Q.If $A=\begin{bmatrix}0 & 1 & c\\ -1 & a & -b\\ 2 & 3 & 0\end{bmatrix}$ is a skew-symmetric matrix then the value of $a+b+c$ =
(A) $1$
(B) $2$
(C) $3$
(D) $4$

Showing the 12 most recent of 40 on this concept.

1. 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.

   1. 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.

   2. 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.

   3. 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.

      Tip

      A quick way: the square of any matrix is symmetric if the matrix itself is skew-symmetric? Actually, for any matrix $M$, $(M^2)^T = (M^T)^2$. Here $M = A+B$, and $M^T = -M$, so $(M^2)^T = (-M)^2 = M^2$, confirming symmetry. So it’s never skew-symmetric unless it’s zero.

   4. Option (D): $A - B$

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

      This exactly matches the condition for skew-symmetry. So $A - B$ is skew-symmetric.

   Watch out

   A common mistake is to think that $AB$ is skew-symmetric because $A$ and $B$ individually are. But the transpose of a product reverses order, and unless $A$ and $B$ commute in a special way (anticommute), $AB$ is not skew-symmetric. Always check the transpose carefully.

   ✓Final answer

   The correct option is (D), since $A - B$ is skew-symmetric.

2. CBSE 2026Set 65/2/11 markMCQ
   Q.If $A = \begin{bmatrix} 1 & a & b \\ -1 & 2 & c \\ 0 & 5 & 3 \end{bmatrix}$ is a symmetric matrix, then the value of $3a + b + c$ is (A) 2 (B) 6 (C) 4 (D) 0

   ›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.

   1. 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 $a = -1$.

   2. 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 $b = 0$.

   3. Compare the $(2,3)$ and $(3,2)$ positions:

      From $A$: the $(2,3)$ entry is $c$.

      From $A^T$: the $(2,3)$ entry is $5$.

      Therefore $c = 5$.

   Note

   The diagonal entries $(1,1) = 1$, $(2,2) = 2$, $(3,3) = 3$ are already equal in $A$ and $A^T$, so they impose no constraints.

   With $a = -1$, $b = 0$, and $c = 5$, we compute:

   $3a + b + c = 3(-1) + 0 + 5 = -3 + 5 = 2$

   ✓Final answer

   The value of $3a + b + c$ is $\boxed{2}$, so the correct option is (A).

3. CBSE 2026Set 65/3/11 markMCQ
   Q.If $A$ and $B$ are skew-symmetric matrices of the same order, then $AB' + BA'$ is a/an: (A) symmetric matrix (B) skew-symmetric matrix (C) null matrix (D) identity matrix

   ›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.

   1. 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.)

   2. Identify the expression to classify:

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

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

   3. 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 $(P^T)^T = P$:

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

   4. Substitute the given conditions into the transposed expression:

      Now, we use the fact that $A^T = -A$ and $B^T = -B$:

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

      $X^T = -BA - AB$

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

   5. Substitute the given conditions into the original expression:

      Let's also express the original matrix $X$ in terms of $A$ and $B$ using the given conditions:

      $X = AB^T + BA^T$

      $X = A(-B) + B(-A)$

      $X = -AB - BA$

      $X = -(AB + BA)$

   6. Compare the transposed expression with the original expression:

      We found:

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

      $X = -(AB + BA)$

      Since matrix addition is commutative, $BA + AB = AB + BA$.

      Therefore, $X^T = -(AB + BA) = X$.

      Since $X^T = X$, the matrix $AB' + BA'$ is a symmetric matrix.

   ✓Final answer

   The expression $AB' + BA'$ is a (A) symmetric matrix.

4. CBSE 2026Set A1 markMCQ
   Q.If $A = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}$ and $A + A' = I$ then $\alpha =$

   (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 $A' = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$.

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

   For this to equal $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ we need $2\cos\alpha = 1$, i.e. $\cos\alpha = \tfrac12$, giving $\alpha = \tfrac{\pi}{3}$.

   ✓Final answer

   (b) $\frac{\pi}{3}$.

5. CBSE 2026Set A1 markMCQ
   Q.$A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} \Rightarrow A + A' =$

   (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 $A' = \begin{bmatrix} 3 & 1 \\ -4 & -1 \end{bmatrix}$.

   $A + A' = \begin{bmatrix} 3+3 & -4+1 \\ 1+(-4) & -1+(-1) \end{bmatrix} = \begin{bmatrix} 6 & -3 \\ -3 & -2 \end{bmatrix}.$

   ✓Final answer

   (a) $\begin{bmatrix} 6 & -3 \\ -3 & -2 \end{bmatrix}$.

