Which of the following properties is/are true for two matrices of suitable orders? (i) (A + B)' = A' + B' (ii)

Question

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)

CBSECBSE Class XII Board 2026MCQ· 1mImportance★★★★★

✓ Free question

Concept understanding — Matrix Transpose

Matrix Transpose

The transpose is one of the simplest yet most useful operations on a matrix: you flip the matrix across its main diagonal, so that its rows become columns and its columns become rows.

The intuition

Picture writing a table of marks with students down the rows and subjects across the columns. If instead you want subjects down the rows and students across the columns, you don't recollect the data — you just turn the table on its side. That turn is the transpose.

The precise definition

If $A = [a_{ij}]$ is a matrix of order $m \times n$ , its transpose, written $A'$ (or $A^{T}$ ), is the $n \times m$ matrix obtained by interchanging rows and columns:

A' = [a_{ji}], \qquad \text{so the } (i,j) \text{ entry of } A' \text{ is the } (j,i) \text{ entry of } A.

The entry in row $i$ , column $j$ of $A$ moves to row $j$ , column $i$ of $A'$ .

A worked look

A = \begin{bmatrix} 2 & 5 & 1 \\ 0 & 3 & 4 \end{bmatrix}_{2\times 3} \qquad\Longrightarrow\qquad A' = \begin{bmatrix} 2 & 0 \\ 5 & 3 \\ 1 & 4 \end{bmatrix}_{3\times 2}.

The first row $(2,5,1)$ of $A$ has become the first column of $A'$ .

Properties you must know

For matrices $A,B$ of suitable orders and a scalar $k$ :

$(A')' = A$ — transposing twice returns the original.
$(kA)' = k\,A'$ — a scalar comes straight through.
$(A+B)' = A' + B'$ — transpose distributes over addition.
$(AB)' = B'A'$ — the reversal law: the transpose of a product reverses the order of the factors.

Watch out

That last rule catches many students: $(AB)' = B'A'$ , not $A'B'$ . The order flips, just as it does for the inverse of a product.

Why it matters

The transpose is the gateway to two important families of matrices you meet next:

a symmetric matrix satisfies $A' = A$ ;
a skew-symmetric matrix satisfies $A' = -A$ .

Both are defined purely through the transpose, so getting comfortable with this flip makes the rest of the chapter far easier.

Matrix Transpose, including the reversal law (AB)' = B'A', is a core definition in the CBSE Class 12 Matrices chapter and a direct prerequisite for understanding symmetric and skew-symmetric matrices later in the same unit. "Properties of transpose of a matrix class 12" is a frequently searched revision topic ahead of both board exams and JEE Main.

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

This is false. The correct property is $(AB)' = B'A'$ — the order of multiplication reverses. The reason is dimensional: if $A$ is $m \times n$ and $B$ is $n \times p$ , then $AB$ is $m \times p$ , so $(AB)'$ is $p \times m$ . For $A'B'$ to be defined, $A'$ would need to be $n \times m$ and $B'$ would be $p \times n$ , which cannot multiply in that order unless $m = p$ . The correct product $B'A'$ has $B'$ as $p \times n$ and $A'$ as $n \times m$ , giving a $p \times m$ result — matching dimensions perfectly.

Watch out

A common mistake is to forget the reversal in the transpose of a product. Always remember: the transpose of a product is the product of the transposes in reverse order.

Statement (iv): $(kAB)' = kB'A'$ This is true. The scalar $k$ is a constant, so it factors out unchanged: $(kAB)' = k(AB)' = k(B'A')$ . The order reversal is the same as in statement (iii), and the scalar simply tags along.

Tip

You can remember the reversal rule by thinking of socks and shoes: you put on socks then shoes, but to undo (transpose) you take off shoes first then socks — the order reverses.

Only statements (i) and (iv) are correct.

✓Final answer

The correct option is (D) (i) and (iv).

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

This is false. The correct property is $(AB)' = B'A'$ — the order of multiplication reverses. The reason is dimensional: if $A$ is $m \times n$ and $B$ is $n \times p$ , then $AB$ is $m \times p$ , so $(AB)'$ is $p \times m$ . For $A'B'$ to be defined, $A'$ would need to be $n \times m$ and $B'$ would be $p \times n$ , which cannot multiply in that order unless $m = p$ . The correct product $B'A'$ has $B'$ as $p \times n$ and $A'$ as $n \times m$ , giving a $p \times m$ result — matching dimensions perfectly.

Watch out

A common mistake is to forget the reversal in the transpose of a product. Always remember: the transpose of a product is the product of the transposes in reverse order.

