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

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

First compute $AB$:

Then $(AB)C$:

Now compute $BC$:

Then $A(BC)$:

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

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

First $B+C$:

Then $A(B+C)$: …

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 $AB$

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: $(-2)\cdot 3 + 1\cdot(-4) = -6 - 4 = -10$

So

Step 2: Compute $(AB)C$

• 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

Step 3: Compute $BC$

• 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: $3\cdot 0 + (-4)\cdot 0 = 0$

So

Step 4: Compute $A(BC)$

• 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

Step 5: Compare

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

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

Step 1: Compute $B + C$

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

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

Method: Verifying associativity and distributivity by computation

Use this for identities like $(AB)C=A(BC)$ and $A(B+C)=AB+AC$.

Steps

Step 1: For associativity, group two different ways.

Compute $AB$ first, then $(AB)C$; separately compute $BC$ first, then $A(BC)$. The order of the matrices stays the same — only the grouping changes.

Step 2: For distributivity, expand both sides. …

Common Mistakes

Mistake 1: Changing the order of the matrices when testing associativity.

Why it's wrong: associativity only regroups the same ordered product ($(AB)C=A(BC)$); it does not let you swap factors. Correct approach: keep the sequence $A,B,C$ and change only the brackets.

Mistake 2: Distributing on the wrong side. …