Q.(i) Show that the matrix $A = \begin{bmatrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{bmatrix}$ is a symmetric matrix.

Showing the 12 most recent of 40 on this concept.

1. CBSE 20241 markMCQ
   Q.If the matrix $A = \begin{bmatrix} 0 & 5 & -7 \\ a & 0 & 3 \\ b & -3 & 0 \end{bmatrix}$ is a skew-symmetric matrix, then the values of ‘a’ and ‘b’ are : (A) $a = 5, b = 3$ (B) $a = 5, b = -7$ (C) $a = -5, b = -7$ (D) $a = -5, b = 7$

   ›Reveal solutionSolution

   For a skew-symmetric matrix, $A^T = -A$. Comparing the given matrix with its transpose forces $a = -5$ and $b = 7$, which matches option (D).

   A skew-symmetric matrix is one where the transpose equals the negative of the original matrix. That is, $A^T = -A$. This condition imposes strict relationships between entries: the diagonal must be all zeros (which is already satisfied here), and for any $i \neq j$, the entry at $(i, j)$ must be the negative of the entry at $(j, i)$. In other words, $a_{ij} = -a_{ji}$.

   This is not a property you guess — it is the definition. So the entire solution flows from writing $A^T$, setting it equal to $-A$, and matching corresponding entries.

   Let’s do it step by step.

   1. Write the given matrix $A$

   $A = \begin{bmatrix} 0 & 5 & -7 \\ a & 0 & 3 \\ b & -3 & 0 \end{bmatrix}$

   2. Find $A^T$ — swap rows and columns:

   $A^T = \begin{bmatrix} 0 & a & b \\ 5 & 0 & -3 \\ -7 & 3 & 0 \end{bmatrix}$

   3. Write $-A$ — multiply every entry of $A$ by $-1$:

   $-A = \begin{bmatrix} 0 & -5 & 7 \\ -a & 0 & -3 \\ -b & 3 & 0 \end{bmatrix}$

   4. Set $A^T = -A$ and compare entry by entry.

      The $(1,1)$ entry gives $0 = 0$, which is fine. Similarly, all diagonal entries are automatically $0$ — that’s always true for skew-symmetric matrices.

      Now compare the $(1,2)$ entry:

      From $A^T$: $a$

      From $-A$: $-5$

      So $a = -5$.

      Compare the $(1,3)$ entry:

      From $A^T$: $b$

      From $-A$: $7$

      So $b = 7$. …

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2. CBSE 2023Set 65/3/11 markMCQ
   Q.$A$ and $B$ are skew-symmetric matrices of same order. $AB$ is symmetric, if :

   (a) $AB = O$

   (b) $AB = -BA$

   (c) $AB = BA$

   (d) $BA = O$

   ›Reveal solutionSolution

   For skew-symmetric matrices $A$ and $B$, their product $AB$ is symmetric if and only if they commute: $AB = BA$.

   Understanding Symmetric and Skew-Symmetric Matrices

   Before diving into the product, recall what these properties mean. A matrix $M$ is symmetric if $M^T = M$ (it equals its own transpose), while it's skew-symmetric if $M^T = -M$ (its transpose is its negative).

   The key insight here is that when we transpose a product of matrices, the order reverses: $(AB)^T = B^T A^T$. This reversal is what makes the interaction between skew-symmetric matrices interesting.

   Finding When $AB$ is Symmetric

   For $AB$ to be symmetric, we need $(AB)^T = AB$. Let's use the properties of $A$ and $B$ to see what this requires.

   1. Start with the symmetry condition for $AB$:

      We want $(AB)^T = AB$.

   2. Apply the transpose rule to the left side:

      Using $(AB)^T = B^T A^T$, our condition becomes:

   $B^T A^T = AB$

   3. Use the skew-symmetric property:

      Since $A$ and $B$ are both skew-symmetric, we have $A^T = -A$ and $B^T = -B$. Substituting these:

   $(-B)(-A) = AB$

   $BA = AB$

   4. Interpret the result:

      The condition $BA = AB$ means that $A$ and $B$ must commute. When two skew-symmetric matrices commute, their product is symmetric. …

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3. CBSE 2024Set 65/3/11 markMCQ
   Q.If A and B are two skew symmetric matrices, then $(AB + BA)$ is: (A) a skew symmetric matrix (B) a symmetric matrix (C) a null matrix (D) an identity matrix

   ›Reveal solutionSolution

   When two skew-symmetric matrices are multiplied, the sum $AB + BA$ is always symmetric because its transpose equals itself. The answer is (B).

