Q.Compute the indicated products.

Concept: Matrix multiplication is defined only when the number of columns in the first matrix equals the number of rows in the second. The product's $(i,j)$ entry is the dot product of row $i$ of the first with column $j$ of the second.

(i) $\begin{bmatrix} a & b \\ -b & a \end{bmatrix} \begin{bmatrix} a & -b \\ b & a \end{bmatrix}$

Row 1: $(a)(a)+(b)(b)=a^2+b^2$, $(a)(-b)+(b)(a)=0$

Row 2: $(-b)(a)+(a)(b)=0$, $(-b)(-b)+(a)(a)=a^2+b^2$

(ii) $\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \begin{bmatrix} 2 & 3 & 4 \end{bmatrix}$

A $3\times1$ times a $1\times3$ gives a $3\times3$ matrix. Each entry is the product of the row element and the column element:

(iii) $\begin{bmatrix} 1 & -2 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix}$

$2\times2$ times $2\times3$ gives $2\times3$. Compute each entry:

Row 1: $(1)(1)+(-2)(2)=-3$, $(1)(2)+(-2)(3)=-4$, $(1)(3)+(-2)(1)=1$

Row 2: $(2)(1)+(3)(2)=8$, $(2)(2)+(3)(3)=13$, $(2)(3)+(3)(1)=9$

(iv) $\begin{bmatrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{bmatrix} \begin{bmatrix} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{bmatrix}$

$3\times3$ times $3\times3$ gives $3\times3$. Compute systematically:

$(1,1)=2\cdot1+3\cdot0+4\cdot3=14$, $(1,2)=2(-3)+3\cdot2+4\cdot0=0$, $(1,3)=2\cdot5+3\cdot4+4\cdot5=42$

$(2,1)=3\cdot1+4\cdot0+5\cdot3=18$, $(2,2)=3(-3)+4\cdot2+5\cdot0=-1$, $(2,3)=3\cdot5+4\cdot4+5\cdot5=56$

$(3,1)=4\cdot1+5\cdot0+6\cdot3=22$, $(3,2)=4(-3)+5\cdot2+6\cdot0=-2$, $(3,3)=4\cdot5+5\cdot4+6\cdot5=70$

(v) $\begin{bmatrix} 2 & 1 \\ 3 & 2 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 1 \\ -1 & 2 & 1 \end{bmatrix}$

$3\times2$ times $2\times3$ gives $3\times3$:

Row 1: $(2)(1)+(1)(-1)=1$, $(2)(0)+(1)(2)=2$, $(2)(1)+(1)(1)=3$

Row 2: $(3)(1)+(2)(-1)=1$, $(3)(0)+(2)(2)=4$, $(3)(1)+(2)(1)=5$

Row 3: $(-1)(1)+(1)(-1)=-2$, $(-1)(0)+(1)(2)=2$, $(-1)(1)+(1)(1)=0$

(vi) $\begin{bmatrix} 3 & -1 & 3 \\ -1 & 0 & 2 \end{bmatrix} \begin{bmatrix} 2 & -3 \\ 1 & 0 \\ 3 & 1 \end{bmatrix}$

$2\times3$ times $3\times2$ gives $2\times2$:

Row 1: $(3)(2)+(-1)(1)+(3)(3)=14$, $(3)(-3)+(-1)(0)+(3)(1)=-6$

Row 2: $(-1)(2)+(0)(1)+(2)(3)=4$, $(-1)(-3)+(0)(0)+(2)(1)=5$

The Core Idea: Why This Works

Matrix multiplication is not just a mechanical shuffle of numbers — it’s a way to combine two linear transformations. But at the arithmetic level, the rule is simple: you can multiply an $m \times n$ matrix $A$ by an $n \times p$ matrix $B$ to get an $m \times p$ product $AB$. The entry in row $i$, column $j$ of $AB$ is the sum of products of corresponding entries from row $i$ of $A$ and column $j$ of $B$.

Let’s apply this to each part.

(i) $\begin{bmatrix} a & b \\ -b & a \end{bmatrix} \begin{bmatrix} a & -b \\ b & a \end{bmatrix}$

Both matrices are $2 \times 2$, so the product is $2 \times 2$.

1. Entry (1,1): Row 1 of first: $[a, b]$, Column 1 of second: $\begin{bmatrix} a \\ b \end{bmatrix}$.

   Dot product: $a \cdot a + b \cdot b = a^2 + b^2$.

