Multiplying matrices

Mul­ti­ply­ing two matri­ces is not done entry by entry. Instead each row of the first matrix is com­bined with each col­umn of the sec­ond. The rule looks strange at first, but it is exactly what is needed to work out totals such as a shop­ping bill, and it is what turns a sys­tem of equa­tions into a sin­gle matrix equa­tion. This les­son explains the rule, when a prod­uct exists, and the laws it does and does not obey.

Row times col­umn#

Start with a row matrix times a col­umn matrix of the same length. Mul­ti­ply the match­ing entries and add:

$$\begin{bmatrix} 3 & 2 & 1 \end{bmatrix}\begin{bmatrix} 52 \\ 110 \\ 180 \end{bmatrix} = \begin{bmatrix} 3 \times 52 + 2 \times 110 + 1 \times 180 \end{bmatrix} = \begin{bmatrix} 556 \end{bmatrix}$$

This is a bill: buy­ing $3$ kg of rice at ₹52, $2$ kg of dal at ₹110 and $1$ kg of oil at ₹180 costs ₹556.

The prod­uct of two matri­ces#

If $A$ is of order $m \times n$ and $B$ is of order $n \times p$, the prod­uct $AB$ is the $m \times p$ matrix whose entry in row $i$ and col­umn $j$ is

$$\text{(row } i \text{ of } A) \times \text{(column } j \text{ of } B)$$

worked out as above: mul­ti­ply match­ing entries and add.

When does $AB$ exist? Only when the num­ber of columns of $A$ equals the num­ber of rows of $B$. A use­ful pic­ture:

$$(m \times \underline{n}) \times (\underline{n} \times p) \to m \times p$$

The inner num­bers must match; the outer num­bers give the order of the prod­uct.

Exam­ple 1#

Find $AB$ and $BA$ for $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}$.

$$AB = \begin{bmatrix} 1 \times 5 + 2 \times 7 & 1 \times 6 + 2 \times 8 \\ 3 \times 5 + 4 \times 7 & 3 \times 6 + 4 \times 8 \end{bmatrix} = \begin{bmatrix} 19 & 22 \\ 43 & 50 \end{bmatrix}$$

$$BA = \begin{bmatrix} 5 \times 1 + 6 \times 3 & 5 \times 2 + 6 \times 4 \\ 7 \times 1 + 8 \times 3 & 7 \times 2 + 8 \times 4 \end{bmatrix} = \begin{bmatrix} 23 & 34 \\ 31 & 46 \end{bmatrix}$$

So $AB \neq BA$. Matrix mul­ti­pli­ca­tion is not com­mu­ta­tive: the order of the fac­tors mat­ters.

Exam­ple 2#

Let $C = \begin{bmatrix} 1 & 2 & 0 \\ 3 & -1 & 4 \end{bmatrix}$ ($2 \times 3$) and $E = \begin{bmatrix} 2 & 1 \\ 0 & 3 \\ -1 & 2 \end{bmatrix}$ ($3 \times 2$). Find $CE$ and $EC$.

$CE$ has order $2 \times 2$:

$$CE = \begin{bmatrix} 2 + 0 + 0 & 1 + 6 + 0 \\ 6 + 0 - 4 & 3 - 3 + 8 \end{bmatrix} = \begin{bmatrix} 2 & 7 \\ 2 & 8 \end{bmatrix}$$

$EC$ has order $3 \times 3$:

$$EC = \begin{bmatrix} 2 + 3 & 4 - 1 & 0 + 4 \\ 0 + 9 & 0 - 3 & 0 + 12 \\ -1 + 6 & -2 - 2 & 0 + 8 \end{bmatrix} = \begin{bmatrix} 5 & 3 & 4 \\ 9 & -3 & 12 \\ 5 & -4 & 8 \end{bmatrix}$$

Here $CE$ and $EC$ do not even have the same order.

Exam­ple 3#

Ravi buys $3$ kg of rice, $2$ kg of dal and $1$ kg of oil. The prices per kg at two shops are given by the columns of $\begin{bmatrix} 52 & 60 \\ 110 & 104 \\ 180 & 170 \end{bmatrix}$ (rows: rice, dal, oil). Where is his shop­ping cheaper?

$$\begin{bmatrix} 3 & 2 & 1 \end{bmatrix}\begin{bmatrix} 52 & 60 \\ 110 & 104 \\ 180 & 170 \end{bmatrix} = \begin{bmatrix} 556 & 558 \end{bmatrix}$$

The bill is ₹556 at the first shop and ₹558 at the sec­ond, so the first shop is cheaper by ₹2.

Laws of mul­ti­pli­ca­tion#

When­ever the prod­ucts exist:

1. $(AB)C = A(BC)$ (asso­cia­tive law)
2. $A(B + C) = AB + AC$ and $(A + B)C = AC + BC$ (dis­trib­u­tive laws)
3. $AI = IA = A$ for a square matrix $A$ and the iden­tity matrix $I$ of the same order
4. $k(AB) = (kA)B = A(kB)$ for a scalar $k$

Two famil­iar laws of num­bers fail for matri­ces:

• In gen­eral $AB \neq BA$ (Exam­ple 1).
• $AB = O$ does not force $A = O$ or $B = O$. For exam­ple,

$$\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} 1 - 1 & -1 + 1 \\ 1 - 1 & -1 + 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$

even though nei­ther fac­tor is the zero matrix.

Pow­ers of a square matrix#

For a square matrix, $A^2 = AA$, $A^3 = A^2A$, and so on.

Exam­ple 4#

If $A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$, show that $A^2 - 4A + 3I = O$.

$$A^2 = \begin{bmatrix} 4 + 1 & 2 + 2 \\ 2 + 2 & 1 + 4 \end{bmatrix} = \begin{bmatrix} 5 & 4 \\ 4 & 5 \end{bmatrix}$$

$$A^2 - 4A + 3I = \begin{bmatrix} 5 & 4 \\ 4 & 5 \end{bmatrix} - \begin{bmatrix} 8 & 4 \\ 4 & 8 \end{bmatrix} + \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$

Prac­tice#

1. If $A$ is $3 \times 4$ and $B$ is $4 \times 2$, does $AB$ exist? What is its order? Does $BA$ exist?
2. Find $\begin{bmatrix} 1 & 2 & 3 \end{bmatrix}\begin{bmatrix} 4 \\ 5 \\ 6 \end{bmatrix}$.
3. Find $\begin{bmatrix} 4 \\ 5 \\ 6 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \end{bmatrix}$.
4. For $A = \begin{bmatrix} 2 & 0 \\ 1 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 4 \\ -1 & 2 \end{bmatrix}$, find $AB$ and $BA$.
5. For $A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 2 \\ 3 & 0 \\ 1 & 1 \end{bmatrix}$, find $AB$.
6. If $A = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}$, show that $A^2 - 2A - 3I = O$.
7. Find $A^2$ for $A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$. What does this show?
8. For $A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$, find $A^2$ and $A^3$.
9. Sita buys $4$ pens and $3$ note­books. Pens cost ₹10 and note­books ₹30 at one shop, and ₹12 and ₹28 at another. Use a matrix prod­uct to find her bill at each shop.

Answers#