If $A = \begin{bmatrix} 2 & 3 & -1 \\ 1 & 4 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 3 \\ 4 & 5 \\ 2 & 1 \end{bmatrix}$, then $AB$ and $BA$ are defined and equal.

If A = 2 3 -1 \1 4 2 and B = 2 3 \4 5 \2 1 , then AB and BA are defined and equal.

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

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Matrix multiplication is defined only when the number of columns in the first matrix equals the number of rows in the second. Here $A$ is $2 \times 3$ and $B$ is $3 \times 2$ , so both $AB$ ( $2 \times 2$ ) and $BA$ ( $3 \times 3$ ) are defined — but they are not equal because their dimensions differ.

Why this question is a trap

Many students see that both products are defined and assume they must be equal. That’s a natural guess — but matrix multiplication is not commutative. Even when both products exist, they rarely give the same result. Here, the dimensions alone tell you they can’t be equal: $AB$ is a $2 \times 2$ matrix, while $BA$ is $3 \times 3$ . Two matrices of different sizes cannot be equal.

Let’s work through the multiplication to confirm.

Step-by-step computation

1. Check compatibility for $AB$

$A$ is $2 \times 3$ (2 rows, 3 columns). $B$ is $3 \times 2$ (3 rows, 2 columns). The inner dimensions match (3 = 3), so $AB$ is defined and will be $2 \times 2$ .

2. Compute $AB$

The entry in row $i$ , column $j$ of $AB$ is the dot product of row $i$ of $A$ with column $j$ of $B$ .

Row 1 of $A$ : $[2, 3, -1]$
Row 2 of $A$ : $[1, 4, 2]$
Column 1 of $B$ : $\begin{bmatrix}2 \\ 4 \\ 2\end{bmatrix}$
Column 2 of $B$ : $\begin{bmatrix}3 \\ 5 \\ 1\end{bmatrix}$

Now:

$(AB)_{11} = 2 \cdot 2 + 3 \cdot 4 + (-1) \cdot 2 = 4 + 12 - 2 = 14$

$(AB)_{12} = 2 \cdot 3 + 3 \cdot 5 + (-1) \cdot 1 = 6 + 15 - 1 = 20$

$(AB)_{21} = 1 \cdot 2 + 4 \cdot 4 + 2 \cdot 2 = 2 + 16 + 4 = 22$

$(AB)_{22} = 1 \cdot 3 + 4 \cdot 5 + 2 \cdot 1 = 3 + 20 + 2 = 25$

AB = \begin{bmatrix} 14 & 20 \\ 22 & 25 \end{bmatrix}

3. Check compatibility for $BA$

$B$ is $3 \times 2$ , $A$ is $2 \times 3$ . Inner dimensions match (2 = 2), so $BA$ is defined and will be $3 \times 3$ .

4. Compute $BA$

Row $i$ of $B$ with column $j$ of $A$ .

Rows of $B$ :

Row 1: $[2, 3]$

Row 2: $[4, 5]$

Row 3: $[2, 1]$

Columns of $A$ :

Col 1: $\begin{bmatrix}2 \\ 1\end{bmatrix}$ , Col 2: $\begin{bmatrix}3 \\ 4\end{bmatrix}$ , Col 3: $\begin{bmatrix}-1 \\ 2\end{bmatrix}$

Now:

$(BA)_{11} = 2 \cdot 2 + 3 \cdot 1 = 4 + 3 = 7$

$(BA)_{12} = 2 \cdot 3 + 3 \cdot 4 = 6 + 12 = 18$

$(BA)_{13} = 2 \cdot (-1) + 3 \cdot 2 = -2 + 6 = 4$

$(BA)_{21} = 4 \cdot 2 + 5 \cdot 1 = 8 + 5 = 13$ …

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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 39 on this concept.

CBSE 2025Set 65/1/11 markMCQ
$A = \begin{bmatrix} 1 & -2 & -1 \\ 0 & 4 & -1 \\ -3 & 2 & 1 \end{bmatrix}$
›Reveal solutionSolution

Matrix multiplication is defined only when the number of columns in the first matrix equals the number of rows in the second. Here, $A$ is $3 \times 3$ , $B$ is $3 \times 1$ , and $C$ is $1 \times 3$ . So $AB$ ( $3\times3$ times $3\times1$ ) is defined, $AC$ ( $3\times3$ times $1\times3$ ) is defined, but $BA$ ( $3\times1$ times $3\times3$ ) is not defined. The correct option is (B) Only AC.

The key idea is simple: you can multiply two matrices only if the inner dimensions match. That is, if the first matrix has size $m \times n$ and the second has size $p \times q$ , the product is defined iff $n = p$ . The resulting matrix then has size $m \times q$ .

Let’s check each product one by one.