Statement (iv): $(kAB)' = kB'A'$ This is true. The scalar $k$ is a constant, so it factors out unchanged: $(kAB)' = k(AB)' = k(B'A')$ . The order reversal is the same as in statement (iii), and the scalar simply tags along.

Tip

You can remember the reversal rule by thinking of socks and shoes: you put on socks then shoes, but to undo (transpose) you take off shoes first then socks — the order reverses.

Only statements (i) and (iv) are correct.

✓Final answer
$The correct option is (D) (i) and (iv).$
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.

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

The interval is $x \in \left[0, \frac{\pi}{2}\right]$ . The angle whose cosine is $\frac{1}{2}$ in this interval is $x = \frac{\pi}{3}$ (since $\cos \frac{\pi}{3} = \frac{1}{2}$ ).

The other standard angle $\frac{\pi}{6}$ gives $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$ , not $\frac{1}{2}$ .

Watch out

A common mistake is to forget that $A'$ is the transpose, not the conjugate transpose. Here all entries are real, so it's fine — but also note that $A'$ is not the same as $A^{-1}$ in general; it just happens to be here because $A$ is orthogonal.

Tip

Recognising $A$ as a rotation matrix immediately tells you that $A + A' = 2\cos x \cdot I$ , because the sum of a rotation and its inverse (transpose) is always a scalar multiple of the identity. This saves you from writing out all four entries.

✓Final answer
$x$
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 order:

$A=[4\ 2\ 3]\Rightarrow A'=\begin{bmatrix}4\\2\\3\end{bmatrix}$ .

✓Final answer
$\begin{bmatrix}4\\2\\3\end{bmatrix}$
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).

Since $A$ has order $2\times3$ , $(A')'$ also has order $2\times3$ .

✓Final answer
$(A')'$
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 $A' = \begin{pmatrix}\cos\alpha & \sin\alpha\\ -\sin\alpha & \cos\alpha\end{pmatrix}$ .

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

For this to equal $I=\begin{pmatrix}1&0\\0&1\end{pmatrix}$ , we need $2\cos\alpha = 1 \Rightarrow \cos\alpha = \dfrac12 \Rightarrow \alpha = \dfrac{\pi}{3}$ .

✓Final answer
$\alpha = \pi/3$
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

$A' = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}.$

Note (B) reverses the order, so it is wrong.

✓Final answer
$\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$
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.

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

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

We are given $A + A' = I = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$ . Comparing entries:

$2\cos\alpha = 1 \implies \cos\alpha = \dfrac12 \implies \alpha = \dfrac{\pi}{3}$

✓Final answer
$\dfrac{\pi}{3}$
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.

For any matrix $A$ , $(A')' = A$ (transposing twice undoes the operation). Since $A$ has order $3\times2$ , so does $(A')'$ .

✓Final answer
$3\times2$
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 $A$ :

$A' = \begin{bmatrix} 2 & 3 \\ \sqrt{2} & -2 \\ 0 & 2/5 \end{bmatrix}.$

✓Final answer
$\begin{bmatrix} 2 & 3 \\ \sqrt{2} & -2 \\ 0 & 2/5 \end{bmatrix}$
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 $B$ ) $= 2\times 4$ .

The transpose of a matrix of order $m\times n$ has order $n\times m$ . So $(AB)'$ has order $4\times 2$ .

✓Final answer
$(AB)' = 4\times 2$
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).

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

Step 2: Transpose AB (rows become columns).

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

✓Final answer
$(AB)' = \begin{bmatrix}2 & 4 & 8\\ 0 & 0 & 0\\ 1 & 2 & 4\end{bmatrix}$
CBSE 2023Set 65/3/11 markMCQ
$x \in \left[0, \frac{\pi}{2}\right]$
(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.

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

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

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.

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:

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

Solve for $\cos x$ .

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

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.)

Watch out

A common mistake is to forget that $A'$ means transpose, not conjugate transpose. In this real matrix, they are the same, but the notation $A'$ is standard for transpose in many Indian exam contexts. Also, do not confuse $\cos x = \frac{\sqrt{3}}{2}$ with $x = \frac{\pi}{3}$ — that's a classic slip. $\cos \frac{\pi}{3} = \frac{1}{2}$ , not $\frac{\sqrt{3}}{2}$ .

Tip

Notice that $A$ is a rotation matrix. Its transpose is its inverse (since it's orthogonal). So $A + A'$ is symmetric and diagonal in this case — a neat property that saves time if you recognise it.

✓Final answer
$x = \frac{\pi}{6}$

Q.Which of the following properties is/are true for two matrices of suitable orders?

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