If $A = \begin{bmatrix} 1 & 1 & -1 \\ 2 & 0 & 3 \\ 3 & -1 & 2 \end{bmatrix}$, $B = \begin{bmatrix} 1 & 3 \\ 0 & 2 \\ -1 & 4 \end{bmatrix}$ and $C = \begin{bmatrix} 1 & 2 & 3 & -4 \\ 2 & 0 & -2 & 1 \end{bmatrix}$, find $A(BC)$, $(AB)C$ and show that $(AB)C = A(BC)$.

If A = 1 1 -1 \2 0 3 \3 -1 2 , B = 1 3 \0 2 \-1 4 and C = 1 2 3 -4 \2 0 -2 1 , find A(BC), (AB)C and show that

Worked Examples · Example 16

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

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Concept understanding — Matrix Multiplication Compatibility

Matrix Multiplication Compatibility: The Inner-Dimensions Rule

You cannot multiply just any two matrices. Multiplication is defined only when their sizes line up in a specific way, and this compatibility check is always the very first step of any product.

The Idea: A Row Meets a Column

When you multiply $A$ by $B$ , you take each row of $A$ and pair it against each column of $B$ , multiply corresponding entries, and add. For that pairing to work, a row of $A$ must have exactly as many entries as a column of $B$ .

Note

Think of a handshake: each finger of one hand must meet a finger of the other. If one hand has $4$ fingers and the other has $3$ , the handshake fails.

The Precise Statement

Let $A$ be $m \times n$ and $B$ be $p \times q$ .

$A \times B$ is defined if and only if $n = p$ — the number of columns of $A$ equals the number of rows of $B$ . The product $C = AB$ then has order $m \times q$ .

Writing the sizes side by side, $(m \times \mathbf{n})(\mathbf{p} \times q)$ , the inner numbers ( $n, p$ ) must match; the outer numbers ( $m, q$ ) give the result's shape.

Why the Rule Exists

Each entry of the product is

c_{ij} = \sum_{k=1}^{n} a_{ik}\, b_{kj}.

Here $k$ runs over the columns of $A$ (up to $n$ ) and the rows of $B$ (up to $p$ ). If $n \neq p$ , the sum runs out of matching terms and is meaningless — that is exactly why compatibility demands $n = p$ .

Watch out

Even when both $AB$ and $BA$ are defined, they usually differ. For $A$ of order $2 \times 3$ and $B$ of order $3 \times 2$ , $AB$ is $2 \times 2$ but $BA$ is $3 \times 3$ — different sizes entirely. Matrix multiplication is not commutative.

Quick Check

$A$	$B$	Defined?	Result
$2 \times 3$	$3 \times 4$	Yes	$2 \times 4$
$2 \times 3$	$2 \times 4$	No	—
$1 \times 4$	$4 \times 1$	Yes	$1 \times 1$

Tip

Before multiplying, write both orders side by side and circle the inner numbers. If they are equal, multiply; if not, stop — the product does not exist.

Matrix Multiplication Compatibility — the rule that the number of columns of the first matrix must equal the number of rows of the second — is one of the first checks taught in the CBSE Class 12 Matrices chapter, and "matrix multiplication rules class 12" is a common search among students preparing for board exams and JEE Main. NCERT's own solved examples emphasize checking this condition before attempting any product.

Concept: Matrix Multiplication Compatibility — multiplication is defined only when the number of columns in the first matrix equals the number of rows in the second. The associative property $(AB)C = A(BC)$ holds whenever all products are defined.

Step 1: Compute $BC$

$B$ is $3 \times 2$ , $C$ is $2 \times 4$ , so $BC$ is $3 \times 4$ .

BC = \begin{bmatrix} 1 & 3 \\ 0 & 2 \\ -1 & 4 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 & -4 \\ 2 & 0 & -2 & 1 \end{bmatrix} = \begin{bmatrix} 1(1)+3(2) & 1(2)+3(0) & 1(3)+3(-2) & 1(-4)+3(1) \\ 0(1)+2(2) & 0(2)+2(0) & 0(3)+2(-2) & 0(-4)+2(1) \\ -1(1)+4(2) & -1(2)+4(0) & -1(3)+4(-2) & -1(-4)+4(1) \end{bmatrix} = \begin{bmatrix} 7 & 2 & -3 & -1 \\ 4 & 0 & -4 & 2 \\ 7 & -2 & -11 & 8 \end{bmatrix}

Step 2: Compute $A(BC)$

$A$ is $3 \times 3$ , $BC$ is $3 \times 4$ , product is $3 \times 4$ .

