Definition and computation of determinants

The determinant assigns to each square matrix a single real number. Its value records crucial information about the matrix: whether it is invertible, how its rows are related, and how it scales volumes when applied as a transformation. The $2 \times 2$ formula is one line; the $3 \times 3$ requires cofactor expansion; for larger matrices the recursive expansion continues. This subtopic shows the computation and prepares the way for the structural properties of the next subtopic.

The $2 \times 2$ determinant

For $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$,

$\det A = |A| = ad - bc.$

The determinant is denoted $\det A$ or $|A|$. (The vertical-bar notation does not mean absolute value here , it is a different symbol.)

Examples:

• $\det\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = 4 - 6 = -2$.
• $\det\begin{pmatrix} 3 & 0 \\ 0 & 5 \end{pmatrix} = 15$.

The $3 \times 3$ determinant

For $A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix}$,

$\det A = a_{11}(a_{22} a_{33} - a_{23} a_{32}) - a_{12}(a_{21} a_{33} - a_{23} a_{31}) + a_{13}(a_{21} a_{32} - a_{22} a_{31}).$

This is expansion along the first row. The alternating signs $+, -, +$ are characteristic. Note that the three pieces are each a multiplication of an entry of row $1$ by a $2 \times 2$ determinant , the determinant of the matrix you get by deleting that entry's row and column. These smaller determinants are called minors (next subtopic).

Worked example

$\det\begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}$ $= 1(5 \cdot 9 - 6 \cdot 8) - 2(4 \cdot 9 - 6 \cdot 7) + 3(4 \cdot 8 - 5 \cdot 7)$ $= 1(45 - 48) - 2(36 - 42) + 3(32 - 35)$ $= -3 + 12 - 9 = 0$.

The determinant being zero tells us the rows are linearly dependent , indeed, $R_3 = 2 R_2 - R_1$.

Expansion along any row or column

A theorem (proved in any linear algebra textbook): the determinant equals the cofactor expansion along any row or column, with the signs $(-1)^{i+j}$ on each cofactor.

This lets you choose the easiest row or column , typically one containing many zeros.

Example: $\det\begin{pmatrix} 1 & 0 & 0 \\ 2 & 5 & 6 \\ 3 & 8 & 9 \end{pmatrix}$ along column $1$:

$= 1 \cdot (45 - 48) - 2 \cdot 0 + 3 \cdot 0 = -3$.

Even simpler: expand along the top row, which has two zeros:

$= 1 \cdot (45 - 48) - 0 + 0 = -3$. Same answer.

Determinants of triangular matrices

For an upper or lower triangular matrix, the determinant equals the product of the diagonal entries. Reason: expanding along a non-diagonal row repeatedly hits zeros.

So $\det\begin{pmatrix} 2 & 7 & 8 \\ 0 & 3 & -1 \\ 0 & 0 & 5 \end{pmatrix} = 2 \cdot 3 \cdot 5 = 30$.

Sarrus' rule for $3 \times 3$ matrices

A mnemonic: write the first two columns again to the right, multiply along the three downward diagonals (with positive sign), and along the three upward diagonals (with negative sign):

$\det A = a_{11} a_{22} a_{33} + a_{12} a_{23} a_{31} + a_{13} a_{21} a_{32} - a_{13} a_{22} a_{31} - a_{11} a_{23} a_{32} - a_{12} a_{21} a_{33}$.

Sarrus' rule applies only to $3 \times 3$ matrices.

Worked examples

Example 1. $\det\begin{pmatrix} 4 & 3 \\ 2 & 5 \end{pmatrix} = 20 - 6 = 14$.

Example 2. $\det\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} = 0 - (-1) = 1$.

Example 3. Find $\det\begin{pmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{pmatrix}$.

Upper triangular; determinant $= 1 \cdot 4 \cdot 6 = 24$.

Example 4. $\det\begin{pmatrix} 2 & 5 & 7 \\ -3 & 1 & 8 \\ 4 & 0 & 6 \end{pmatrix}$ along row 3 (has a zero):

$= 4(5 \cdot 8 - 7 \cdot 1) - 0 + 6(2 \cdot 1 - 5 \cdot (-3))$

$= 4 \cdot 33 + 6 \cdot 17 = 132 + 102 = 234$.

Example 5. Solve $\det\begin{pmatrix} x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x \end{pmatrix} = 0$ for $x$.

Expand: $x(x^2 - 1) - 1(x - 1) + 1(1 - x) = x^3 - x - x + 1 + 1 - x = x^3 - 3x + 2 = (x - 1)^2 (x + 2)$.

Roots: $x = 1$ (double) and $x = -2$.

Example 6. Compute $\det\begin{pmatrix} \sin\theta & \cos\theta \\ -\cos\theta & \sin\theta \end{pmatrix}$.

$\sin^2 \theta + \cos^2 \theta = 1$. A rotation determinant.

Try it yourself

1. Compute $\det\begin{pmatrix} 1 & 2 \\ 5 & 7 \end{pmatrix}$.
2. Compute $\det\begin{pmatrix} -3 & 4 \\ 2 & -1 \end{pmatrix}$.
3. Compute $\det\begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 10 \end{pmatrix}$.
4. Compute $\det\begin{pmatrix} 0 & 2 & 1 \\ 3 & 0 & 4 \\ 5 & 6 & 0 \end{pmatrix}$.
5. Compute $\det\begin{pmatrix} a & b & c \\ b & c & a \\ c & a & b \end{pmatrix}$ in terms of $a, b, c$.
6. Show that $\det\begin{pmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{pmatrix} = (b - a)(c - a)(c - b)$. (Vandermonde.)
7. Find $x$ if $\det\begin{pmatrix} x & 3 \\ 2 & x \end{pmatrix} = 0$.
8. Show that the determinant of $I$ is $1$.
9. Show that the determinant of a skew-symmetric matrix of odd order is $0$.
10. Compute $\det\begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{pmatrix}$.
11. Compute $\det\begin{pmatrix} 2 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 2 \end{pmatrix}$.
12. Solve $\det\begin{pmatrix} x & 2 & 3 \\ 2 & x & 3 \\ 3 & 3 & x \end{pmatrix} = 0$ for $x$.
13. Compute $\det\begin{pmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{pmatrix}$.
14. Compute $\det\begin{pmatrix} 1 & a & b + c \\ 1 & b & c + a \\ 1 & c & a + b \end{pmatrix}$.

Pitfalls / Tricks

• The signs alternate $+, -, +, -, \ldots$ along any row or column; do not forget the minus on the second term.
• Sarrus' rule is only for $3 \times 3$ matrices , do not try to extend it.
• Determinant of a triangular matrix is the product of diagonals , use this whenever possible.
• A zero row or zero column forces the determinant to be zero.
• Expanding along a row or column with many zeros saves enormous arithmetic.

Next we list the properties that turn computation into a strategy game.