If $A = \begin{bmatrix} 1 & 2 \\ -2 & 1 \end{bmatrix}$, $B = \begin{bmatrix} 2 & 3 \\ 3 & -4 \end{bmatrix}$ and $C = \begin{bmatrix} 1 & 0 \\ -1 & 0 \end{bmatrix}$, verify: (i) $(AB)C = A(BC)$ (ii) $A(B + C) = AB + AC$.

If A = 1 2 \-2 1 , B = 2 3 \3 -4 and C = 1 0 \-1 0 , verify: (i) (AB)C = A(BC) (ii) A(B + C) = AB + AC.

NCERT Exemplar · Q22

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

(i)

(AB)C = A(BC)

(ii)

A(B + C) = AB + AC

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Concept: Matrix multiplication is associative and distributive over addition, provided the dimensions are compatible.

(i) Associativity: $(AB)C = A(BC)$

First compute $AB$ :

AB = \begin{bmatrix} 1 & 2 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 3 & -4 \end{bmatrix} = \begin{bmatrix} 1\cdot2+2\cdot3 & 1\cdot3+2\cdot(-4) \\ -2\cdot2+1\cdot3 & -2\cdot3+1\cdot(-4) \end{bmatrix} = \begin{bmatrix} 8 & -5 \\ -1 & -10 \end{bmatrix}

Then $(AB)C$ :

(AB)C = \begin{bmatrix} 8 & -5 \\ -1 & -10 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} 8\cdot1+(-5)\cdot(-1) & 8\cdot0+(-5)\cdot0 \\ -1\cdot1+(-10)\cdot(-1) & -1\cdot0+(-10)\cdot0 \end{bmatrix} = \begin{bmatrix} 13 & 0 \\ 9 & 0 \end{bmatrix}

Now compute $BC$ :

BC = \begin{bmatrix} 2 & 3 \\ 3 & -4 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} 2\cdot1+3\cdot(-1) & 2\cdot0+3\cdot0 \\ 3\cdot1+(-4)\cdot(-1) & 3\cdot0+(-4)\cdot0 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 7 & 0 \end{bmatrix}

Then $A(BC)$ :

A(BC) = \begin{bmatrix} 1 & 2 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 7 & 0 \end{bmatrix} = \begin{bmatrix} 1\cdot(-1)+2\cdot7 & 1\cdot0+2\cdot0 \\ -2\cdot(-1)+1\cdot7 & -2\cdot0+1\cdot0 \end{bmatrix} = \begin{bmatrix} 13 & 0 \\ 9 & 0 \end{bmatrix}

Since both equal $\begin{bmatrix} 13 & 0 \\ 9 & 0 \end{bmatrix}$ , associativity holds.

(ii) Distributivity: $A(B + C) = AB + AC$

First $B+C$ :

B+C = \begin{bmatrix} 2 & 3 \\ 3 & -4 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} 3 & 3 \\ 2 & -4 \end{bmatrix}

Then $A(B+C)$ : …

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Matrix multiplication is associative and distributive over addition, just like ordinary multiplication — but only when the dimensions are compatible. For these given $2\times 2$ matrices, we verify both properties by direct computation: $(AB)C = A(BC)$ and $A(B+C) = AB + AC$ both hold.

Why This Works

Matrix multiplication is not commutative — $AB \neq BA$ in general — but it is associative and distributive. These properties are not automatic; they depend on the dimensions lining up correctly. Here, all three matrices are $2\times 2$ , so every product we write is defined and yields another $2\times 2$ matrix.

The associative law $(AB)C = A(BC)$ means we can group the multiplication any way we like, as long as the order of the matrices stays the same. The distributive law $A(B+C) = AB + AC$ means we can multiply a sum inside or add the products after — again, provided the dimensions match.

We'll verify both by computing each side separately and checking they are identical.

