Q.If the order of matrix A is $3 \times 2$, then the order of matrix A' will be _________.

Showing the 12 most recent of 62 on this concept.

1. CBSE 2026Set 65/1/11 markMCQ
   Q.Which of the following properties is/are true for two matrices of suitable orders?

   (i) $(A + B)' = A' + B'$

   (ii) $(A - B)' = B' - A'$

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

   (iv) $(kAB)' = kB'A'$ ($k$ is a scalar) (A)

   (i) only (B) (i),

   (ii) and

   (iii) (C)

   (i) and

   (ii) (D)

   (i) and (iv)

   ›Reveal solutionSolution

   The transpose of a sum is the sum of transposes, and the transpose of a product reverses the order. Only statements (i) and (iv) are correct.

   The transpose operation flips a matrix over its diagonal — rows become columns and columns become rows. The key intuition is that transposition distributes over addition but reverses the order of multiplication. This reversal is not arbitrary; it comes from the fact that when you multiply two matrices and then transpose, the dimensions must still match, which forces the order swap.

   Let’s check each statement carefully.

   1. Statement (i): $(A + B)' = A' + B'$

      This is true. Transposition is a linear operation — adding two matrices and then transposing gives the same result as transposing each first and then adding. Element-wise, the $(i,j)$ entry of $(A+B)'$ is $a_{ji} + b_{ji}$, which is exactly the $(i,j)$ entry of $A' + B'$.

   2. Statement (ii): $(A - B)' = B' - A'$

      This is false. The correct property is $(A - B)' = A' - B'$, because transposition distributes over subtraction just as it does over addition. The given expression has the order swapped, which is wrong. For example, take $A = \begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}$ and $B = \begin{pmatrix}0 & 0 \\ 1 & 0\end{pmatrix}$; the left side gives $\begin{pmatrix}1 & -1 \\ 0 & 0\end{pmatrix}' = \begin{pmatrix}1 & 0 \\ -1 & 0\end{pmatrix}$, while the right side gives $\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix} - \begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix} = \begin{pmatrix}-1 & 1 \\ 0 & 0\end{pmatrix}$, which are not equal.

   3. Statement (iii): $(AB)' = A'B'$ …

   🔒 Start your 14-day free trial to see the full solution →
2. 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. …

   🔒 Start your 14-day free trial to see the full solution →
3. 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 $b = 0$. …

   🔒 Start your 14-day free trial to see the full solution →
4. CBSE 2026Set 65/2/11 markMCQ
   Q.If $A = \begin{bmatrix} \cos x & -\sin x \\ \sin x & \cos x \end{bmatrix}$ and $A + A' = I$, then the value of $x \in \left[0, \frac{\pi}{2}\right]$ is (A) $0$ (B) $\frac{\pi}{4}$ (C) $\frac{\pi}{3}$ (D) $\frac{\pi}{2}$

   ›Reveal solutionSolution

   The key idea is that $A$ is a rotation matrix, $A'$ is its transpose (which is also its inverse), and the condition $A + A' = I$ forces the diagonal sum $2\cos x = 1$, giving $x = \frac{\pi}{3}$.

   We are given a $2 \times 2$ matrix $A$ that depends on an angle $x$. The matrix $A$ is a classic rotation matrix: it rotates a vector in the plane by angle $x$ counterclockwise. Its transpose $A'$ is simply the rotation by $-x$ (clockwise), which is also the inverse of $A$.

   The condition $A + A' = I$ means that when we add the matrix and its transpose, we get the identity matrix. This is a direct equation in the entries of the matrices.

   Let’s write it out step by step.

   1. Write $A$ and $A'$ explicitly.

      $A = \begin{bmatrix} \cos x & -\sin x \\ \sin x & \cos x \end{bmatrix}$.

      The transpose $A'$ swaps rows and columns:

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

   2. Add them entrywise.

      $A + A' = \begin{bmatrix} \cos x + \cos x & -\sin x + \sin x \\ \sin x + (-\sin x) & \cos x + \cos x \end{bmatrix} = \begin{bmatrix} 2\cos x & 0 \\ 0 & 2\cos x \end{bmatrix}$.

      Notice the off-diagonal terms cancel perfectly: $-\sin x + \sin x = 0$ and $\sin x - \sin x = 0$. So the sum is a diagonal matrix with both diagonal entries equal to $2\cos x$.

   3. Set this equal to $I$.

      The identity matrix $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.

      So we require:

      $\begin{bmatrix} 2\cos x & 0 \\ 0 & 2\cos x \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.

      This gives a single equation from the diagonal entries: $2\cos x = 1$.

   4. Solve for $x$ in the given interval.

      $2\cos x = 1 \implies \cos x = \frac{1}{2}$. …

   🔒 Start your 14-day free trial to see the full solution →
5. 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 $(P^T)^T = P$:

      $X^T = B A^T + A B^T$ …

   🔒 Start your 14-day free trial to see the full solution →
6. CBSE 2026Set A1 markMCQ
   Q.$A = [4\ 2\ 3] \Rightarrow A' =$

   (a) $[4\ 2\ 3]$

   (b) $\begin{bmatrix} 3 \\ 2 \\ 4 \end{bmatrix}$

   (c) $[3\ 2\ 4]$

   (d) $\begin{bmatrix} 4 \\ 2 \\ 3 \end{bmatrix}$

   ›Reveal solutionSolution

   Transpose turns the row $[4\ 2\ 3]$ into the column $\begin{bmatrix}4\\2\\3\end{bmatrix}$.

   The transpose $A'$ (or $A^T$) interchanges rows and columns, keeping each entry's value. A $1\times3$ row matrix becomes a $3\times1$ column matrix with the entries listed top-to-bottom in the same ord …

   🔒 Start your 14-day free trial to see the full solution →
7. 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 $A' = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$.

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

   🔒 Start your 14-day free trial to see the full solution →
8. 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 $A' = \begin{bmatrix} 3 & 1 \\ -4 & -1 \end{bmatrix}$. …

   🔒 Start your 14-day free trial to see the full solution →
9. 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.

   …

   🔒 Start your 14-day free trial to see the full solution →
10. CBSE 2026Set ANNUAL1 markMCQ
    Q.If the order of the matrix A is 2×3 then the order of the matrix (A')' is:

    (a) 2×3

    (b) 3×2

    (c) 2×2

    (d) 3×3

    ›Reveal solutionSolution

    The transpose of the transpose of a matrix is the matrix itself, so $(A')' = A$, which has the same order as $A$.

    For any matrix $A$, the property $(A')' = A$ always holds (taking the transpose twice returns the original matrix).

    …

    🔒 Start your 14-day free trial to see the full solution →
11. CBSE 2026Set ANNUAL1 markMCQ
    Q.If A = [[cos α, −sin α], [sin α, cos α]], then A + A' = I if the value of α is

    (a) π/6

    (b) π/3

    (c) π

    (d) 3π/2

    ›Reveal solutionSolution

    Adding $A$ to its transpose cancels the sine terms and leaves a diagonal matrix of $2\cos\alpha$; setting this equal to $I$ pins down $\alpha$.

    $A = \begin{pmatrix}\cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha\end{pmatrix}$, so $A' = \begin{pmatrix}\cos\alpha & \sin\alpha\\ -\sin\alpha & \cos\alpha\end{pmatrix}$.

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

    …

    🔒 Start your 14-day free trial to see the full solution →
12. 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 $A$.

    …

    🔒 Start your 14-day free trial to see the full solution →