2. Entry (1,2): Row 1 of first: $[a, b]$, Column 2 of second: $\begin{bmatrix} -b \\ a \end{bmatrix}$.

   Dot product: $a(-b) + b(a) = -ab + ab = 0$.

3. Entry (2,1): Row 2 of first: $[-b, a]$, Column 1 of second: $\begin{bmatrix} a \\ b \end{bmatrix}$.

   Dot product: $(-b)(a) + a(b) = -ab + ab = 0$.

4. Entry (2,2): Row 2 of first: $[-b, a]$, Column 2 of second: $\begin{bmatrix} -b \\ a \end{bmatrix}$.

   Dot product: $(-b)(-b) + a(a) = b^2 + a^2 = a^2 + b^2$.

So the product is $\begin{bmatrix} a^2+b^2 & 0 \\ 0 & a^2+b^2 \end{bmatrix}$.

(ii) $\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \begin{bmatrix} 2 & 3 & 4 \end{bmatrix}$

First matrix is $3 \times 1$, second is $1 \times 3$. The product is $3 \times 3$.

1. The first matrix has only one column; the second has only one row. So each entry $(i,j)$ is simply the product of the $i$-th entry of the first matrix and the $j$-th entry of the second.

2. Row 1 of first: $[1]$, Column 1 of second: $[2]$ → $1 \cdot 2 = 2$

   Row 1, Col 2: $1 \cdot 3 = 3$

   Row 1, Col 3: $1 \cdot 4 = 4$

3. Row 2: $[2]$ times each column entry: $2 \cdot 2 = 4$, $2 \cdot 3 = 6$, $2 \cdot 4 = 8$

4. Row 3: $[3]$ times each: $3 \cdot 2 = 6$, $3 \cdot 3 = 9$, $3 \cdot 4 = 12$

Thus the product is $\begin{bmatrix} 2 & 3 & 4 \\ 4 & 6 & 8 \\ 6 & 9 & 12 \end{bmatrix}$.

(iii) $\begin{bmatrix} 1 & -2 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix}$

First is $2 \times 2$, second is $2 \times 3$, so product is $2 \times 3$.

1. Row 1, Col 1: $[1, -2]$ dot $\begin{bmatrix} 1 \\ 2 \end{bmatrix} = 1(1) + (-2)(2) = 1 - 4 = -3$

2. Row 1, Col 2: $[1, -2]$ dot $\begin{bmatrix} 2 \\ 3 \end{bmatrix} = 1(2) + (-2)(3) = 2 - 6 = -4$

3. Row 1, Col 3: $[1, -2]$ dot $\begin{bmatrix} 3 \\ 1 \end{bmatrix} = 1(3) + (-2)(1) = 3 - 2 = 1$

4. Row 2, Col 1: $[2, 3]$ dot $\begin{bmatrix} 1 \\ 2 \end{bmatrix} = 2(1) + 3(2) = 2 + 6 = 8$

5. Row 2, Col 2: $[2, 3]$ dot $\begin{bmatrix} 2 \\ 3 \end{bmatrix} = 2(2) + 3(3) = 4 + 9 = 13$

6. Row 2, Col 3: $[2, 3]$ dot $\begin{bmatrix} 3 \\ 1 \end{bmatrix} = 2(3) + 3(1) = 6 + 3 = 9$

So the product is $\begin{bmatrix} -3 & -4 & 1 \\ 8 & 13 & 9 \end{bmatrix}$.

(iv) $\begin{bmatrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{bmatrix} \begin{bmatrix} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{bmatrix}$

Both are $3 \times 3$, so product is $3 \times 3$. We compute each entry systematically.

Let $A$ be the first matrix, $B$ the second.

Row 1 of $A$: $[2, 3, 4]$

• Col 1 of $B$: $\begin{bmatrix} 1 \\ 0 \\ 3 \end{bmatrix}$ → $2(1) + 3(0) + 4(3) = 2 + 0 + 12 = 14$
• Col 2: $\begin{bmatrix} -3 \\ 2 \\ 0 \end{bmatrix}$ → $2(-3) + 3(2) + 4(0) = -6 + 6 + 0 = 0$
• Col 3: $\begin{bmatrix} 5 \\ 4 \\ 5 \end{bmatrix}$ → $2(5) + 3(4) + 4(5) = 10 + 12 + 20 = 42$

Row 2 of $A$: $[3, 4, 5]$

• Col 1: $3(1) + 4(0) + 5(3) = 3 + 0 + 15 = 18$
• Col 2: $3(-3) + 4(2) + 5(0) = -9 + 8 + 0 = -1$
• Col 3: $3(5) + 4(4) + 5(5) = 15 + 16 + 25 = 56$

Row 3 of $A$: $[4, 5, 6]$

• Col 1: $4(1) + 5(0) + 6(3) = 4 + 0 + 18 = 22$
• Col 2: $4(-3) + 5(2) + 6(0) = -12 + 10 + 0 = -2$
• Col 3: $4(5) + 5(4) + 6(5) = 20 + 20 + 30 = 70$

Thus the product is $\begin{bmatrix} 14 & 0 & 42 \\ 18 & -1 & 56 \\ 22 & -2 & 70 \end{bmatrix}$.