(i) Verifying $(AB)C = A(BC)$

Step 1: Compute $AB$

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

Multiply row by column:

First row, first column: $1\cdot 2 + 2\cdot 3 = 2 + 6 = 8$
First row, second column: $1\cdot 3 + 2\cdot(-4) = 3 - 8 = -5$
Second row, first column: $(-2)\cdot 2 + 1\cdot 3 = -4 + 3 = -1$
Second row, second column: $(-2)\cdot 3 + 1\cdot(-4) = -6 - 4 = -10$

AB = \begin{bmatrix} 8 & -5 \\ -1 & -10 \end{bmatrix}

Step 2: Compute $(AB)C$

(AB)C = \begin{bmatrix} 8 & -5 \\ -1 & -10 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -1 & 0 \end{bmatrix}

First row, first column: $8\cdot 1 + (-5)\cdot(-1) = 8 + 5 = 13$
First row, second column: $8\cdot 0 + (-5)\cdot 0 = 0$
Second row, first column: $(-1)\cdot 1 + (-10)\cdot(-1) = -1 + 10 = 9$
Second row, second column: $(-1)\cdot 0 + (-10)\cdot 0 = 0$

Thus

(AB)C = \begin{bmatrix} 13 & 0 \\ 9 & 0 \end{bmatrix}

Step 3: Compute $BC$

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

First row, first column: $2\cdot 1 + 3\cdot(-1) = 2 - 3 = -1$
First row, second column: $2\cdot 0 + 3\cdot 0 = 0$
Second row, first column: $3\cdot 1 + (-4)\cdot(-1) = 3 + 4 = 7$
Second row, second column: $3\cdot 0 + (-4)\cdot 0 = 0$

BC = \begin{bmatrix} -1 & 0 \\ 7 & 0 \end{bmatrix}

Step 4: Compute $A(BC)$

A(BC) = \begin{bmatrix} 1 & 2 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 7 & 0 \end{bmatrix}

First row, first column: $1\cdot(-1) + 2\cdot 7 = -1 + 14 = 13$
First row, second column: $1\cdot 0 + 2\cdot 0 = 0$
Second row, first column: $(-2)\cdot(-1) + 1\cdot 7 = 2 + 7 = 9$
Second row, second column: $(-2)\cdot 0 + 1\cdot 0 = 0$

Thus

A(BC) = \begin{bmatrix} 13 & 0 \\ 9 & 0 \end{bmatrix}

Step 5: Compare

Both $(AB)C$ and $A(BC)$ equal $\begin{bmatrix} 13 & 0 \\ 9 & 0 \end{bmatrix}$ . Associativity holds.

Watch out

A common mistake is to try to multiply $A(BC)$ by first computing $AB$ and then multiplying by $C$ — but that's exactly $(AB)C$ , not $A(BC)$ . The order of multiplication matters: in $A(BC)$ , you must multiply $B$ and $C$ first. Here, because of associativity, both give the same result, but the process is different.

(ii) Verifying $A(B + C) = AB + AC$

Step 1: Compute $B + C$

B + C = \begin{bmatrix} 2 & 3 \\ 3 & -4 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} 2+1 & 3+0 \\ 3+(-1) & -4+0 \end{bmatrix} = \begin{bmatrix} 3 & 3 \\ 2 & -4 \end{bmatrix}

Step 2: Compute $A(B + C)$

A(B + C) = \begin{bmatrix} 1 & 2 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} 3 & 3 \\ 2 & -4 \end{bmatrix}

First row, first column: $1\cdot 3 + 2\cdot 2 = 3 + 4 = 7$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
$A+B = \begin{bmatrix} 2 & 1 & 2 \\ 1 & 2 & 0 \\ 0 & 2 & 2 \end{bmatrix},\ AB = \begin{bmatrix} 1 & 2 & 2 \\ 1 & 1 & 0 \\ 1 & 2 & 1 \end{bmatrix}$
›Reveal solutionSolution
The expression $A^2 + B(A+B)$ simplifies to $(A+B)^2 - AB$ by using the distributive property of matrix multiplication. We calculate $(A+B)^2$ and then subtract $AB$ to find the result $\boxed{\begin{bmatrix} 4 & 6 & 6 \\ 3 & 4 & 2 \\ 1 & 6 & 3 \end{bmatrix}}$ .

