Q.If $A=\begin{bmatrix}1&-2&2\\3&1&-1\end{bmatrix}$ and $B=\begin{bmatrix}2&4\\1&2\\3&-1\end{bmatrix}$, show that $(AB)^T=B^TA^T$.

Showing the 12 most recent of 22 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): $(AB)' = A'B'$ …

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2. 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.

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

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3. 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 …

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4. 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).

   …

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

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

   …

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6. CBSE 2025Set E1 markMCQ
   Q.$A = \begin{bmatrix} 1 & 2 & 3 \end{bmatrix} \Rightarrow A' =$

   (a) $\begin{bmatrix} 1 & 2 & 3 \end{bmatrix}$

   (b) $\begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix}$

   (c) $\begin{bmatrix} 3 & 2 & 1 \end{bmatrix}$

   (d) $\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$

   ›Reveal solutionSolution

   Transposing the row $\begin{bmatrix} 1 & 2 & 3 \end{bmatrix}$ gives the column $\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$.

   The transpose $A'$ turns rows into columns while keeping the entries in the same order. So the $1\times3$ matrix $A = \begin{bmatrix} 1 & 2 & 3 \end{bmatrix}$ becomes the $3\times1$ matrix …

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7. CBSE 2025Set A1 markMCQ
   Q.If $A = \begin{bmatrix}\cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha\end{bmatrix}$ and $A + A' = I$, then the value of $\alpha$ is:

   (a) $\dfrac{\pi}{6}$

   (b) $\pi$

   (c) $\dfrac{3\pi}{2}$

   (d) $\dfrac{\pi}{3}$

   ›Reveal solutionSolution

   Compute $A'$ (transpose), add to $A$, and match to the identity matrix.

   $A = \begin{bmatrix}\cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha\end{bmatrix}, \qquad A' = \begin{bmatrix}\cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha\end{bmatrix}$

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

   …

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8. CBSE 2025Set ANNUAL1 mark
   Q.If order of the matrix A is 3×2 then order of matrix (A')' is ______.

   ›Reveal solutionSolution

   Taking the transpose twice returns the original matrix, so its order is unchanged.

   …

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9. CBSE 2024Set D1 markMCQ
   Q.If $A = \begin{bmatrix} 2 & \sqrt{2} & 0 \\ 3 & -2 & \frac{2}{5} \end{bmatrix}$ then $A' =$

   (a) $\begin{bmatrix} 2 & -3 \\ \sqrt{2} & 2 \\ 0 & 2/5 \end{bmatrix}$

   (b) $\begin{bmatrix} 2 & 3 \\ \sqrt{2} & -2 \\ 0 & 2/5 \end{bmatrix}$

   (c) $\begin{bmatrix} 3 & 2 \\ -2 & \sqrt{2} \\ -2/5 & 0 \end{bmatrix}$

   (d) $\begin{bmatrix} 3 & -2 & 2/5 \\ 2 & \sqrt{2} & 0 \end{bmatrix}$

   ›Reveal solutionSolution

   The transpose $A'$ turns each row of $A$ into a column.

   $A = \begin{bmatrix} 2 & \sqrt{2} & 0 \\ 3 & -2 & \frac{2}{5} \end{bmatrix}$ is $2\times3$, so $A'$ is $3\times2$ with columns equal to the rows of $A$: …

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10. CBSE 2024Set ANNUAL1 markMCQ
    Q.If $A$ is a matrix of order $2 \times 3$ and $B$ is a matrix of order $3 \times 4$, then the order of $(AB)'$ is

    (a) $2 \times 3$

    (b) $2 \times 4$

    (c) $4 \times 2$

    (d) $3 \times 4$

    ›Reveal solutionSolution

    Order of $AB$ is (rows of $A$) $\times$ (columns of $B$); a transpose swaps rows and columns.

    $A$ is $2\times 3$ and $B$ is $3\times 4$. Since the number of columns of $A$ (=3) equals the number of rows of $B$ (=3), $AB$ is defined and has order (rows of $A$) $\times$ (columns of $B$) $= 2\times 4$.

    …

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11. CBSE 2024Set ANNUAL1 mark
    Q.If $A = \begin{bmatrix}1\\2\\4\end{bmatrix}_{3\times1}$ and $B = \begin{bmatrix}2 & 0 & 1\end{bmatrix}_{1\times3}$, then find the matrix (AB)', where (AB)' is the transpose of matrix (AB).

    ›Reveal solutionSolution

    First multiply A (3×1) and B (1×3) to get a 3×3 matrix, then transpose it.

    Given $A = \begin{bmatrix}1\\2\\4\end{bmatrix}_{3\times1}$ and $B = \begin{bmatrix}2 & 0 & 1\end{bmatrix}_{1\times3}$.

    Step 1: Compute AB (3×3).

    $AB = \begin{bmatrix}1\\2\\4\end{bmatrix}\begin{bmatrix}2 & 0 & 1\end{bmatrix} = \begin{bmatrix}1(2) & 1(0) & 1(1)\\ 2(2) & 2(0) & 2(1)\\ 4(2) & 4(0) & 4(1)\end{bmatrix} = \begin{bmatrix}2 & 0 & 1\\ 4 & 0 & 2\\ 8 & 0 & 4\end{bmatrix}$

    …

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12. CBSE 2023Set 65/3/11 markMCQ
    Q.For what value of $x \in \left[0, \frac{\pi}{2}\right]$, is $A + A' = \sqrt{3}\, I$, where $A = \begin{bmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{bmatrix}$ ?

    (a) $\frac{\pi}{3}$

    (b) $\frac{\pi}{6}$

    (c) $0$

    (d) $\frac{\pi}{2}$

    ›Reveal solutionSolution

    The key idea is to compute $A + A'$ (where $A'$ is the transpose of $A$), set it equal to $\sqrt{3}\,I$, and solve for $x$ in the given interval. The value is $x = \frac{\pi}{6}$.

    We are given a matrix $A$ that depends on $x$, and we need to find $x$ in $\left[0, \frac{\pi}{2}\right]$ such that $A + A' = \sqrt{3}\, I$. Here $A'$ means the transpose of $A$.

    The problem tests two things: understanding matrix transpose and solving a simple trigonometric equation. Let's break it down.

    1. Write down $A$ and its transpose $A'$.

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

       The transpose swaps rows and columns, so

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

    2. Add $A$ and $A'$.

       Adding element-wise:

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

       So $A + A' = (2\cos x) \, I$, where $I$ is the $2\times 2$ identity matrix.

    3. Set this equal to $\sqrt{3}\, I$.

       We have $(2\cos x) \, I = \sqrt{3} \, I$.

       Since the identity matrix is non-zero, we can equate the scalar multipliers:

       $2\cos x = \sqrt{3}$.

    4. Solve for $\cos x$.

       $\cos x = \frac{\sqrt{3}}{2}$.

    5. Find $x$ in $\left[0, \frac{\pi}{2}\right]$.

       The angle whose cosine is $\frac{\sqrt{3}}{2}$ in the first quadrant is $x = \frac{\pi}{6}$.

       (Recall: $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$ and $\frac{\pi}{6}$ lies in the given interval.) …

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