A(BC) = \begin{bmatrix} 1 & 1 & -1 \\ 2 & 0 & 3 \\ 3 & -1 & 2 \end{bmatrix} \begin{bmatrix} 7 & 2 & -3 & -1 \\ 4 & 0 & -4 & 2 \\ 7 & -2 & -11 & 8 \end{bmatrix} = \begin{bmatrix} 1(7)+1(4)-1(7) & 1(2)+1(0)-1(-2) & 1(-3)+1(-4)-1(-11) & 1(-1)+1(2)-1(8) \\ 2(7)+0(4)+3(7) & 2(2)+0(0)+3(-2) & 2(-3)+0(-4)+3(-11) & 2(-1)+0(2)+3(8) \\ 3(7)-1(4)+2(7) & 3(2)-1(0)+2(-2) & 3(-3)-1(-4)+2(-11) & 3(-1)-1(2)+2(8) \end{bmatrix} = \begin{bmatrix} 4 & 4 & 4 & -7 \\ 35 & -2 & -39 & 22 \\ 31 & 2 & -27 & 11 \end{bmatrix}

Step 3: Compute $AB$

$A$ is $3 \times 3$ , $B$ is $3 \times 2$ , product is $3 \times 2$ .

AB = \begin{bmatrix} 1 & 1 & -1 \\ 2 & 0 & 3 \\ 3 & -1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ 0 & 2 \\ -1 & 4 \end{bmatrix} = \begin{bmatrix} 1(1)+1(0)-1(-1) & 1(3)+1(2)-1(4) \\ 2(1)+0(0)+3(-1) & 2(3)+0(2)+3(4) \\ 3(1)-1(0)+2(-1) & 3(3)-1(2)+2(4) \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ -1 & 18 \\ 1 & 15 \end{bmatrix}

Step 4: Compute $(AB)C$

$AB$ is $3 \times 2$ , $C$ is $2 \times 4$ , product is $3 \times 4$ .

(AB)C = \begin{bmatrix} 2 & 1 \\ -1 & 18 \\ 1 & 15 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 & -4 \\ 2 & 0 & -2 & 1 \end{bmatrix} = \begin{bmatrix} 2(1)+1(2) & 2(2)+1(0) & 2(3)+1(-2) & 2(-4)+1(1) \\ -1(1)+18(2) & -1(2)+18(0) & -1(3)+18(-2) & -1(-4)+18(1) \\ 1(1)+15(2) & 1(2)+15(0) & 1(3)+15(-2) & 1(-4)+15(1) \end{bmatrix} = \begin{bmatrix} 4 & 4 & 4 & -7 \\ 35 & -2 & -39 & 22 \\ 31 & 2 & -27 & 11 \end{bmatrix}

Since $A(BC)$ and $(AB)C$ both equal the same matrix, associativity is verified.

✓Final answer

A(BC) = (AB)C = \boxed{\begin{bmatrix} 4 & 4 & 4 & -7 \\ 35 & -2 & -39 & 22 \\ 31 & 2 & -27 & 11 \end{bmatrix}}

Matrix multiplication is associative: $(AB)C = A(BC)$ always holds when the products are defined. For these matrices, both sides equal a $3 \times 4$ matrix, confirming the property.

The key idea here is that matrix multiplication is associative — but that's not just a rule to memorise; it follows from how multiplication is defined. When you multiply matrices, you're really composing linear transformations, and composition of functions is always associative. So $(AB)C$ and $A(BC)$ must give the same result, provided the dimensions allow both products.

Let's check the dimensions first. $A$ is $3 \times 3$ , $B$ is $3 \times 2$ , $C$ is $2 \times 4$ . For $BC$ , we multiply a $3 \times 2$ by a $2 \times 4$ — the inner dimensions match (2), so $BC$ is $3 \times 4$ . Then $A(BC)$ is $3 \times 3$ times $3 \times 4$ , giving $3 \times 4$ . For $(AB)C$ , $AB$ is $3 \times 3$ times $3 \times 2$ , giving $3 \times 2$ ; then multiplying by $C$ ( $2 \times 4$ ) gives $3 \times 4$ . Both paths yield a $3 \times 4$ matrix, so the product is defined both ways.

Now let's compute both sides step by step.

Compute $BC$ first.

$B$ is $3 \times 2$ , $C$ is $2 \times 4$ . The $(i,j)$ entry of $BC$ is the dot product of row $i$ of $B$ with column $j$ of $C$ .

Row 1 of $B$ : $[1, 3]$ .

