Product of Matrices

This lesson introduces the product of two matrices: when the product exists, the row-into-column rule for computing it, and the laws matrix multiplication obeys, followed by several worked questions from Exercise 3.2.

When can two matrices be multiplied

To find the product $AB$, the number of columns of $A$ must equal the number of rows of $B$. If $A$ has order $m \times n$ and $B$ has order $n \times p$, then the product $AB$ exists and has order $m \times p$.

To multiply, take row into column (not row into row): multiply the first row of $A$ by the first column of $B$ element by element and add up, then the first row by the second column, and so on. Move to the second row of $A$ and repeat against each column of $B$.

First example

Let $A = \begin{bmatrix} 3 & 4 & 5 \\ 1 & 2 & 5 \end{bmatrix}$ (order $2 \times 3$) and $B = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$ (order $3 \times 1$).

Since the columns of $A$ match the rows of $B$, the product has order $2 \times 1$: $AB = \begin{bmatrix} 3(1) + 4(2) + 5(3) \\ 1(1) + 2(2) + 5(3) \end{bmatrix} = \begin{bmatrix} 3 + 8 + 15 \\ 1 + 4 + 15 \end{bmatrix} = \begin{bmatrix} 26 \\ 20 \end{bmatrix}$

Multiplying two square matrices

Let $P = \begin{bmatrix} 1 & 3 \\ 2 & 5 \end{bmatrix}$ and $Q = \begin{bmatrix} 4 & 5 \\ 2 & 6 \end{bmatrix}$, both $2 \times 2$, so $PQ$ is $2 \times 2$.

$PQ = \begin{bmatrix} 1(4) + 3(2) & 1(5) + 3(6) \\ 2(4) + 5(2) & 2(5) + 5(6) \end{bmatrix} = \begin{bmatrix} 10 & 23 \\ 18 & 40 \end{bmatrix}$

Laws of matrix multiplication

• Not commutative. In general $AB \neq BA$.
• Associative. $(AB)C = A(BC)$.
• Distributive. $A(B \pm C) = AB \pm AC$ and $(A \pm B)C = AC \pm BC$.
• Multiplicative identity. For every square matrix there is an identity matrix of the same order that acts as the multiplicative identity. $I_1, I_2, I_3$ are the identities for square matrices of order $1, 2, 3$ respectively, so for a $2 \times 2$ matrix the identity is $I_2$.

Exercise 3.2: worked questions

Question 1

Compute $\begin{bmatrix} a & b \\ -b & a \end{bmatrix} \begin{bmatrix} a & b \\ -b & a \end{bmatrix}$. The product is $2 \times 2$: $\begin{bmatrix} a(a) + b(-b) & a(b) + b(a) \\ -b(a) + a(-b) & -b(b) + a(a) \end{bmatrix} = \begin{bmatrix} a^2 + b^2 & 0 \\ 0 & a^2 + b^2 \end{bmatrix}$ The off-diagonal entries are $ab - ab = 0$ and $-ab + ab = 0$.

Question 2

Multiply the column matrix by the row matrix. Here $A = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$ is $3 \times 1$ and $B = \begin{bmatrix} 2 & 3 & 4 \end{bmatrix}$ is $1 \times 3$, so the answer is $3 \times 3$. Each entry is one element of $A$ times one element of $B$ (do not add): $AB = \begin{bmatrix} 2 & 3 & 4 \\ 4 & 6 & 8 \\ 6 & 9 & 12 \end{bmatrix}$

Question 4

Let $A = \begin{bmatrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{bmatrix}$, both $3 \times 3$, so $AB$ is $3 \times 3$: $AB = \begin{bmatrix} 2 + 0 + 12 & -6 + 6 + 0 & 10 + 12 + 20 \\ 3 + 0 + 15 & -9 + 8 + 0 & 15 + 16 + 25 \\ 4 + 0 + 18 & -12 + 10 + 0 & 20 + 20 + 30 \end{bmatrix} = \begin{bmatrix} 14 & 0 & 42 \\ 18 & -1 & 56 \\ 22 & -2 & 70 \end{bmatrix}$

Question 6

Let $A = \begin{bmatrix} 3 & -1 & 3 \\ -1 & 0 & 2 \end{bmatrix}$ ($2 \times 3$) and $B = \begin{bmatrix} 2 & -3 \\ 1 & 0 \\ 3 & 1 \end{bmatrix}$ ($3 \times 2$), so $AB$ is $2 \times 2$: $AB = \begin{bmatrix} 6 - 1 + 9 & -9 + 0 + 3 \\ -2 + 0 + 6 & 3 + 0 + 2 \end{bmatrix} = \begin{bmatrix} 14 & -6 \\ 4 & 5 \end{bmatrix}$

Key takeaways

• The product $AB$ exists only when the columns of $A$ equal the rows of $B$; if $A$ is $m \times n$ and $B$ is $n \times p$, then $AB$ is $m \times p$.
• Multiply row into column: pair each row of the first matrix with each column of the second, multiply matching elements, and add.
• Matrix multiplication is associative and distributive but not commutative, and every square matrix has an identity matrix of the same order as its multiplicative identity.