(v) $\begin{bmatrix} 2 & 1 \\ 3 & 2 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 1 \\ -1 & 2 & 1 \end{bmatrix}$

First is $3 \times 2$, second is $2 \times 3$, so product is $3 \times 3$.

Row 1 of first: $[2, 1]$

• Col 1 of second: $\begin{bmatrix} 1 \\ -1 \end{bmatrix}$ → $2(1) + 1(-1) = 2 - 1 = 1$
• Col 2: $\begin{bmatrix} 0 \\ 2 \end{bmatrix}$ → $2(0) + 1(2) = 0 + 2 = 2$
• Col 3: $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ → $2(1) + 1(1) = 2 + 1 = 3$

Row 2 of first: $[3, 2]$

• Col 1: $3(1) + 2(-1) = 3 - 2 = 1$
• Col 2: $3(0) + 2(2) = 0 + 4 = 4$
• Col 3: $3(1) + 2(1) = 3 + 2 = 5$

Row 3 of first: $[-1, 1]$

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

So the product is $\begin{bmatrix} 1 & 2 & 3 \\ 1 & 4 & 5 \\ -2 & 2 & 0 \end{bmatrix}$.

(vi) $\begin{bmatrix} 3 & -1 & 3 \\ -1 & 0 & 2 \end{bmatrix} \begin{bmatrix} 2 & -3 \\ 1 & 0 \\ 3 & 1 \end{bmatrix}$

First is $2 \times 3$, second is $3 \times 2$, so product is $2 \times 2$.

Row 1 of first: $[3, -1, 3]$

• Col 1 of second: $\begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix}$ → $3(2) + (-1)(1) + 3(3) = 6 - 1 + 9 = 14$
• Col 2: $\begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix}$ → $3(-3) + (-1)(0) + 3(1) = -9 + 0 + 3 = -6$

Row 2 of first: $[-1, 0, 2]$

• Col 1: $(-1)(2) + 0(1) + 2(3) = -2 + 0 + 6 = 4$
• Col 2: $(-1)(-3) + 0(0) + 2(1) = 3 + 0 + 2 = 5$

Thus the product is $\begin{bmatrix} 14 & -6 \\ 4 & 5 \end{bmatrix}$.

Method: Computing Matrix Products of Various Shapes

Use this for problems asking for several indicated products. The same two-part routine — check compatibility, then apply the row-by-column rule — works regardless of the matrices' shapes.

Steps

Step 1: Check compatibility and read off the product's order.

For $A$ of order $m\times n$ and $B$ of order $p\times q$, $AB$ exists only if $n=p$; the product is then $m\times q$. This tells you the shape before any arithmetic (e.g. a $3\times1$ times a $1\times3$ gives a full $3\times3$, not a scalar).

Step 2: Apply the row-by-column rule to each entry.

$(AB)_{ij}=\sum_k a_{ik}b_{kj}$ — dot row $i$ of $A$ with column $j$ of $B$. Fill the product one entry at a time.

Step 3: Verify shape and keep order.

Confirm the finished product has the predicted order, and never swap the factors, since $AB\neq BA$ in general.

Common Mistakes

Mistake 1: Multiplying without checking the inner dimensions.

Why it's wrong: $AB$ is only defined when columns of $A$ equal rows of $B$; otherwise the product does not exist. Correct approach: verify $n=p$ first, then compute an $m\times q$ result.

Mistake 2: Expecting a column-times-row product to be a scalar.

Why it's wrong: a $3\times1$ times a $1\times3$ has matching inner dimension $1$, giving a $3\times3$ matrix, not a single number. Correct approach: use the outer dimensions ($m\times q$) for the product's order.

Mistake 3: Pairing rows with rows (or columns with columns).

Why it's wrong: each entry is row $i$ of the first with column $j$ of the second; mixing this up gives wrong entries. Correct approach: strictly dot rows of $A$ against columns of $B$.