Q.If $A = \begin{bmatrix} 0 & 6 & 7 \\ -6 & 0 & 8 \\ 7 & -8 & 0 \end{bmatrix}$, $B = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 2 \\ 1 & 2 & 0 \end{bmatrix}$, $C = \begin{bmatrix} 2 \\ -2 \\ 3 \end{bmatrix}$. Calculate $AC$, $BC$ and $(A + B)C$. Also, verify that $(A + B)C = AC + BC$.

Concept: Distributive property of matrix multiplication over addition.

Step 1 – Compute $AC$

$AC = \begin{bmatrix} 0 & 6 & 7 \\ -6 & 0 & 8 \\ 7 & -8 & 0 \end{bmatrix} \begin{bmatrix} 2 \\ -2 \\ 3 \end{bmatrix} = \begin{bmatrix} 0\cdot2 + 6\cdot(-2) + 7\cdot3 \\ -6\cdot2 + 0\cdot(-2) + 8\cdot3 \\ 7\cdot2 + (-8)\cdot(-2) + 0\cdot3 \end{bmatrix} = \begin{bmatrix} 0 -12 + 21 \\ -12 + 0 + 24 \\ 14 + 16 + 0 \end{bmatrix} = \begin{bmatrix} 9 \\ 12 \\ 30 \end{bmatrix}$

Step 2 – Compute $BC$

$BC = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 2 \\ 1 & 2 & 0 \end{bmatrix} \begin{bmatrix} 2 \\ -2 \\ 3 \end{bmatrix} = \begin{bmatrix} 0\cdot2 + 1\cdot(-2) + 1\cdot3 \\ 1\cdot2 + 0\cdot(-2) + 2\cdot3 \\ 1\cdot2 + 2\cdot(-2) + 0\cdot3 \end{bmatrix} = \begin{bmatrix} 0 -2 + 3 \\ 2 + 0 + 6 \\ 2 -4 + 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 8 \\ -2 \end{bmatrix}$

Step 3 – Compute $(A+B)C$

First, $A+B = \begin{bmatrix} 0 & 7 & 8 \\ -5 & 0 & 10 \\ 8 & -6 & 0 \end{bmatrix}$. Then

$(A+B)C = \begin{bmatrix} 0\cdot2 + 7\cdot(-2) + 8\cdot3 \\ -5\cdot2 + 0\cdot(-2) + 10\cdot3 \\ 8\cdot2 + (-6)\cdot(-2) + 0\cdot3 \end{bmatrix} = \begin{bmatrix} 0 -14 + 24 \\ -10 + 0 + 30 \\ 16 + 12 + 0 \end{bmatrix} = \begin{bmatrix} 10 \\ 20 \\ 28 \end{bmatrix}$

Step 4 – Verify

$AC + BC = \begin{bmatrix} 9 \\ 12 \\ 30 \end{bmatrix} + \begin{bmatrix} 1 \\ 8 \\ -2 \end{bmatrix} = \begin{bmatrix} 10 \\ 20 \\ 28 \end{bmatrix}$, which matches $(A+B)C$.

The core idea here is the distributive property of matrix multiplication: for matrices of compatible sizes,

$(A + B)C = AC + BC$. This is not just a rule to memorise — it follows from the fact that matrix multiplication is defined entry‑wise as a sum of products, and addition of matrices is entry‑wise. So when you multiply a sum of matrices by a column vector, each entry in the result is a sum of two separate dot products, which can be rearranged.

We are given three matrices: $A$ (a $3 \times 3$ skew‑symmetric matrix), $B$ (another $3 \times 3$ matrix), and $C$ (a $3 \times 1$ column vector). All multiplications are defined because the number of columns in $A$ and $B$ (3) matches the number of rows in $C$ (3).

Let’s compute step by step.

1. Compute $AC$

$A$ is $3 \times 3$, $C$ is $3 \times 1$, so $AC$ is $3 \times 1$.

• First entry (row 1 of $A$ dot $C$):

  $(0)(2) + (6)(-2) + (7)(3) = 0 - 12 + 21 = 9$

• Second entry (row 2 of $A$ dot $C$):

  $(-6)(2) + (0)(-2) + (8)(3) = -12 + 0 + 24 = 12$

• Third entry (row 3 of $A$ dot $C$):

  $(7)(2) + (-8)(-2) + (0)(3) = 14 + 16 + 0 = 30$

So

2. Compute $BC$

• First entry: $(0)(2) + (1)(-2) + (1)(3) = 0 - 2 + 3 = 1$
• Second entry: $(1)(2) + (0)(-2) + (2)(3) = 2 + 0 + 6 = 8$
• Third entry: $(1)(2) + (2)(-2) + (0)(3) = 2 - 4 + 0 = -2$

So

3. Compute $(A + B)C$

First, add $A$ and $B$ entry‑wise:

Now multiply by $C$:

• First entry: $(0)(2) + (7)(-2) + (8)(3) = 0 - 14 + 24 = 10$
• Second entry: $(-5)(2) + (0)(-2) + (10)(3) = -10 + 0 + 30 = 20$
• Third entry: $(8)(2) + (-6)(-2) + (0)(3) = 16 + 12 + 0 = 28$

So

4. Verify $(A + B)C = AC + BC$

We already have $AC = \begin{bmatrix} 9 \\ 12 \\ 30 \end{bmatrix}$ and $BC = \begin{bmatrix} 1 \\ 8 \\ -2 \end{bmatrix}$.

Add them entry‑wise:

This matches $(A + B)C$ exactly.

Method: Verifying the distributive law $(A+B)C = AC + BC$

Use this when asked to confirm distribution of matrix multiplication over addition.

Steps

Step 1: Compute the left side by adding first, then multiplying.

Form $A + B$ entry-wise, then multiply by $C$ (row-by-column).

Step 2: Compute the right side by multiplying first, then adding.

Find $AC$ and $BC$ separately, then add them entry-wise.

Step 3: Compare the two results.

They must be identical, confirming distributivity.

Common Mistakes

Mistake 1: Computing $(A+B)C$ as $AC \cdot BC$ or some product.

Why it's wrong: distribution gives a sum $AC + BC$, not a product of the two. Correct approach: add $AC$ and $BC$.

Mistake 2: Adding $A$ and $B$ after multiplying by $C$ on the left side incorrectly.

Why it's wrong: $(A+B)C$ means add $A,B$ first, then multiply once by $C$. Correct approach: respect the grouping — one multiplication on the left, two-then-add on the right.

Mistake 3: Sign errors in the column-vector dot products.

Why it's wrong: entries like $(6)(-2)+(7)(3)$ need careful signs. Correct approach: track the sign of every term.