Free ACT Math Number and Quantity

ACT Math · Number and Quantity · free lesson

Matrices

The idea, then worked examples solved step by step, then practice questions whose numbers change every time you try them.

The idea

A table of numbers with a size

A matrix is a rectangle of numbers set out in rows and columns, inside brackets. Its size is the number of rows, then the number of columns. A matrix with 2 rows and 3 columns is a $2 \times 3$ matrix, read "2 by 3". Rows go across, columns go down.

Here is a $2 \times 3$ matrix:

\begin{aligned} M &= {\begin{bmatrix} 4 & -1 & 0 \\ 2 & 7 & 5 \end{bmatrix}} \end{aligned}

Each number is an entry. The entry in row 2, column 3 is $5$ . The picture marks that row and that column; the entry sits where they cross.

Adding, subtracting and multiplying by a number

Two matrices of the same size are added by adding the entries in the same places. Subtracting works the same way. Multiplying a matrix by a number, a scalar, multiplies every entry by it.

\begin{aligned} A + B &= {\begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix}} + {\begin{bmatrix} 1 & -2 \\ 0 & 4 \end{bmatrix}} = {\begin{bmatrix} 3 & -1 \\ 5 & 7 \end{bmatrix}} \end{aligned}

\begin{aligned} 3A &= 3{\begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix}} = {\begin{bmatrix} 6 & 3 \\ 15 & 9 \end{bmatrix}} \end{aligned}

Matrices of different sizes cannot be added: some entries would have no partner.

Multiplying two matrices: row times column

To multiply $A$ by $B$ , take a row of $A$ and a column of $B$ , multiply the entries in pairs and add. That sum is one entry of $AB$ : row 1 of $A$ with column 2 of $B$ gives the entry in row 1, column 2. Watch the four entries of $AB$ filled one at a time.

The engine works the whole product, and the product the other way round:

\begin{aligned} P &= \operatorname{matMul}(A, B) \\ &= \begin{bmatrix} 2 & 0 \\ 5 & 2 \end{bmatrix} \\ Q &= \operatorname{matMul}(B, A) \\ &= \begin{bmatrix} -8 & -5 \\ 20 & 12 \end{bmatrix} \end{aligned}

The two products are different. For matrices, $AB$ and $BA$ are usually not equal, so the order matters.

Rule.

A product $AB$ exists only when the number of columns of $A$ equals the number of rows of $B$ . An $m \times n$ matrix times an $n \times p$ matrix gives an $m \times p$ matrix: the inside numbers must match, and the outside numbers give the size.

So a $2 \times 3$ matrix times a $3 \times 4$ matrix is a $2 \times 4$ matrix, but a $3 \times 4$ matrix times a $2 \times 3$ matrix does not exist.

The determinant of a 2 by 2 matrix

For a $2 \times 2$ matrix with rows $a, b$ and $c, d$ , the determinant is $ad - bc$ : the product down the main diagonal minus the product down the other one.

For $A$ above, with rows $2$ , $1$ and $5$ , $3$ :

\begin{aligned} D &= ad - bc \\ &= 2 \times 3 - 1 \times 5 \\ &= 1 \\ E &= \operatorname{matDet}(A) \\ &= 1 \end{aligned}

When the determinant is 0, the matrix has no inverse. The picture shows why. Take the matrix with rows $2, 1$ and $c, 3$ . Its two columns are drawn as arrows, and the size of its determinant, $2(3) - 1(c)$ , is the area of the shaded shape they make. Drag $c$ : at $c = 6$ the two arrows line up, the shape goes flat, and the determinant is 0.

A system of equations as a matrix equation

The system $2x + y = 4$ and $5x + 3y = 11$ can be written as one matrix equation. The numbers in front of $x$ and $y$ make the coefficient matrix, the unknowns make a column, and the right sides make another column:

\begin{aligned} {\begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix}}\begin{bmatrix} x \\ y \end{bmatrix} &= \begin{bmatrix} 4 \\ 11 \end{bmatrix} \end{aligned}

Multiplying row by column on the left gives back the two equations. The engine solves the matrix equation:

\begin{aligned} s &= \operatorname{matSolve}(A, r) \\ &= [1, 2] \end{aligned}

So $x = {1}$ and $y = {2}$ . The two lines meet at that point:

Note.

On the ACT a matrix question is usually short: the size of a product, one entry of a sum or a product, a scalar multiple, a $2 \times 2$ determinant, or which matrix equation stands for a system. Write the sizes down first, then work only the entries asked for.

Worked examples

Each example is one of the kinds of matrix question the ACT asks. The numbers in the working are found by the engine.

Example.

1. The size of a product. Matrix $A$ is $2 \times 3$ and matrix $B$ is $3 \times 4$ . Does $AB$ exist, and what is its size? Does $BA$ exist?

For $AB$ , write the sizes side by side: $(2 \times 3)(3 \times 4)$ . The inside numbers, 3 and 3, match, so $AB$ exists. The outside numbers give its size: $2 \times 4$ .

For $BA$ the order is $(3 \times 4)(2 \times 3)$ . The inside numbers, 4 and 2, do not match, so $BA$ does not exist. Here are two such matrices; the engine multiplies them in the order that works:

\begin{aligned} P &= \operatorname{matMul}(A, B) \\ &= \begin{bmatrix} 5 & 2 & 2 & 3 \\ 5 & 5 & 0 & 2 \end{bmatrix} \end{aligned}

The product has 2 rows and 4 columns, as the sizes said.

