If $A = \begin{bmatrix}0 & -1 & 2\\ 4 & 3 & -4\end{bmatrix}$ and $B = \begin{bmatrix}4 & 0\\ 1 & 3\\ 2 & 6\end{bmatrix}$, then verify that: (i) $(A')' = A$ (ii) $(AB)' = B'A'$ (iii) $(kA)' = (kA')$.

> [!TLDR] > Matrix transpose flips rows and columns. For any matrix, transposing twice returns the original; the transpose of a product reverses the order; and scalar multiplication commutes with transposition. Here we verify all three properties explicitly for the given $2\times3$ matrix $A$ and $3\times2$ matrix $B$. ## The Core Idea The transpose of a matrix $M$, written $M'$ or $M^T$, is obtained by swapping its rows and columns: the entry at position $(i,j)$ in $M$ becomes the entry at $(j,i)$ in $M'$. This simple operation has beautiful algebraic properties that make it indispensable in linear algebra. Rather than memorising them, notice why each makes sense: - **Double transpose**: Flipping rows and columns twice brings you back to the original arrangement — like turning a page over and back. - **Product transpose**: When you multiply $A$ and $B$, the rows of $A$ combine with columns of $B$. Transposing the product means those combinations become columns of $A'$ combining with rows of $B'$ — which is exactly $B'A'$ in reverse order. - **Scalar multiplication**: Scaling every entry by $k$ and then transposing is the same as transposing first and then scaling — multiplication by a number doesn't care about the arrangement. Let's verify each with the given matrices. --- ## Verification ### (i) $(A')' = A$ **Step 1: Find $A'$.** $A$ is $2\times3$: $$A = \begin{bmatrix}0 & -1 & 2\\ 4 & 3 & -4\end{bmatrix}$$ Transpose: row 1 becomes column 1, row 2 becomes column 2. $$A' = \begin{bmatrix}0 & 4\\ -1 & 3\\ 2 & -4\end{bmatrix}$$ **Step 2: Transpose $A'$.** Now $A'$ is $3\times2$. Transpose it: row 1 becomes column 1, row 2 becomes column 2, row 3 becomes column 3. $$(A')' = \begin{bmatrix}0 & -1 & 2\\ 4 & 3 & -4\end{bmatrix}$$ This is exactly $A$. So $(A')' = A$ holds. > [!TIP] > The double-transpose property is the algebraic equivalent of "undoing" an operation — it's why we say transposition is an **involution**. --- ### (ii) $(AB)' = B'A'$ **Step 1

If A = 0 -1 2\4 3 -4 and B = 4 0\1 3\2 6 , then verify that: (i) (A')' = A (ii) (AB)' = B'A' (iii) (kA)' = (kA

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A = \begin{bmatrix}0 & -1 & 2\\ 4 & 3 & -4\end{bmatrix}

(i)

(A')' = A

(ii)

(AB)' = B'A'

(iii)

(kA)' = (kA')

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Matrix transpose flips rows and columns. For any matrix, transposing twice returns the original; the transpose of a product reverses the order; and scalar multiplication commutes with transposition. Here we verify all three properties explicitly for the given $2\times3$ matrix $A$ and $3\times2$ matrix $B$ .

The Core Idea

The transpose of a matrix $M$ , written $M'$ or $M^T$ , is obtained by swapping its rows and columns: the entry at position $(i,j)$ in $M$ becomes the entry at $(j,i)$ in $M'$ . This simple operation has beautiful algebraic properties that make it indispensable in linear algebra. Rather than memorising them, notice why each makes sense:

Double transpose: Flipping rows and columns twice brings you back to the original arrangement — like turning a page over and back.
Product transpose: When you multiply $A$ and $B$ , the rows of $A$ combine with columns of $B$ . Transposing the product means those combinations become columns of $A'$ combining with rows of $B'$ — which is exactly $B'A'$ in reverse order.
Scalar multiplication: Scaling every entry by $k$ and then transposing is the same as transposing first and then scaling — multiplication by a number doesn't care about the arrangement.

Let's verify each with the given matrices.

Verification

(i) $(A')' = A$

Step 1: Find $A'$ .

$A$ is $2\times3$ :

A = \begin{bmatrix}0 & -1 & 2\\ 4 & 3 & -4\end{bmatrix}

Transpose: row 1 becomes column 1, row 2 becomes column 2.

A' = \begin{bmatrix}0 & 4\\ -1 & 3\\ 2 & -4\end{bmatrix}

Step 2: Transpose $A'$ .

Now $A'$ is $3\times2$ . Transpose it: row 1 becomes column 1, row 2 becomes column 2, row 3 becomes column 3.

(A')' = \begin{bmatrix}0 & -1 & 2\\ 4 & 3 & -4\end{bmatrix}

This is exactly $A$ . So $(A')' = A$ holds.

Tip

The double-transpose property is the algebraic equivalent of "undoing" an operation — it's why we say transposition is an involution.

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

Step 1: Compute $AB$ .

$A$ is $2\times3$ , $B$ is $3\times2$ , so $AB$ will be $2\times2$ .

