Q.Show that

Key idea: Matrix multiplication is not commutative — the order of multiplication matters, and the products are generally different even when both are defined.

(i) Compute $AB$ and $BA$ for $A = \begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix}$, $B = \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}$.

$AB = \begin{bmatrix} 5(2)+(-1)(3) & 5(1)+(-1)(4) \\ 6(2)+7(3) & 6(1)+7(4) \end{bmatrix} = \begin{bmatrix} 7 & 1 \\ 33 & 34 \end{bmatrix}$

$BA = \begin{bmatrix} 2(5)+1(6) & 2(-1)+1(7) \\ 3(5)+4(6) & 3(-1)+4(7) \end{bmatrix} = \begin{bmatrix} 16 & 5 \\ 39 & 25 \end{bmatrix}$

Since $\begin{bmatrix} 7 & 1 \\ 33 & 34 \end{bmatrix} \neq \begin{bmatrix} 16 & 5 \\ 39 & 25 \end{bmatrix}$, the products are unequal.

(ii) Compute $CD$ and $DC$ for $C = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}$, $D = \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix}$.

$CD = \begin{bmatrix} 1(-1)+2(0)+3(2) & 1(1)+2(-1)+3(3) & 1(0)+2(1)+3(4) \\ 0(-1)+1(0)+0(2) & 0(1)+1(-1)+0(3) & 0(0)+1(1)+0(4) \\ 1(-1)+1(0)+0(2) & 1(1)+1(-1)+0(3) & 1(0)+1(1)+0(4) \end{bmatrix} = \begin{bmatrix} 5 & 8 & 14 \\ 0 & -1 & 1 \\ -1 & 0 & 1 \end{bmatrix}$ …

Why Matrix Multiplication Isn't Like Number Multiplication

When you multiply two numbers, $3 \times 5$ always equals $5 \times 3$. But matrices are different. Each entry in a matrix product is a dot product of a row from the first matrix with a column from the second. Swap the matrices, and you're pairing entirely different rows with entirely different columns — so the result changes.

This is the core idea: compatibility for multiplication requires the number of columns in the first matrix to equal the number of rows in the second. When you reverse the order, that condition may still hold (as it does here), but the actual arithmetic is completely different.

Part (i): $2 \times 2$ Matrices

Let

$A = \begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix}$,

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

Both are $2 \times 2$, so $AB$ and $BA$ are both defined and also $2 \times 2$.

Step 1: Compute $AB$

Multiply $A$ (rows) by $B$ (columns):

• Entry $(1,1)$: row 1 of $A$ $\cdot$ column 1 of $B$ $= (5)(2) + (-1)(3) = 10 - 3 = 7$
• Entry $(1,2)$: row 1 of $A$ $\cdot$ column 2 of $B$ $= (5)(1) + (-1)(4) = 5 - 4 = 1$
• Entry $(2,1)$: row 2 of $A$ $\cdot$ column 1 of $B$ $= (6)(2) + (7)(3) = 12 + 21 = 33$
• Entry $(2,2)$: row 2 of $A$ $\cdot$ column 2 of $B$ $= (6)(1) + (7)(4) = 6 + 28 = 34$

So

$AB = \begin{bmatrix} 7 & 1 \\ 33 & 34 \end{bmatrix}$.

Step 2: Compute $BA$

Now multiply $B$ (rows) by $A$ (columns):

• Entry $(1,1)$: row 1 of $B$ $\cdot$ column 1 of $A$ $= (2)(5) + (1)(6) = 10 + 6 = 16$
• Entry $(1,2)$: row 1 of $B$ $\cdot$ column 2 of $A$ $= (2)(-1) + (1)(7) = -2 + 7 = 5$
• Entry $(2,1)$: row 2 of $B$ $\cdot$ column 1 of $A$ $= (3)(5) + (4)(6) = 15 + 24 = 39$
• Entry $(2,2)$: row 2 of $B$ $\cdot$ column 2 of $A$ $= (3)(-1) + (4)(7) = -3 + 28 = 25$

So

$BA = \begin{bmatrix} 16 & 5 \\ 39 & 25 \end{bmatrix}$.

Step 3: Compare

$AB = \begin{bmatrix} 7 & 1 \\ 33 & 34 \end{bmatrix}$

$BA = \begin{bmatrix} 16 & 5 \\ 39 & 25 \end{bmatrix}$

Every single entry is different. Clearly $AB \neq BA$.

Part (ii): $3 \times 3$ Matrices

Let

$P = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}$,

$Q = \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix}$.

Both are $3 \times 3$, so $PQ$ and $QP$ are both $3 \times 3$.

Step 1: Compute $PQ$

Multiply $P$ (rows) by $Q$ (columns). We'll do it entry by entry.

Row 1 of $P$: $[1, 2, 3]$

• Column 1 of $Q$: $[-1, 0, 2]^T$ $(1)(-1) + (2)(0) + (3)(2) = -1 + 0 + 6 = 5$
• Column 2 of $Q$: $[1, -1, 3]^T$ $(1)(1) + (2)(-1) + (3)(3) = 1 - 2 + 9 = 8$
• Column 3 of $Q$: $[0, 1, 4]^T$ $(1)(0) + (2)(1) + (3)(4) = 0 + 2 + 12 = 14$

Row 2 of $P$: $[0, 1, 0]$

• Column 1: $(0)(-1) + (1)(0) + (0)(2) = 0$
• Column 2: $(0)(1) + (1)(-1) + (0)(3) = -1$
• Column 3: $(0)(0) + (1)(1) + (0)(4) = 1$

Row 3 of $P$: $[1, 1, 0]$

• Column 1: $(1)(-1) + (1)(0) + (0)(2) = -1$
• Column 2: $(1)(1) + (1)(-1) + (0)(3) = 1 - 1 = 0$
• Column 3: $(1)(0) + (1)(1) + (0)(4) = 1$

So

$PQ = \begin{bmatrix} 5 & 8 & 14 \\ 0 & -1 & 1 \\ -1 & 0 & 1 \end{bmatrix}$.

Step 2: Compute $QP$

Now multiply $Q$ (rows) by $P$ (columns).

Row 1 of $Q$: $[-1, 1, 0]$

• Column 1 of $P$: $[1, 0, 1]^T$ …

Method: Showing two matrix products are unequal (non-commutativity)

To demonstrate $AB \neq BA$, compute both products in full and point to at least one entry that differs — a single mismatched entry is enough.

Steps

Step 1: Check both products are defined

Confirm the inner dimensions match for each order; for square matrices of the same size both $AB$ and $BA$ exist and share that size.

Step 2: Compute each product with the row-by-column rule …

Common Mistakes

Mistake 1: Assuming that because both products are defined they must be equal

Why it's wrong: matrix multiplication is generally non-commutative; being defined in both orders says nothing about equality. Correct approach: actually compute both products and compare.

Mistake 2: Reversing the row-by-column rule …