   Understanding Skew-Symmetric Matrices

   A matrix $M$ is skew-symmetric when $M^T = -M$. This means every element above the diagonal is the negative of its mirror below the diagonal, and all diagonal entries must be zero.

   The key insight here is that matrix transpose reverses the order of multiplication: $(PQ)^T = Q^T P^T$. When we combine this property with the defining property of skew-symmetric matrices, we can determine what happens to expressions like $AB + BA$.

   Step-by-Step Solution

   1. Start with what we know

   Given that $A$ and $B$ are both skew-symmetric:

   $A^T = -A \quad \text{and} \quad B^T = -B$

   2. Take the transpose of the entire expression

   To determine the nature of $AB + BA$, we need to find $(AB + BA)^T$ and see how it relates to the original:

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

   3. Apply the transpose reversal rule

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

   $(AB)^T = B^T A^T \quad \text{and} \quad (BA)^T = A^T B^T$

   So:

   $(AB + BA)^T = B^T A^T + A^T B^T$

   4. Substitute the skew-symmetric property

   Since $A^T = -A$ and $B^T = -B$:

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

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

   5. Recognize the result

   Notice that $BA + AB = AB + BA$ (addition is commutative). Therefore:

   $(AB + BA)^T = AB + BA$ …

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

      $X^T = B A^T + A B^T$ …

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5. CBSE 2025Set 65/4/11 markMCQ
   Q.The matrix $\begin{bmatrix} 0 & 1 & -2 \\ -1 & 0 & -7 \\ 2 & 7 & 0 \end{bmatrix}$ is a : (A) diagonal matrix (B) symmetric matrix (C) skew symmetric matrix (D) scalar matrix

   ›Reveal solutionSolution

   A matrix is skew-symmetric when $A^T = -A$, meaning each element satisfies $a_{ij} = -a_{ji}$ and the diagonal is all zeros. Checking the given matrix confirms it is skew-symmetric.

   Every square matrix can be uniquely decomposed into the sum of a symmetric part and a skew-symmetric part. Understanding these two types is fundamental because they capture different geometric behaviors—symmetric matrices represent self-adjoint operators, while skew-symmetric matrices represent infinitesimal rotations.

   A symmetric matrix satisfies $A^T = A$, meaning the matrix equals its own transpose. Visually, elements are mirrored across the main diagonal: $a_{ij} = a_{ji}$ for all $i, j$.

   A skew-symmetric matrix (also called antisymmetric) satisfies $A^T = -A$. This means:

   • Elements are negatives of their mirror images: $a_{ij} = -a_{ji}$
   • The diagonal must be all zeros (since $a_{ii} = -a_{ii}$ implies $a_{ii} = 0$)

   Let's examine the given matrix systematically.

   $A = \begin{bmatrix} 0 & 1 & -2 \\ -1 & 0 & -7 \\ 2 & 7 & 0 \end{bmatrix}$

   1. Check the diagonal

   The diagonal elements are $a_{11} = 0$, $a_{22} = 0$, $a_{33} = 0$. All zeros—this is necessary (but not sufficient) for skew-symmetry. It immediately rules out diagonal and scalar matrices, which require non-zero diagonal entries (at least for scalar matrices, all diagonal entries must be equal and typically non-zero).

   2. Check symmetry vs. skew-symmetry

   Compare corresponding off-diagonal pairs:

   • $a_{12} = 1$ and $a_{21} = -1$: we have $a_{21} = -a_{12}$ ✓
   • $a_{13} = -2$ and $a_{31} = 2$: we have $a_{31} = -a_{13}$ ✓
   • $a_{23} = -7$ and $a_{32} = 7$: we have $a_{32} = -a_{23}$ ✓

   Every element satisfies $a_{ij} = -a_{ji}$.

   3. Verify by computing the transpose …

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6. CBSE 2024Set 65/2/11 markMCQ
   Q.$Assertion\ (A):$ 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.

   ›Reveal solutionSolution

   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

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

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

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

   2. Simplify using the transpose properties:

      • $(B')' = B$ (transpose of transpose returns the original)
      • $A' = A$ (since $A$ is symmetric)

      Therefore:

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

   3. 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 $(B'AB)' = +B'AB$.