AB = \begin{bmatrix}0 & -1 & 2\\ 4 & 3 & -4\end{bmatrix} \begin{bmatrix}4 & 0\\ 1 & 3\\ 2 & 6\end{bmatrix}

Compute entry by entry:

$(1,1)$ : $0\cdot4 + (-1)\cdot1 + 2\cdot2 = 0 - 1 + 4 = 3$
$(1,2)$ : $0\cdot0 + (-1)\cdot3 + 2\cdot6 = 0 - 3 + 12 = 9$
$(2,1)$ : $4\cdot4 + 3\cdot1 + (-4)\cdot2 = 16 + 3 - 8 = 11$
$(2,2)$ : $4\cdot0 + 3\cdot3 + (-4)\cdot6 = 0 + 9 - 24 = -15$

So:

AB = \begin{bmatrix}3 & 9\\ 11 & -15\end{bmatrix}

Step 2: Transpose $AB$ .

(AB)' = \begin{bmatrix}3 & 11\\ 9 & -15\end{bmatrix}

Step 3: Compute $B'A'$ .

First find $B'$ (transpose of $3\times2$ $B$ gives $2\times3$ ):

B' = \begin{bmatrix}4 & 1 & 2\\ 0 & 3 & 6\end{bmatrix}

We already have $A'$ from part (i):

A' = \begin{bmatrix}0 & 4\\ -1 & 3\\ 2 & -4\end{bmatrix}

Now multiply $B'$ ( $2\times3$ ) by $A'$ ( $3\times2$ ), result is $2\times2$ :

$(1,1)$ : $4\cdot0 + 1\cdot(-1) + 2\cdot2 = 0 - 1 + 4 = 3$
$(1,2)$ : $4\cdot4 + 1\cdot3 + 2\cdot(-4) = 16 + 3 - 8 = 11$
$(2,1)$ : $0\cdot0 + 3\cdot(-1) + 6\cdot2 = 0 - 3 + 12 = 9$ …

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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 22 on this concept.

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.

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 $A' + B'$ .

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.

Statement (iii): $(AB)' = A'B'$ …
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CBSE 2020Set 65/1/11 mark
$3 \times 2$
(OR)
$A square matrix A will be a skew-symmetric matrix, if _________.$
›Reveal solutionSolution
$A'$ has order $2\times3$ .

$A$ is skew-symmetric iff $A'=-A$ .

Part (a)

The transpose $A'$ of a matrix is formed by interchanging its rows and columns: an entry in row $i$ , column $j$ of $A$ moves to row $j$ , column $i$ of $A'$ . So an $m\times n$ matrix has an $n\times m$ transpose. With $A$ of order $3\times2$ ( $m=3,n=2$ ): …
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CBSE 2026Set 65/2/11 markMCQ
$A = \begin{bmatrix} \cos x & -\sin x \\ \sin x & \cos x \end{bmatrix}$
›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.

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:

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

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 $2\cos x$ .

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: $2\cos x = 1$ .

Solve for $x$ in the given interval.

$2\cos x = 1 \implies \cos x = \frac{1}{2}$ . …
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CBSE 2026Set A1 markMCQ
$A = [4\ 2\ 3] \Rightarrow A' =$
(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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CBSE 2026Set ANNUAL1 markMCQ
$Q. If the order of the matrix A is 2\times3 then the order of the matrix (A')' is:$
(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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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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CBSE 2025Set E1 markMCQ
$A = \begin{bmatrix} 1 & 2 & 3 \end{bmatrix} \Rightarrow A' =$
(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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CBSE 2025Set A1 markMCQ
$A = \begin{bmatrix}\cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha\end{bmatrix}$
(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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CBSE 2025Set ANNUAL1 mark
$Q. If order of the matrix A is 3\times2 then order of matrix (A')' is ______.$
›Reveal solutionSolution

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

…

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CBSE 2024Set D1 markMCQ
$A = \begin{bmatrix} 2 & \sqrt{2} & 0 \\ 3 & -2 & \frac{2}{5} \end{bmatrix}$
(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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CBSE 2024Set ANNUAL1 markMCQ
$A$
(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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CBSE 2024Set ANNUAL1 mark
$A = \begin{bmatrix}1\\2\\4\end{bmatrix}_{3\times1}$
›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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Q.If $A = \begin{bmatrix}0 & -1 & 2\\ 4 & 3 & -4\end{bmatrix}$ and $B = \begin{bmatrix}4 & 0\\ 1 & 3\\ 2 & 6\end{bmatrix}$ , then verify that:

Concept understanding — Matrix Transpose

Matrix Transpose

The intuition

The precise definition

A worked look

Properties you must know

The Core Idea

Verification

(i) $(A')' = A$

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

Method: Verifying a transpose identity by computing both sides

Steps

Common Mistakes

Part (a)

Q.If A=[0−1243−4]A = \begin{bmatrix}0 & -1 & 2\\ 4 & 3 & -4\end{bmatrix}A=[04​−13​2−4​] and B=[401326]B = \begin{bmatrix}4 & 0\\ 1 & 3\\ 2 & 6\end{bmatrix}B=​412​036​​, then verify that:

The Core Idea

Verification

(i) (A′)′=A(A')' = A(A′)′=A

(ii) (AB)′=B′A′(AB)' = B'A'(AB)′=B′A′

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Q.If $A = \begin{bmatrix}0 & -1 & 2\\ 4 & 3 & -4\end{bmatrix}$ and $B = \begin{bmatrix}4 & 0\\ 1 & 3\\ 2 & 6\end{bmatrix}$ , then verify that:

(i) $(A')' = A$

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

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