Example.

2. A scalar multiple and a difference. Find $3A - 2B$ for the matrices below.

\begin{aligned} A &= {\begin{bmatrix} 4 & -1 \\ 2 & 5 \end{bmatrix}}, B = {\begin{bmatrix} 3 & 2 \\ -1 & 6 \end{bmatrix}} \end{aligned}

Multiply every entry of $A$ by 3 and every entry of $B$ by 2, then subtract entry by entry:

\begin{aligned} U &= \operatorname{matScale}(A, 3) \\ &= \begin{bmatrix} 12 & -3 \\ 6 & 15 \end{bmatrix} \\ V &= \operatorname{matScale}(B, 2) \\ &= \begin{bmatrix} 6 & 4 \\ -2 & 12 \end{bmatrix} \\ W &= \operatorname{matSub}(U, V) \\ &= \begin{bmatrix} 6 & -7 \\ 8 & 3 \end{bmatrix} \end{aligned}

So $3A - 2B = {\begin{bmatrix} 6 & -7 \\ 8 & 3 \end{bmatrix}}$ . Check one entry by hand: the top left is $3(4) - 2(3)$ , which is $6$ .

Example.

3. One entry of a product. For the matrices below, find the entry in row 2, column 1 of $AB$ .

\begin{aligned} A &= {\begin{bmatrix} 2 & -3 \\ 4 & 1 \end{bmatrix}}, B = {\begin{bmatrix} 5 & 0 \\ -2 & 6 \end{bmatrix}} \end{aligned}

Row 2 of $A$ is $4$ , $1$ . Column 1 of $B$ is $5$ , $-2$ . Multiply in pairs and add:

\begin{aligned} z &= eg + fh \\ &= 4 \times 5 + 1 \times (-2) \\ &= 18 \end{aligned}

The entry is $18$ . The whole product shows it in row 2, column 1:

\begin{aligned} P &= \operatorname{matMul}(A, B) \\ &= \begin{bmatrix} 16 & -18 \\ 18 & 6 \end{bmatrix} \end{aligned}

Multiplying only the two entries that sit in that place, $4$ and $-2$ , gives $-8$ : a common slip, and not how a product works.

Example.

4. From a situation. A school sells adult tickets for 9 dollars, student tickets for 5 dollars and child tickets for 3 dollars. The table shows the tickets sold for two plays. Use matrices to find the money taken at each play.

Play	Adult	Student	Child
Friday	120	210	40
Saturday	150	180	65

Put the counts in a $2 \times 3$ matrix $T$ (a row for each play) and the prices in a $3 \times 1$ column $C$ . The inside numbers match, so $TC$ is $2 \times 1$ : one amount for each play.

\begin{aligned} M &= \operatorname{matMul}(T, C) \\ &= \begin{bmatrix} 2250 \\ 2445 \end{bmatrix} \end{aligned}

Friday took $2250$ dollars and Saturday took $2445$ dollars. The bars show the two amounts and the part from each kind of ticket.

Example.

5. When there is no inverse. For what value of $k$ does the matrix $\begin{bmatrix} k & 4 \\ 3 & 6 \end{bmatrix}$ have no inverse?

A matrix has no inverse when its determinant is 0. The determinant is $6k - 4(3)$ . Set it equal to 0 and solve:

\begin{aligned} 6k - 4 \times 3 &= 0 \\ k &= 2 \end{aligned}

Check: with $k = 2$ the determinant $D$ is

\begin{aligned} D &= \operatorname{matDet}(M) \\ &= 0 \end{aligned}

So $k = 2$ . The rows $2$ , $4$ and $3$ , $6$ are then in the same ratio: the second row is $\frac{3}{2}$ times the first.

Example.

6. A system as a matrix equation. Write the system $3x - y = 7$ and $x + 2y = 7$ as a matrix equation, and solve it.

The coefficients of $x$ and $y$ go in the rows of the matrix, one row for each equation, in the same order as the right sides:

\begin{aligned} {\begin{bmatrix} 3 & -1 \\ 1 & 2 \end{bmatrix}}\begin{bmatrix} x \\ y \end{bmatrix} &= \begin{bmatrix} 7 \\ 7 \end{bmatrix} \end{aligned}

The determinant is not 0, so there is exactly one solution. The engine finds it:

\begin{aligned} d &= \operatorname{matDet}(A) \\ &= 7 \\ v &= \operatorname{matSolve}(A, r) \\ &= [3, 2] \end{aligned}

The first number is $x$ and the second is $y$ . Check in both equations:

\begin{aligned} x &= v[0] \\ &= 3 \\ y &= v[1] \\ &= 2 \\ L &= 3x - y \\ &= 3 \times 3 - 2 \\ &= 7 \\ R &= x + 2y \\ &= 3 + 2 \times 2 \\ &= 7 \end{aligned}

Both give 7. The two lines cross at that point:

Practice

Answer ACT-style questions on matrices. Each answer is checked, with the full solution, and every try brings new numbers. Practice needs a free account: it keeps your progress.

Practise matrices with new numbers