also have $A = 0$ .

Watch out

A common mistake is to think a diagonal matrix or a unit matrix could work. Check: the unit matrix $I$ satisfies $I = I^T$ (symmetric), but $I = -I^T$ would require $I = -I$ , which is false unless every entry is zero. So only the null matrix survives.

3. Interpret the result.

The only matrix that is both symmetric and skew-symmetric is the null matrix (all entries zero). No other matrix — unit, diagonal, or row — can satisfy both conditions unless it is identically zero. …
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Concept understanding — Symmetric And Skew Symmetric Matrices

Symmetric and Skew-Symmetric Matrices

Symmetric matrix

Skew-symmetric matrix

Key facts

Testing the Assertion

Why This Question Tests a Core Definition

Step-by-Step Reasoning

Testing the Assertion

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