Column 1 of $C$ : $\begin{bmatrix}1 \\ 2\end{bmatrix}$ → $1\cdot1 + 3\cdot2 = 1 + 6 = 7$

Column 2: $\begin{bmatrix}2 \\ 0\end{bmatrix}$ → $1\cdot2 + 3\cdot0 = 2$

Column 3: $\begin{bmatrix}3 \\ -2\end{bmatrix}$ → $1\cdot3 + 3\cdot(-2) = 3 - 6 = -3$

Column 4: $\begin{bmatrix}-4 \\ 1\end{bmatrix}$ → $1\cdot(-4) + 3\cdot1 = -4 + 3 = -1$

So first row of $BC$ : $[7, 2, -3, -1]$ .

Row 2 of $B$ : $[0, 2]$ .

Column 1: $0\cdot1 + 2\cdot2 = 4$

Column 2: $0\cdot2 + 2\cdot0 = 0$

Column 3: $0\cdot3 + 2\cdot(-2) = -4$

Column 4: $0\cdot(-4) + 2\cdot1 = 2$

Second row: $[4, 0, -4, 2]$ .

Row 3 of $B$ : $[-1, 4]$ .

Column 1: $(-1)\cdot1 + 4\cdot2 = -1 + 8 = 7$

Column 2: $(-1)\cdot2 + 4\cdot0 = -2$

Column 3: $(-1)\cdot3 + 4\cdot(-2) = -3 - 8 = -11$

Column 4: $(-1)\cdot(-4) + 4\cdot1 = 4 + 4 = 8$

Third row: $[7, -2, -11, 8]$ .

So

BC = \begin{bmatrix} 7 & 2 & -3 & -1 \\ 4 & 0 & -4 & 2 \\ 7 & -2 & -11 & 8 \end{bmatrix}

Now compute $A(BC)$ .

$A$ is $3 \times 3$ , $BC$ is $3 \times 4$ . Multiply row $i$ of $A$ by each column of $BC$ .

Row 1 of $A$ : $[1, 1, -1]$ .

Column 1 of $BC$ : $\begin{bmatrix}7 \\ 4 \\ 7\end{bmatrix}$ → $1\cdot7 + 1\cdot4 + (-1)\cdot7 = 7 + 4 - 7 = 4$

Column 2: $\begin{bmatrix}2 \\ 0 \\ -2\end{bmatrix}$ → $1\cdot2 + 1\cdot0 + (-1)\cdot(-2) = 2 + 0 + 2 = 4$

Column 3: $\begin{bmatrix}-3 \\ -4 \\ -11\end{bmatrix}$ → $1\cdot(-3) + 1\cdot(-4) + (-1)\cdot(-11) = -3 - 4 + 11 = 4$

Column 4: $\begin{bmatrix}-1 \\ 2 \\ 8\end{bmatrix}$ → $1\cdot(-1) + 1\cdot2 + (-1)\cdot8 = -1 + 2 - 8 = -7$

First row of $A(BC)$ : $[4, 4, 4, -7]$ .

Row 2 of $A$ : $[2, 0, 3]$ .

Column 1: $2\cdot7 + 0\cdot4 + 3\cdot7 = 14 + 0 + 21 = 35$

Column 2: $2\cdot2 + 0\cdot0 + 3\cdot(-2) = 4 + 0 - 6 = -2$

Column 3: $2\cdot(-3) + 0\cdot(-4) + 3\cdot(-11) = -6 + 0 - 33 = -39$

Column 4: $2\cdot(-1) + 0\cdot2 + 3\cdot8 = -2 + 0 + 24 = 22$

Second row: $[35, -2, -39, 22]$ .

Row 3 of $A$ : $[3, -1, 2]$ .

Column 1: $3\cdot7 + (-1)\cdot4 + 2\cdot7 = 21 - 4 + 14 = 31$

Column 2: $3\cdot2 + (-1)\cdot0 + 2\cdot(-2) = 6 + 0 - 4 = 2$

Column 3: $3\cdot(-3) + (-1)\cdot(-4) + 2\cdot(-11) = -9 + 4 - 22 = -27$

Column 4: $3\cdot(-1) + (-1)\cdot2 + 2\cdot8 = -3 - 2 + 16 = 11$

Third row: $[31, 2, -27, 11]$ .

So

A(BC) = \begin{bmatrix} 4 & 4 & 4 & -7 \\ 35 & -2 & -39 & 22 \\ 31 & 2 & -27 & 11 \end{bmatrix}

Now compute $AB$ first.

$A$ is $3 \times 3$ , $B$ is $3 \times 2$ . Multiply row $i$ of $A$ by each column of $B$ .

Row 1 of $A$ : $[1, 1, -1]$ .

