Determinant of a Matrix: 2×2 and 3×3

The determinant of a matrix is a single number calculated from a square matrix. For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ , the determinant is $ad - bc$ . For a $3 \times 3$ matrix, expand along a row using $2 \times 2$ determinants. A determinant of 0 means the matrix has no inverse.

Before this lesson: You should know Matrix Multiplication, Solving Systems with Augmented Matrices and Multiplying and Dividing Integers.

What a Determinant Tells You

Only square matrices have determinants. The determinant is written $\det(A)$ or with straight bars, $\lvert A \rvert$ . It packs a lot of information into one number.

If the determinant is 0, the matrix has no inverse, and a system with that coefficient matrix has no unique solution. If it is not 0, the system has exactly one solution. In geometry, the absolute value of a $2 \times 2$ determinant is the area scale factor of the matrix, and determinants also give the areas of triangles from their vertices.

For a 2 × 2 matrix, multiply down the main diagonal, multiply up the other diagonal, and subtract: ad − bc.

Determinant Formulas

2 × 2 matrices

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

3 × 3 matrices: expand along row 1

\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = a\begin{vmatrix} e & f \\ h & i \end{vmatrix} - b\begin{vmatrix} d & f \\ g & i \end{vmatrix} + c\begin{vmatrix} d & e \\ g & h \end{vmatrix}

Each small determinant comes from crossing out the row and column of the entry in front of it. The signs alternate $+, -, +$ . A matrix with 0s on one side of the main diagonal is triangular, and its determinant is just the product of the diagonal entries.

Worked Examples

Example 1: A 2 × 2 determinant

Find $\det \begin{bmatrix} 4 & 3 \\ 2 & 5 \end{bmatrix}$ .

4 \cdot 5 = 20

Multiply down the main diagonal.

3 \cdot 2 = 6

Multiply up the other diagonal.

20 - 6 = 14

Subtract.

Answer: 14

Example 2: Negative entries

Find $\det \begin{bmatrix} -2 & 7 \\ 3 & -1 \end{bmatrix}$ .

(-2)(-1) = 2

Multiply the main diagonal.

7 \cdot 3 = 21

Multiply the other diagonal.

2 - 21 = -19

Subtract.

Answer: −19. Determinants can be negative.

Example 3: A 3 × 3 determinant

Find $\det \begin{bmatrix} 2 & 1 & 3 \\ 0 & -1 & 4 \\ 1 & 2 & 0 \end{bmatrix}$ .

\begin{vmatrix} -1 & 4 \\ 2 & 0 \end{vmatrix} = (-1)(0) - 4(2) = 0 - 8 = -8

For the 2, cross out row 1 and column 1.

\begin{vmatrix} 0 & 4 \\ 1 & 0 \end{vmatrix} = 0(0) - 4(1) = -4

For the 1, cross out row 1 and column 2.

\begin{vmatrix} 0 & -1 \\ 1 & 2 \end{vmatrix} = 0(2) - (-1)(1) = 0 + 1 = 1

For the 3, cross out row 1 and column 3.

2(-8) - 1(-4) + 3(1)

Combine with signs $+, -, +$ .

-16 + 4 + 3 = -9

Multiply.

Answer: −9

Example 4: Solve for an unknown entry

For what value of $x$ is $\begin{bmatrix} x & 6 \\ 2 & 3 \end{bmatrix}$ singular (no inverse)?

3x - 6 \cdot 2 = 0

A matrix is singular when its determinant is 0.

3x - 12 = 0

Multiply.

3x = 12

Add 12 to both sides.

x = 4

Divide by 3.

Answer: $x = 4$ . Then row 1, $4, 6$ , is exactly 2 times row 2, $2, 3$ .

Example 5: Word problem with a triangular park

On a city map with a grid in kilometers, a triangular park has corners at $(1, 2)$ , $(5, 3)$ and $(2, 7)$ . Find its area.

\begin{bmatrix} 1 & 2 & 1 \\ 5 & 3 & 1 \\ 2 & 7 & 1 \end{bmatrix}

Use the formula $\text{Area} = \frac{1}{2}\lvert \det \rvert$ , with one row per corner and a column of 1s.

1\begin{vmatrix} 3 & 1 \\ 7 & 1 \end{vmatrix} - 2\begin{vmatrix} 5 & 1 \\ 2 & 1 \end{vmatrix} + 1\begin{vmatrix} 5 & 3 \\ 2 & 7 \end{vmatrix}

Expand along row 1.

Find each small determinant: $3 - 7 = -4$ , $5 - 2 = 3$ and $35 - 6 = 29$

1(-4) - 2(3) + 1(29) = -4 - 6 + 29 = 19

Combine.

\frac{1}{2}\lvert 19 \rvert = 9.5

Take half the absolute value.

Answer: 9.5 square kilometers

Common Mistakes

Mistake	Example of the error	Fix
Adding the diagonal products	$4 \cdot 5 + 3 \cdot 2 = 26$	Subtract: $ad - bc = 20 - 6 = 14$ .
Subtracting in the wrong order	$bc - ad = 6 - 20 = -14$	Main diagonal first: $ad - bc$ .
Forgetting the alternating signs	$2(-8) + 1(-4) + 3(1)$	The middle term gets a minus: $2(-8) - 1(-4) + 3(1)$ .
Sign slips with negatives	$(-2)(-1) = -2$	A negative times a negative is positive: $2$ .
Taking a determinant of a non-square matrix	Finding $\det$ of a $2 \times 3$ matrix	Only square matrices have determinants.