1. Check $AB$

$A$ is $3 \times 3$ (3 rows, 3 columns).

$B$ is $3 \times 1$ (3 rows, 1 column).

The inner dimensions: $3$ (columns of $A$ ) and $3$ (rows of $B$ ) are equal. So $AB$ is defined. The result will be a $3 \times 1$ matrix.

2. Check $AC$

$A$ is $3 \times 3$ .

$C$ is $1 \times 3$ (1 row, 3 columns).

Inner dimensions: $3$ (columns of $A$ ) and $1$ (rows of $C$ ) — these are not equal. So $AC$ is not defined. …
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CBSE 20241 markMCQ
$(\mathrm{A}+\mathrm{B})^{2}=\mathrm{A}^{2}+\mathrm{B}^{2}$
›Reveal solutionSolution
The given equation $(A+B)^2 = A^2 + B^2$ forces the cross terms to cancel, which means $AB = -BA$ . The correct option is (B).

Why This Works: The Core Idea

Matrix multiplication is not commutative — $AB$ is generally not equal to $BA$ . When you expand $(A+B)^2$ , you get $A^2 + AB + BA + B^2$ . The given condition says this equals $A^2 + B^2$ , so the middle terms $AB + BA$ must vanish. That gives $AB = -BA$ , a condition called anti-commutativity.

Watch out

A common mistake is to assume $AB = O$ (zero matrix) from $AB + BA = O$ . But that’s only one possibility — the matrices could be non-zero and still satisfy $AB = -BA$ . For example, take $A = \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}$ and $B = \begin{pmatrix}0 & 0 \\ 1 & 0\end{pmatrix}$ ; then $AB = \begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}$ and $BA = \begin{pmatrix}0 & 0 \\ 0 & 1\end{pmatrix}$ , so $AB = -BA$ holds but neither product is zero.

Step-by-Step Reasoning

Expand the square Since $A$ and $B$ are square matrices of the same order, we can multiply them. The distributive law holds for matrices, so:

$(A+B)^2 = (A+B)(A+B) = A^2 + AB + BA + B^2$

Apply the given condition The problem states:

$(A+B)^2 = A^2 + B^2$
Substituting the expansion:
$A^2 + AB + BA + B^2 = A^2 + B^2$

Cancel the common terms Subtract $A^2 + B^2$ from both sides:

$AB + BA = O$
where $O$ is the zero matrix of the same order.

Interpret the result The equation $AB + BA = O$ is equivalent to:

$AB = -BA$
This is the definition of anti-commuting matrices. It does not force $AB$ or $BA$ to be zero individually — only that they are negatives of each other. …
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CBSE 2026Set 65/2/11 markMCQ
$A$
(i) $(A + B)(A - B) = A^2 - B^2$
(ii) $AB = BA$
(iii) $(A + B)^2 = A^2 + AB + BA + B^2$
(iv) $AB = 0 \Rightarrow A = 0$ or $B = 0$ (A) Only
(i) and
(iii) (B) Only
(ii) and
(iii) (C) Only
(iii) (D) Only
(iii) and (iv)
›Reveal solutionSolution
Matrix multiplication is generally not commutative ( $AB \neq BA$ ), which means many algebraic identities from scalar arithmetic do not hold for matrices. Only statement (iii) is always true, making (C) the correct option.

Concept and Intuition

When we work with numbers (scalars), we are used to properties like $ab = ba$ (commutativity) and $ab = 0 \Rightarrow a = 0 \text{ or } b = 0$ . However, matrices behave differently. The most crucial distinction is that matrix multiplication is generally not commutative. This means that for two matrices $A$ and $B$ , $AB$ is usually not equal to $BA$ . This single property is the root cause for why many familiar algebraic identities, which rely on terms like $AB$ and $BA$ cancelling or combining, do not hold true for matrices.

Let's examine each statement with this fundamental understanding in mind.

Step-by-step Evaluation

Evaluate statement (i): $(A + B)(A - B) = A^2 - B^2$ To check if this is always true, we expand the left-hand side using the distributive property of matrix multiplication over addition, which does hold for matrices:

$(A + B)(A - B) = A(A - B) + B(A - B)$ $= A \cdot A - A \cdot B + B \cdot A - B \cdot B$ $= A^2 - AB + BA - B^2$
For this expression to be equal to $A^2 - B^2$, we would need the terms $-AB + BA$ to be zero. This implies $BA = AB$. However, as discussed, matrix multiplication is generally not commutative, meaning $AB \neq BA$ in most cases. > [!WARNING] > This is a classic pitfall! The identity $(x+y)(x-y) = x^2 - y^2$ is true for scalars because $xy = yx$. For matrices, this is only true if $A$ and $B$ commute. Therefore, statement (i) is not always true.