Column 1 of $B$ : $\begin{bmatrix}1 \\ 0 \\ -1\end{bmatrix}$ → $1\cdot1 + 1\cdot0 + (-1)\cdot(-1) = 1 + 0 + 1 = 2$

Column 2: $\begin{bmatrix}3 \\ 2 \\ 4\end{bmatrix}$ → $1\cdot3 + 1\cdot2 + (-1)\cdot4 = 3 + 2 - 4 = 1$

First row of $AB$ : $[2, 1]$ .

Row 2 of $A$ : $[2, 0, 3]$ .

Column 1: $2\cdot1 + 0\cdot0 + 3\cdot(-1) = 2 + 0 - 3 = -1$

Column 2: $2\cdot3 + 0\cdot2 + 3\cdot4 = 6 + 0 + 12 = 18$

Second row: $[-1, 18]$ .

Row 3 of $A$ : $[3, -1, 2]$ .

Column 1: $3\cdot1 + (-1)\cdot0 + 2\cdot(-1) = 3 + 0 - 2 = 1$

Column 2: $3\cdot3 + (-1)\cdot2 + 2\cdot4 = 9 - 2 + 8 = 15$

Third row: $[1, 15]$ .

So

AB = \begin{bmatrix} 2 & 1 \\ -1 & 18 \\ 1 & 15 \end{bmatrix}

Now compute $(AB)C$ .

$AB$ is $3 \times 2$ , $C$ is $2 \times 4$ . Multiply row $i$ of $AB$ by each column of $C$ .

Row 1 of $AB$ : $[2, 1]$ .

Column 1 of $C$ : $\begin{bmatrix}1 \\ 2\end{bmatrix}$ → $2\cdot1 + 1\cdot2 = 2 + 2 = 4$

Column 2: $\begin{bmatrix}2 \\ 0\end{bmatrix}$ → $2\cdot2 + 1\cdot0 = 4$

Column 3: $\begin{bmatrix}3 \\ -2\end{bmatrix}$ → $2\cdot3 + 1\cdot(-2) = 6 - 2 = 4$

Column 4: $\begin{bmatrix}-4 \\ 1\end{bmatrix}$ → $2\cdot(-4) + 1\cdot1 = -8 + 1 = -7$

First row: $[4, 4, 4, -7]$ .

Row 2 of $AB$ : $[-1, 18]$ .

Column 1: $(-1)\cdot1 + 18\cdot2 = -1 + 36 = 35$

Column 2: $(-1)\cdot2 + 18\cdot0 = -2$

Column 3: $(-1)\cdot3 + 18\cdot(-2) = -3 - 36 = -39$

Column 4: $(-1)\cdot(-4) + 18\cdot1 = 4 + 18 = 22$

Second row: $[35, -2, -39, 22]$ .

Row 3 of $AB$ : $[1, 15]$ .

Column 1: $1\cdot1 + 15\cdot2 = 1 + 30 = 31$

Column 2: $1\cdot2 + 15\cdot0 = 2$

Column 3: $1\cdot3 + 15\cdot(-2) = 3 - 30 = -27$

Column 4: $1\cdot(-4) + 15\cdot1 = -4 + 15 = 11$

Third row: $[31, 2, -27, 11]$ .

So

(AB)C = \begin{bmatrix} 4 & 4 & 4 & -7 \\ 35 & -2 & -39 & 22 \\ 31 & 2 & -27 & 11 \end{bmatrix}

Compare the two results. $A(BC)$ and $(AB)C$ are identical matrices — every entry matches exactly. This confirms the associative property of matrix multiplication for these specific matrices.

Watch out

A common mistake is to assume that because matrix multiplication is not commutative ( $AB \neq BA$ in general), it might not be associative either. That's wrong — associativity always holds when the products are defined. Don't confuse commutativity with associativity.

Tip

You could have saved time by noticing that both products must be equal by the associative law, so computing just one side would suffice. But the problem explicitly asks you to compute both and show they're equal, so the full work is needed.

✓Final answer

A(BC)

←Previous section · 3.4.6Properties of Multiplication of Matrices →Next question · 3.4.6Example 17 — If A = 0 6 7 \-6 0 8 \7 -8 0 , B = 0 1 1 \1 0 2 \1 2 0 , C = 2 \-2 \3 …

Q.If $A = \begin{bmatrix} 1 & 1 & -1 \\ 2 & 0 & 3 \\ 3 & -1 & 2 \end{bmatrix}$ , $B = \begin{bmatrix} 1 & 3 \\ 0 & 2 \\ -1 & 4 \end{bmatrix}$ and $C = \begin{bmatrix} 1 & 2 & 3 & -4 \\ 2 & 0 & -2 & 1 \end{bmatrix}$ , find $A(BC)$ , $(AB)C$ and show that $(AB)C = A(BC)$ .

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

Method: Verifying associativity $(AB)C = A(BC)$

Steps

Common Mistakes

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