Math Core

Every square matrix has a single number attached to it, its determinant, that answers a surprising number of questions at once. Is the matrix invertible? Does the system $AX = B$ have exactly one solution? What is the area of the triangle with these three vertices? In the last lesson you met the $2 \times 2$ determinant $ad - bc$ inside the inverse formula. This lesson extends it to $3 \times 3$ matrices, lists its most useful properties and puts it to work.

The 2 × 2 determinant

Definition

Determinant of a 2 × 2 matrix

The determinant of $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is

\det A = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc.

Vertical bars around the entries mean "determinant of", not absolute value; a determinant can be negative.

A handy picture: multiply down the main diagonal, then subtract the product up the other diagonal. For example,

\begin{vmatrix} 5 & -2 \\ 3 & 4 \end{vmatrix} = 5(4) - (-2)(3) = 20 + 6 = 26.

The 3 × 3 determinant by cofactor expansion

For a $3 \times 3$ matrix, the determinant is built out of $2 \times 2$ determinants. The minor $M_{ij}$ of an entry is the determinant left over when you cross out that entry's row and column. The cofactor is the minor with a sign attached: $C_{ij} = (-1)^{i+j}M_{ij}$ . The signs follow a checkerboard pattern:

\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}

Cofactor expansion

To find the determinant of a $3 \times 3$ matrix, choose any row or column. Multiply each entry in it by its cofactor, and add. Expanding along the first row,

\begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} = a_{11}\begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} - a_{12}\begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix} + a_{13}\begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix}.

Every row and every column gives the same answer, so choose the one with the most zeros.

Worked example: A 3 × 3 determinant

Evaluate $\det A$ for $A = \begin{bmatrix} 2 & -1 & 3 \\ 0 & 4 & 1 \\ 5 & 2 & -2 \end{bmatrix}$ .

Expand along the first row, remembering the signs $+,\,-,\,+$ :

\begin{aligned} \det A &= 2\begin{vmatrix} 4 & 1 \\ 2 & -2 \end{vmatrix} - (-1)\begin{vmatrix} 0 & 1 \\ 5 & -2 \end{vmatrix} + 3\begin{vmatrix} 0 & 4 \\ 5 & 2 \end{vmatrix} \\ &= 2(-8 - 2) + 1(0 - 5) + 3(0 - 20) \\ &= -20 - 5 - 60 \\ &= -85. \end{aligned}

As a check, expand down the first column, which has a zero: $2(-10) - 0 + 5\begin{vmatrix} -1 & 3 \\ 4 & 1 \end{vmatrix} = -20 + 5(-1 - 12) = -20 - 65 = -85$ . ✓

Common mistake

The most common error in cofactor expansion is dropping a sign. The middle term of a first-row expansion is subtracted, and when that entry is itself negative, as with the $-1$ above, the two negatives make a plus. Write the checkerboard signs down before you start.

Properties that save work

You rarely need to expand a large determinant from scratch. These facts, stated for an $n \times n$ matrix $A$ , cover most situations:

Property	Effect
Triangular matrix (all zeros above or below the diagonal)	$\det A$ is the product of the diagonal entries
Swap two rows	the determinant changes sign
Multiply one row by $k$	the determinant is multiplied by $k$
Add a multiple of one row to another	the determinant does not change
A row of zeros, or two equal rows	$\det A = 0$
Scale the whole matrix	$\det(kA) = k^n \det A$
Products	$\det(AB) = \det A \cdot \det B$

The scaling rule surprises many students: $2A$ doubles every row, so for a $3 \times 3$ matrix the determinant is multiplied by $2 \cdot 2 \cdot 2 = 8$ , not by $2$ .

Most important of all:

Determinants and invertibility

A square matrix $A$ is invertible if and only if $\det A \ne 0$ . Equivalently, the system $AX = B$ has exactly one solution for every $B$ exactly when $\det A \ne 0$ . When $A$ is invertible, $\det(A^{-1}) = \dfrac{1}{\det A}$ .

Area with determinants

The absolute value of a $2 \times 2$ determinant is an area. If a parallelogram has sides given by the vectors $\langle a, c\rangle$ and $\langle b, d\rangle$ , its area is $\left| ad - bc \right|$ . A triangle with the same two sides is half of that parallelogram.

To find the area of a triangle with vertices $P_1(x_1, y_1)$ , $P_2(x_2, y_2)$ and $P_3(x_3, y_3)$ , form the two side vectors from $P_1$ and take half the absolute determinant:

\text{Area} = \frac{1}{2}\left|\,\begin{vmatrix} x_2 - x_1 & y_2 - y_1 \\ x_3 - x_1 & y_3 - y_1 \end{vmatrix}\,\right|.

Worked example: Area of a triangle

Find the area of the triangle with vertices $(1, 1)$ , $(5, 2)$ and $(2, 6)$ .

From $(1, 1)$ , the sides are $\langle 4, 1\rangle$ and $\langle 1, 5\rangle$ .

\text{Area} = \frac{1}{2}\left|\,\begin{vmatrix} 4 & 1 \\ 1 & 5 \end{vmatrix}\,\right| = \frac{1}{2}\left| 20 - 1 \right| = \frac{19}{2}.

The triangle has area 19/2 = 9.5 square units.Open in grapher →

Cramer's rule

Determinants also give a formula for the solution of a square system with $\det A \ne 0$ . For the system $ax + by = e$ , $cx + dy = f$ , let

D = \begin{vmatrix} a & b \\ c & d \end{vmatrix}, \qquad D_x = \begin{vmatrix} e & b \\ f & d \end{vmatrix}, \qquad D_y = \begin{vmatrix} a & e \\ c & f \end{vmatrix}.

$D_x$ replaces the $x$ -column with the constants, and $D_y$ replaces the $y$ -column. Then $x = \dfrac{D_x}{D}$ and $y = \dfrac{D_y}{D}$ . The same pattern works for three equations with $3 \times 3$ determinants.

Worked example: Cramer's rule

Solve $3x - 2y = 4$ and $5x + y = 11$ .

D = \begin{vmatrix} 3 & -2 \\ 5 & 1 \end{vmatrix} = 3 + 10 = 13, \quad D_x = \begin{vmatrix} 4 & -2 \\ 11 & 1 \end{vmatrix} = 4 + 22 = 26, \quad D_y = \begin{vmatrix} 3 & 4 \\ 5 & 11 \end{vmatrix} = 33 - 20 = 13.

So $x = \dfrac{26}{13} = 2$ and $y = \dfrac{13}{13} = 1$ . Check: $3(2) - 2(1) = 4$ and $5(2) + 1 = 11$ . ✓

Tip

Cramer's rule is fastest when you need only one of the variables: compute $D$ and a single numerator determinant, and you are done.