2. Evaluate statement (ii): $AB = BA$

This statement claims that matrix multiplication is always commutative. This is false. Matrix multiplication is generally not commutative. We can easily find counterexamples.

Consider:
$A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$
Then:
$AB = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 \cdot 1 + 1 \cdot 1 & 1 \cdot 0 + 1 \cdot 1 \\ 0 \cdot 1 + 1 \cdot 1 & 0 \cdot 0 + 1 \cdot 1 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$
And:
$BA = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 \cdot 1 + 0 \cdot 0 & 1 \cdot 1 + 0 \cdot 1 \\ 1 \cdot 1 + 1 \cdot 0 & 1 \cdot 1 + 1 \cdot 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$
Since $AB \neq BA$, statement (ii) is not always true.

3. Evaluate statement (iii): $(A + B)^2 = A^2 + AB + BA + B^2$

Let's expand the left-hand side:
$(A + B)^2 = (A + B)(A + B)$
Again, using the distributive property:
$= A(A + B) + B(A + B)$ $= A \cdot A + A \cdot B + B \cdot A + B \cdot B$
$$= A^2 + AB + BA + B^2$$ …
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CBSE 2026Set A1 markMCQ
$A = [1\ 2\ 3\ 4]$
(a) $[30]$
(b) $[10]$
(c) $[20]$
(d) $[40]$
›Reveal solutionSolution

A $1\times4$ row times a $4\times1$ column is the dot product $=30$ .

…

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CBSE 2026Set ANNUAL1 mark
$A = \begin{bmatrix}1 & -2 & 3\\ -4 & 2 & 5\end{bmatrix}$
›Reveal solutionSolution

Multiply the $2\times3$ matrix $A$ by the $3\times2$ matrix $B$ row-by-column.

$AB_{11}=1(2)+(-2)(4)+3(2)=2-8+6=0$

$AB_{12}=1(3)+(-2)(5)+3(1)=3-10+3=-4$

$AB_{21}=-4(2)+2(4)+5(2)=-8+8+10=10$

$AB_{22}=-4(3)+2(5)+5(1)=-12+10+5=3$ …

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CBSE 2026Set ANNUAL1 markMCQ
$A = \begin{bmatrix} 2 & 4 \\ 1 & 3 \end{bmatrix}$
(a) $\begin{bmatrix} 4 & 16 \\ 9 & 1 \end{bmatrix}$
(b) $\begin{bmatrix} 2 & 12 \\ 6 & 2 \end{bmatrix}$
(c) $\begin{bmatrix} 16 & 13 \\ 9 & 12 \end{bmatrix}$
(d) None of these
›Reveal solutionSolution

Multiply $A$ by itself using row-by-column matrix multiplication; the result matches none of the printed options.

Given $A = \begin{bmatrix} 2 & 4 \\ 1 & 3 \end{bmatrix}$ .

$A^2 = A \cdot A = \begin{bmatrix} 2 & 4 \\ 1 & 3 \end{bmatrix}\begin{bmatrix} 2 & 4 \\ 1 & 3 \end{bmatrix}$

Entry $(1,1)$ : $2(2)+4(1) = 4+4=8$

Entry $(1,2)$ : $2(4)+4(3) = 8+12=20$

Entry $(2,1)$ : $1(2)+3(1) = 2+3=5$

Entry $(2,2)$ : $1(4)+3(3) = 4+9=13$

…

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CBSE 2026Set ANNUAL1 mark
$AB$
›Reveal solutionSolution

Multiply the two $2\times2$ matrices row-by-column; every entry of the product turns out to be $0$ .

Given $A = \begin{bmatrix} 0 & -1 \\ 0 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 5 \\ 0 & 0 \end{bmatrix}$ .

$AB = \begin{bmatrix} 0 & -1 \\ 0 & 2 \end{bmatrix}\begin{bmatrix} 3 & 5 \\ 0 & 0 \end{bmatrix}$

Entry $(1,1)$ : $0(3)+(-1)(0) = 0$

Entry $(1,2)$ : $0(5)+(-1)(0) = 0$

Entry $(2,1)$ : $0(3)+2(0) = 0$

Entry $(2,2)$ : $0(5)+2(0) = 0$

…

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CBSE 2026Set ANNUAL1 mark
$A=[1\ \ 2\ \ 5\ \ 7]$
›Reveal solutionSolution
$A_{1\times4}B_{4\times1}\to 1\times1$ ; $B_{4\times1}A_{1\times4}\to 4\times4$ .

$A=[1\ 2\ 5\ 7]$ has order $1\times4$ . $B=\begin{bmatrix}6\\2\\4\\8\end{bmatrix}$ has order $4\times1$ .

