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Methods for Computing Determinants

Computing determinants is a basic skill in linear algebra, and mastering several different methods can raise your efficiency in problem solving.

The Expansion Method

Expansion Along a Row

Method: choose a row and expand the determinant along that row.

$|A| = \sum_{j=1}^n (-1)^{i+j} a_{ij} M_{ij}$

where $M_{ij}$ is the minor of the entry $a_{ij}$ .

Principles for choosing the row:

Choose a row with more zero entries
Choose a row whose entries are smaller
Choose a row that makes the computation easier

Example: Compute $\begin{vmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{vmatrix}$

Solution: expand along the second row (which contains a zero entry): $|A| = 0 \times (-1)^{2+1} M_{21} + 1 \times (-1)^{2+2} M_{22} + 4 \times (-1)^{2+3} M_{23}$ $= 0 - M_{21} + M_{22} - 4M_{23}$ $= M_{22} - 4M_{23}$ $= \begin{vmatrix} 1 & 3 \\ 5 & 0 \end{vmatrix} - 4 \begin{vmatrix} 1 & 2 \\ 5 & 6 \end{vmatrix}$ $= (1 \times 0 - 3 \times 5) - 4(1 \times 6 - 2 \times 5)$ $= -15 - 4(-4) = -15 + 16 = 1$

Expansion Along a Column

Method: choose a column and expand the determinant along that column.

$|A| = \sum_{i=1}^n (-1)^{i+j} a_{ij} M_{ij}$

Example: Compute $\begin{vmatrix} 2 & 0 & 1 \\ 1 & 3 & 0 \\ 0 & 2 & 4 \end{vmatrix}$

Solution: expand along the second column (which contains a zero entry): $|A| = 0 \times (-1)^{1+2} M_{12} + 3 \times (-1)^{2+2} M_{22} + 2 \times (-1)^{3+2} M_{32}$ $= -0 \times M_{12} + 3 \times M_{22} - 2 \times M_{32}$ $= 3M_{22} - 2M_{32}$ $= 3 \begin{vmatrix} 2 & 1 \\ 0 & 4 \end{vmatrix} - 2 \begin{vmatrix} 2 & 1 \\ 1 & 0 \end{vmatrix}$ $= 3(2 \times 4 - 1 \times 0) - 2(2 \times 0 - 1 \times 1)$ $= 3(8) - 2(-1) = 24 + 2 = 26$

The Property Method

Simplification with the Basic Properties

Strategy:

Use the row (column) addition property to turn the determinant into upper-triangular or lower-triangular form
Use the row (column) swapping property to move rows (columns) with many zeros into a position that is convenient for expansion
Use the property of factoring out a common factor to simplify the computation

Example: Compute $\begin{vmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{vmatrix}$

Solution:

Add $-2$ times the first row to the second row: $\begin{vmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 3 & 6 & 9 \end{vmatrix}$
Add $-3$ times the first row to the third row: $\begin{vmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{vmatrix}$
Since one row consists entirely of zeros, the determinant is zero.

The Block Method

When it applies: the determinant can be split into blocks, and some of the blocks are zero or the identity matrix.

For $\begin{vmatrix} A & B \\ 0 & D \end{vmatrix}$ , the value is $|A| \cdot |D|$
For $\begin{vmatrix} A & 0 \\ C & D \end{vmatrix}$ , the value is $|A| \cdot |D|$

Example: Compute $\begin{vmatrix} 1 & 2 & 0 & 0 \\ 3 & 4 & 0 & 0 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 3 & 4 \end{vmatrix}$

Solution: This is a block-diagonal matrix: $|A| = \begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} \cdot \begin{vmatrix} 2 & 1 \\ 3 & 4 \end{vmatrix}$ $= (1 \times 4 - 2 \times 3) \times (2 \times 4 - 1 \times 3)$ $= (4 - 6) \times (8 - 3) = (-2) \times 5 = -10$

Special Methods

Vandermonde Determinants

A determinant of the form $\begin{vmatrix} 1 & 1 & \cdots & 1 \\ x_1 & x_2 & \cdots & x_n \\ x_1^2 & x_2^2 & \cdots & x_n^2 \\ \vdots & \vdots & \ddots & \vdots \\ x_1^{n-1} & x_2^{n-1} & \cdots & x_n^{n-1} \end{vmatrix}$ .

$|V| = \prod_{1 \leq i < j \leq n} (x_j - x_i)$

Example: Compute $\begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 9 \end{vmatrix}$

Solution: $|V| = (2-1)(3-1)(3-2) = 1 \times 2 \times 1 = 2$

Circulant Determinants

A determinant of the form $\begin{vmatrix} a & b & c \\ c & a & b \\ b & c & a \end{vmatrix}$ .