Related Methods

Diagonal shortcut for 3 × 3

Copy the first two columns to the right of the matrix. Add the three down-right diagonal products, then subtract the three up-right diagonal products. This shortcut works only for $3 \times 3$ matrices.

Expand along a row with zeros

You can expand along any row or column, using a checkerboard of signs. Choosing a row or column with 0s means fewer small determinants to compute.

What comes next

Determinants appear in the formula for an inverse matrix and in Cramer’s rule, which solves a system using only determinants.

Practice Problems

Find $\det \begin{bmatrix} 3 & 1 \\ 2 & 4 \end{bmatrix}$ .
Find $\det \begin{bmatrix} 5 & -2 \\ 6 & -3 \end{bmatrix}$ .
Find $\det \begin{bmatrix} 0 & 7 \\ -4 & 9 \end{bmatrix}$ .
Find $\det \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$ .
Find $\det \begin{bmatrix} 3 & 0 & 2 \\ 1 & 4 & -1 \\ 0 & 2 & 5 \end{bmatrix}$ .
Find $\det \begin{bmatrix} 2 & 0 & 0 \\ 5 & 3 & 0 \\ 1 & 4 & 6 \end{bmatrix}$ .
Solve for $x$ : $\det \begin{bmatrix} x & 2 \\ 5 & 4 \end{bmatrix} = 2$ .
For what values of $k$ is $\begin{bmatrix} k & 6 \\ 2 & k \end{bmatrix}$ singular?
Find the area of the triangle with vertices $(0, 0)$ , $(6, 0)$ and $(2, 5)$ .
A plot of land is a triangle with corners at $(-1, 1)$ , $(4, -2)$ and $(3, 5)$ on a grid where each unit is 100 meters. Find its area in square meters.

Show answers

Main diagonal: $3 \cdot 4 = 12$ ; other diagonal: $1 \cdot 2 = 2$

12 - 2 = 10

Subtract.

Answer: 10

Main diagonal: $5(-3) = -15$ ; other diagonal: $(-2)(6) = -12$

-15 - (-12) = -15 + 12 = -3

Subtract.

Answer: −3

Main diagonal: $0 \cdot 9 = 0$ ; other diagonal: $7(-4) = -28$

0 - (-28) = 28

Subtract.

Answer: 28

1(5 \cdot 9 - 6 \cdot 8) - 2(4 \cdot 9 - 6 \cdot 7) + 3(4 \cdot 8 - 5 \cdot 7)

Expand along row 1.

= 1(45 - 48) - 2(36 - 42) + 3(32 - 35) = -3 + 12 - 9

Find each 2-by-2 determinant.

Answer: 0 (the matrix is singular)

3(4 \cdot 5 - (-1)(2)) - 0(\ldots) + 2(1 \cdot 2 - 4 \cdot 0)

Expand by cofactors.

= 3(20 + 2) - 0 + 2(2 - 0) = 66 + 4

Find each 2-by-2 determinant.

Answer: 70

The matrix is triangular, so multiply the diagonal

2 \cdot 3 \cdot 6 = 36

The matrix is triangular, so multiply the diagonal.

Answer: 36

4x - 2 \cdot 5 = 4x - 10

Find the determinant.

4x - 10 = 2

Set equal to 2.

4x = 12

Add 10 to both sides.

Answer: $x = 3$

k \cdot k - 6 \cdot 2 = k^2 - 12

Find the determinant.

k^2 = 12

Set equal to 0.

k = \pm\sqrt{12} = \pm 2\sqrt{3}

Take square roots (see simplifying square roots)

Answer: $k = \pm 2\sqrt{3} \approx \pm 3.46$

Matrix rows $0, 0, 1$ ; $6, 0, 1$ ; $2, 5, 1$

0 - 0 + 1(6 \cdot 5 - 0 \cdot 2) = 30

Expand along row 1.

\frac{1}{2}\lvert 30 \rvert = 15

Find the area: half the absolute value of the determinant.

Answer: 15 square units

10.

Matrix rows $-1, 1, 1$ ; $4, -2, 1$ ; $3, 5, 1$

-1(-2 - 5) - 1(4 - 3) + 1(20 - (-6)) = 7 - 1 + 26 = 32

Expand along row 1 and evaluate.

Area: $\frac{1}{2}(32) = 16$ square units, and each square unit is $100 \cdot 100 = 10{,}000$ m²
Answer: $16 \cdot 10{,}000 = 160{,}000$ square meters

FAQ

How do you find the determinant of a 2 × 2 matrix?

Multiply the entries on the main diagonal, multiply the entries on the other diagonal, and subtract: $ad - bc$ .

What does a determinant of 0 mean?

The matrix is singular: it has no inverse. A system of equations with that coefficient matrix has either no solution or infinitely many solutions.

Can a determinant be negative?

Yes. A negative determinant is fine. For area, you take the absolute value, since area can’t be negative.

Which matrices have determinants?

Only square matrices, like $2 \times 2$ and $3 \times 3$ . A matrix with a different number of rows and columns has no determinant.

Summary

A $2 \times 2$ determinant is $ad - bc$ .
Expand a $3 \times 3$ determinant along a row with signs $+, -, +$ , using $2 \times 2$ determinants.
A triangular matrix’s determinant is the product of its diagonal.
A determinant of 0 means no inverse and no unique solution.
Half the absolute value of a determinant gives a triangle’s area from its vertices.

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