$AB$ : $(1\times4)(4\times1)$ — inner dimensions ( $4$ ) agree, result order $1\times1$ . …
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CBSE 2025Set E1 markMCQ
$\begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix} \cdot \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} =$
(a) $\begin{bmatrix} 7 & 11 \\ 33 & 34 \end{bmatrix}$
(b) $\begin{bmatrix} 7 & 1 \\ 33 & 34 \end{bmatrix}$
(c) $\begin{bmatrix} 7 & 1 \\ 34 & 33 \end{bmatrix}$
(d) $\begin{bmatrix} 16 & 5 \\ 39 & 25 \end{bmatrix}$
›Reveal solutionSolution
Row-by-column multiplication gives $\begin{bmatrix} 7 & 1 \\ 33 & 34 \end{bmatrix}$ .

Multiply $\begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix}\begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}$ entry by entry:

$(1,1): 5\cdot2 + (-1)\cdot3 = 10 - 3 = 7$

$(1,2): 5\cdot1 + (-1)\cdot4 = 5 - 4 = 1$

$(2,1): 6\cdot2 + 7\cdot3 = 12 + 21 = 33$ …
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CBSE 2025Set E1 markMCQ
$\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} =$
(a) $\begin{bmatrix} 1 & 0 \\ 0 & 4 \end{bmatrix}$
(b) $\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$
(c) $\begin{bmatrix} 1 & 2 \\ 0 & 4 \end{bmatrix}$
(d) $\begin{bmatrix} 1 & 2 \\ 3 & 0 \end{bmatrix}$
›Reveal solutionSolution

$AI = A$ , so the product is the original matrix.

The second matrix $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ is the $2\times2$ identity $I$ . For any matrix $A$ , $AI = A$ . Hence …

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CBSE 2025Set E1 markMCQ
$\begin{bmatrix} 6 & 5 \end{bmatrix}\begin{bmatrix} -1 \\ 1 \end{bmatrix} =$
(a) $\begin{bmatrix} -6 & 5 \end{bmatrix}$
(b) $\begin{bmatrix} -6 \\ 5 \end{bmatrix}$
(c) $[-1]$
(d) $[1]$
›Reveal solutionSolution

$[6\ 5]\begin{bmatrix} -1 \\ 1 \end{bmatrix} = 6(-1)+5(1) = -1$ , a $1\times1$ matrix.

A $1\times2$ matrix times a $2\times1$ matrix gives a $1\times1$ matrix: …

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CBSE 2025Set E1 markMCQ
$\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} =$
(a) $\begin{bmatrix} 4 & 8 \\ 0 & 16 \end{bmatrix}$
(b) $\begin{bmatrix} 5 & 2 \\ 3 & 8 \end{bmatrix}$
(c) $\begin{bmatrix} 4 & 8 \\ 12 & 16 \end{bmatrix}$
(d) $\begin{bmatrix} 4 & 12 \\ 8 & 16 \end{bmatrix}$
›Reveal solutionSolution

$\begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} = 4I$ , so the product is $4\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ .

Since $\begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} = 4I$ ,

$\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}(4I) = 4\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 4 & 8 \\ 12 & 16 \end{bmatrix}.$ …

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←Previous questionQ93 — AA' is always a symmetric matrix for any matrix A.→Next questionQ95 — If A is skew symmetric matrix, then A² is a symmetric matrix.

Q.If $A = \begin{bmatrix} 2 & 3 & -1 \\ 1 & 4 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 3 \\ 4 & 5 \\ 2 & 1 \end{bmatrix}$ , then $AB$ and $BA$ are defined and equal.

Concept understanding — Matrix Multiplication Compatibility

Matrix Multiplication Compatibility: The Inner-Dimensions Rule

The Idea: A Row Meets a Column

The Precise Statement

Why the Rule Exists

Quick Check

Why this question is a trap

Step-by-step computation

Method: Deciding Whether $AB$ and $BA$ Can Be Equal (Shape-First)

Steps

Common Mistakes

Why This Works: The Core Idea

Step-by-Step Reasoning

Concept and Intuition

Step-by-step Evaluation

Q.If A=[23−1142]A = \begin{bmatrix} 2 & 3 & -1 \\ 1 & 4 & 2 \end{bmatrix}A=[21​34​−12​] and B=[234521]B = \begin{bmatrix} 2 & 3 \\ 4 & 5 \\ 2 & 1 \end{bmatrix}B=​242​351​​, then ABABAB and BABABA are defined and equal.

Why this question is a trap

Step-by-step computation

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Q.If $A = \begin{bmatrix} 2 & 3 & -1 \\ 1 & 4 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 3 \\ 4 & 5 \\ 2 & 1 \end{bmatrix}$ , then $AB$ and $BA$ are defined and equal.

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