$|A| = a^3 + b^3 + c^3 - 3abc$

Example: Compute $\begin{vmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1 \end{vmatrix}$

Solution: $|A| = 1^3 + 2^3 + 3^3 - 3 \times 1 \times 2 \times 3$ $= 1 + 8 + 27 - 18 = 18$

Computational Techniques

Inspection

Technique:

Observe whether the determinant has a special structure
Look for rows or columns with many zero entries
Look for proportional rows or columns
Look for a structure that can be split into blocks

The Recurrence Method

Applies when: the n-th order determinant is related to the (n-1)-th order determinant.

Example: Compute $\begin{vmatrix} 2 & 1 & 0 & 0 \\ 1 & 2 & 1 & 0 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 1 & 2 \end{vmatrix}$

Solution: Expand along the first row: $|A| = 2 \begin{vmatrix} 2 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 2 \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 1 & 2 \end{vmatrix}$

Continuing the expansion yields a recurrence relation.

Exercises

Exercise 1

Compute the determinant $\begin{vmatrix} 2 & 1 & 0 \\ 1 & 3 & 2 \\ 0 & 1 & 4 \end{vmatrix}$ .

Reference Answer

determinantdeterminant calculationexpansion method

Approach: Compute it by expanding along the first row.

Steps:

$|A| = 2 \begin{vmatrix} 3 & 2 \\ 1 & 4 \end{vmatrix} - 1 \begin{vmatrix} 1 & 2 \\ 0 & 4 \end{vmatrix} + 0 \begin{vmatrix} 1 & 3 \\ 0 & 1 \end{vmatrix}$
$= 2(3 \times 4 - 2 \times 1) - 1(1 \times 4 - 2 \times 0) + 0$
$= 2(12 - 2) - 1(4) = 2(10) - 4 = 20 - 4 = 16$

Answer: the value of the determinant is 16.

Exercise 2

Use the property method to compute $\begin{vmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 1 & 1 \end{vmatrix}$ .

Reference Answer

determinantdeterminant calculationexpansion method

Approach: Use the row addition property to simplify the computation.

Steps:

Add $-2$ times the first row to the second row: $\begin{vmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 1 & 1 & 1 \end{vmatrix}$
Since the second row consists entirely of zeros, the determinant is zero.

Answer: the value of the determinant is 0.

Exercise 3

Compute the Vandermonde determinant $\begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 4 & 16 \end{vmatrix}$ .

Reference Answer

determinantdeterminant calculationexpansion methodVandermonde determinant

Approach: Use the formula for the Vandermonde determinant.

Steps:

$x_1 = 1, x_2 = 2, x_3 = 4$
$|V| = (2-1)(4-1)(4-2) = 1 \times 3 \times 2 = 6$

Answer: the value of the determinant is 6.

Exercise 4

Compute the block determinant $\begin{vmatrix} 1 & 2 & 0 & 0 \\ 3 & 4 & 0 & 0 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 3 & 4 \end{vmatrix}$ .

Reference Answer

determinantdeterminant calculationexpansion methodproperties of determinants

Approach: Use the properties of block determinants.

Steps:

This is a block-diagonal matrix
$|A| = \begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} \cdot \begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix}$
$= (1 \times 4 - 2 \times 3)^2 = (4 - 6)^2 = (-2)^2 = 4$

Answer: the value of the determinant is 4.

Exercise 5

Compute the circulant determinant $\begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix}$ .

Reference Answer

determinantdeterminant calculationexpansion methodcirculant determinant

Approach: Use the formula for the circulant determinant.

Steps:

$a = b = c = 1$
$|A| = 1^3 + 1^3 + 1^3 - 3 \times 1 \times 1 \times 1 = 3 - 3 = 0$

Answer: the value of the determinant is 0.

Summary

Symbols Used in This Article

Symbol	Type	Pronunciation / Note	Meaning in This Article
$\lvert A\rvert$	Mathematical symbol	determinant of A	The value of the determinant of the matrix $A$
$M_{ij}$	Mathematical symbol	minor M i j	The minor of the entry $a_{ij}$ , namely the sub-determinant obtained after deleting row $i$ and column $j$
$(-1)^{i+j}$	Mathematical symbol	sign factor	The algebraic sign factor that appears when expanding along a row (column)
$\lvert V\rvert$	Mathematical symbol	determinant of V	A Vandermonde determinant
$\prod_{1 \leq i < j \leq n} (x_j - x_i)$	Mathematical symbol	product over i less than j	The product formula for the Vandermonde determinant

The Geometric Meaning of the Determinant

A Zero Determinant and Degeneracy

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Methods for Computing Determinants