6. CBSE 2026Set ANNUAL1 markMCQ
   Q.If $A = \begin{bmatrix} 4 & x+2 \\ 2x-3 & x+1 \end{bmatrix}$ is symmetric matrix then $x =$

   (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.

   For a $2\times2$ matrix to be symmetric, the off-diagonal entries must match:

   $x+2 = 2x-3$

   $2+3 = 2x-x$

   $x = 5$.

   ✓Final answer

   (c) $5$.

7. 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 $A$.

   (By contrast, $A'=-A$ defines a skew-symmetric matrix, and $\det A=0$ merely means $A$ is singular — unrelated to symmetry.)

   ✓Final answer

   $A' = A$. (Option a)

8. CBSE 2026Set ANNUAL1 markMCQ
   Q.If $A$ and $B$ are symmetric matrices of same order, then $AB - BA$ is 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 $(AB-BA)^T=(AB)^T-(BA)^T=B^TA^T-A^TB^T=BA-AB=-(AB-BA)$.

   A matrix equal to the negative of its transpose is skew-symmetric.

   ✓Final answer

   (a) Skew symmetric matrix.

9. CBSE 2026Set ANNUAL1 mark
   Q.Show that the matrix $B'AB$ is symmetric if $A$ is symmetric.

   ›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 $(B')'=B$:

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

   Since $(B'AB)'=B'AB$, the matrix $B'AB$ is symmetric.

   ✓Final answer

   $(B'AB)'=B'AB$; hence $B'AB$ is symmetric when $A$ is symmetric.

10. CBSE 2026Set SEM31 markMCQ
    Q.The values of $a$, $b$ and $c$ for which the matrix $\begin{bmatrix} 1 & a+b-c & a+b+c \\ 1 & 2 & a-b+c \\ 9 & 5 & 3 \end{bmatrix}$ will be symmetric are

    (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:

    • $(1,2)=(2,1)$: $a+b-c = 1$.
    • $(1,3)=(3,1)$: $a+b+c = 9$.
    • $(2,3)=(3,2)$: $a-b+c = 5$.

    Subtract the first from the second: $2c = 8 \Rightarrow c = 4$, and $a+b = 5$.

    The third gives $a - b + 4 = 5 \Rightarrow a - b = 1$.

    Solving $a+b=5,\ a-b=1$: $a = 3,\ b = 2$.

    ✓Final answer

    $a=3,\ b=2,\ c=4$ — option (a).

11. CBSE 2025Set 65/2/11 markMCQ
    Q.If $A = \begin{bmatrix} 1 & 12 & 4y \\ 6x & 5 & 2x \\ 8x & 4 & 6 \end{bmatrix}$ is a symmetric matrix, then $(2x + y)$ is: (A) $-8$ (B) $0$ (C) $6$ (D) $8$

    ›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 $x = 2$:

    $4y = 8(2) = 16$

    $y = 4$

    3. Verify the $(2,3)$ and $(3,2)$ entries:

    From $A$: the $(2,3)$ entry is $2x = 2(2) = 4$.

    From $A$: the $(3,2)$ entry is $4$.

    These match, confirming our solution is consistent.

    Tip

    In a symmetric matrix, you only need to check the upper (or lower) triangular part against its mirror image — the diagonal always matches itself.

    4. Calculate $(2x + y)$:

    $2x + y = 2(2) + 4 = 4 + 4 = 8$

    ✓Final answer

    The correct option is (D) $8$.

12. 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.

    Tip

    You can also see this entry-wise: for any $i, j$, symmetry says $a_{ij} = a_{ji}$, and skew-symmetry says $a_{ij} = -a_{ji}$. Combining gives $a_{ij} = -a_{ij}$, so $2a_{ij}=0$, hence $a_{ij}=0$ for all $i, j$. Every entry must be zero.

    4. Check the options quickly.

    • (A) Unit Matrix: symmetric but not skew-symmetric (since $I \neq -I$).
    • (B) Diagonal Matrix: symmetric, but a non-zero diagonal entry $d$ would need $d = -d$, so only the zero diagonal matrix works — which is just the null matrix.
    • (C) Null Matrix: satisfies both trivially.
    • (D) Row Matrix: a row matrix can be symmetric only if it’s $1\times1$, and then the same logic forces its single entry to be zero.

    Only the null matrix fits.

    $A \text{ is both symmetric and skew-symmetric} \iff A = 0$

    ---

    ✓Final answer

    The correct option is (C) Null Matrix.