   4. 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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7. 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. …

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8. 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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9. CBSE 2023Set 65/2/11 markMCQ
   Q.If $\begin{bmatrix} 2 & 0 \\ 5 & 4 \end{bmatrix} = P + Q$, where $P$ is a symmetric and $Q$ is a skew symmetric matrix, then $Q$ is equal to:

   (a) $\begin{bmatrix} 2 & \frac{5}{2} \\ \frac{5}{2} & 4 \end{bmatrix}$

   (b) $\begin{bmatrix} 0 & -\frac{5}{2} \\ \frac{5}{2} & 0 \end{bmatrix}$

   (c) $\begin{bmatrix} 0 & \frac{5}{2} \\ -\frac{5}{2} & 0 \end{bmatrix}$

   (d) $\begin{bmatrix} 2 & -\frac{5}{2} \\ \frac{5}{2} & 4 \end{bmatrix}$

   ›Reveal solutionSolution

   Any square matrix can be uniquely expressed as the sum of a symmetric and a skew-symmetric matrix. We find the skew-symmetric part $Q$ by calculating $Q = \frac{1}{2}(A - A^T)$, which results in $\boxed{\begin{bmatrix} 0 & -\frac{5}{2} \\ \frac{5}{2} & 0 \end{bmatrix}}$.

   Matrices possess a fascinating property: any square matrix can be uniquely decomposed into the sum of a symmetric matrix and a skew-symmetric matrix. This decomposition is not just a mathematical curiosity; it's a fundamental concept used in various areas, including physics and engineering, for simplifying matrix analysis.

   Let's first recall what symmetric and skew-symmetric matrices are:

   • A matrix $P$ is symmetric if $P^T = P$. This means its elements are symmetric about the main diagonal ($p_{ij} = p_{ji}$).
   • A matrix $Q$ is skew-symmetric if $Q^T = -Q$. This implies that its diagonal elements must be zero ($q_{ii} = -q_{ii} \implies 2q_{ii} = 0 \implies q_{ii} = 0$) and off-diagonal elements satisfy $q_{ij} = -q_{ji}$.

   Now, consider any square matrix $A$. We want to express it as $A = P + Q$, where $P$ is symmetric and $Q$ is skew-symmetric.

   If we take the transpose of this equation, we get $A^T = (P + Q)^T = P^T + Q^T$.

   Since $P$ is symmetric, $P^T = P$.

   Since $Q$ is skew-symmetric, $Q^T = -Q$.

   So, $A^T = P - Q$.

   We now have a system of two linear matrix equations:

   1. $A = P + Q$
   2. $A^T = P - Q$

   Adding these two equations:

   $A + A^T = (P + Q) + (P - Q) = 2P$

   This gives us $P = \frac{1}{2}(A + A^T)$.

   You can verify that this $P$ is indeed symmetric: $P^T = \left(\frac{1}{2}(A + A^T)\right)^T = \frac{1}{2}(A^T + (A^T)^T) = \frac{1}{2}(A^T + A) = P$.

   Subtracting the second equation from the first:

   $A - A^T = (P + Q) - (P - Q) = 2Q$

   This gives us $Q = \frac{1}{2}(A - A^T)$.

   You can verify that this $Q$ is indeed skew-symmetric: $Q^T = \left(\frac{1}{2}(A - A^T)\right)^T = \frac{1}{2}(A^T - (A^T)^T) = \frac{1}{2}(A^T - A) = -\frac{1}{2}(A - A^T) = -Q$.

   For any square matrix $A$, its unique decomposition into a symmetric matrix $P$ and a skew-symmetric matrix $Q$ is given by:

   $P = \frac{1}{2}(A + A^T)$

   $Q = \frac{1}{2}(A - A^T)$

   The problem asks for the skew-symmetric matrix $Q$. We will use the formula $Q = \frac{1}{2}(A - A^T)$.

   Here are the steps to find $Q$:

   1. Identify the given matrix $A$.

      We are given $A = \begin{bmatrix} 2 & 0 \\ 5 & 4 \end{bmatrix}$.

   2. Find the transpose of $A$, denoted as $A^T$.

      To find the transpose, we swap the rows and columns of $A$.

      $A^T = \begin{bmatrix} 2 & 5 \\ 0 & 4 \end{bmatrix}$.

   3. Calculate the difference $A - A^T$. …

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10. 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 $b = 0$. …

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

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

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

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