The core idea here is to simplify the given expression $A^2 + B(A+B)$ using the properties of matrix algebra, specifically the distributive property of matrix multiplication over addition. We are given $A+B$ and $AB$ , so we should try to express the target expression in terms of these known quantities.

First, let's expand $B(A+B)$ :

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

So, the expression we need to evaluate becomes:

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

Now, let's recall the expansion of $(A+B)^2$ for matrices:
$A$
Watch out

It is crucial to remember that matrix multiplication is generally not commutative, meaning $AB \neq BA$ . Therefore, $(A+B)^2$ is not equal to $A^2 + 2AB + B^2$ unless $AB=BA$ .

Comparing the expression we need, $A^2 + BA + B^2$ , with the expansion of $(A+B)^2$ , which is $A^2 + AB + BA + B^2$ , we can see a direct relationship.

If we subtract $AB$ from $(A+B)^2$ , we get:

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

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

$= A^2 + 0 + BA + B^2$

$= A^2 + BA + B^2$

This shows that $A^2 + B(A+B) = (A+B)^2 - AB$ . This simplification is key because we are given $A+B$ and $AB$ directly.

Now, we can proceed with the calculations:

Calculate $(A+B)^2$ :

We are given $A+B = \begin{bmatrix} 2 & 1 & 2 \\ 1 & 2 & 0 \\ 0 & 2 & 2 \end{bmatrix}$ .

$(A+B)^2 = (A+B)(A+B) = \begin{bmatrix} 2 & 1 & 2 \\ 1 & 2 & 0 \\ 0 & 2 & 2 \end{bmatrix} \begin{bmatrix} 2 & 1 & 2 \\ 1 & 2 & 0 \\ 0 & 2 & 2 \end{bmatrix}$

Let's perform the matrix multiplication:

First row:

$(2)(2) + (1)(1) + (2)(0) = 4 + 1 + 0 = 5$

$(2)(1) + (1)(2) + (2)(2) = 2 + 2 + 4 = 8$

$(2)(2) + (1)(0) + (2)(2) = 4 + 0 + 4 = 8$

Second row:

$(1)(2) + (2)(1) + (0)(0) = 2 + 2 + 0 = 4$

$(1)(1) + (2)(2) + (0)(2) = 1 + 4 + 0 = 5$ …
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Q.If $A = \begin{bmatrix} 1 & 2 \\ -2 & 1 \end{bmatrix}$ , $B = \begin{bmatrix} 2 & 3 \\ 3 & -4 \end{bmatrix}$ and $C = \begin{bmatrix} 1 & 0 \\ -1 & 0 \end{bmatrix}$ , verify:

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 Works

(i) Verifying $(AB)C = A(BC)$

(ii) Verifying $A(B + C) = AB + AC$

Method: Verifying associativity and distributivity by computation

Steps

Common Mistakes

Q.If A=[12−21]A = \begin{bmatrix} 1 & 2 \\ -2 & 1 \end{bmatrix}A=[1−2​21​], B=[233−4]B = \begin{bmatrix} 2 & 3 \\ 3 & -4 \end{bmatrix}B=[23​3−4​] and C=[10−10]C = \begin{bmatrix} 1 & 0 \\ -1 & 0 \end{bmatrix}C=[1−1​00​], verify:

Why This Works

(i) Verifying (AB)C=A(BC)(AB)C = A(BC)(AB)C=A(BC)

(ii) Verifying A(B+C)=AB+ACA(B + C) = AB + ACA(B+C)=AB+AC

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

(i) Verifying $(AB)C = A(BC)$

(ii) Verifying $A(B + C) = AB + AC$

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