This is a beta course, so its structure, chapters, and examples may continue to change.

Applications of Determinants

The determinant holds an important place not only in mathematical theory but also has wide-ranging uses in practical applications.

Systems of Linear Equations

Cramer’s Rule

Cramer's Rule

For the n-variable linear system $Ax = b$ , if the determinant $|A|$ of the coefficient matrix $A$ satisfies $|A| \neq 0$ , then the system has a unique solution:

$x_i = \frac{|A_i|}{|A|}, \quad i = 1, 2, \ldots, n$

where $A_i$ is the matrix obtained from $A$ by replacing its $i$ -th column with the constant vector $b$ .

Proof

Proof:

If $|A| \neq 0$ , then $A$ is invertible
The solution of the system is $x = A^{-1}b$
Using the adjugate formula: $A^{-1} = \frac{1}{|A|} \text{adj}(A)$
This gives $x_i = \frac{|A_i|}{|A|}$

Example: Solve the system: $\begin{cases} 2x + y = 5 \\ x + 3y = 7 \end{cases}$

Solution:

$|A| = \begin{vmatrix} 2 & 1 \\ 1 & 3 \end{vmatrix} = 2 \times 3 - 1 \times 1 = 5$
$|A_1| = \begin{vmatrix} 5 & 1 \\ 7 & 3 \end{vmatrix} = 5 \times 3 - 1 \times 7 = 8$
$|A_2| = \begin{vmatrix} 2 & 5 \\ 1 & 7 \end{vmatrix} = 2 \times 7 - 5 \times 1 = 9$
$x = \frac{8}{5} = \frac{8}{5}$ , $y = \frac{9}{5} = \frac{9}{5}$

Homogeneous Linear Systems

The Condition for a Homogeneous Linear System to Have a Nonzero Solution

A homogeneous linear system $Ax = 0$ has a nonzero solution if and only if $|A| = 0$ .

Proof

Proof:

If $|A| \neq 0$ , then $A$ is invertible and the unique solution is $x = 0$
If $|A| = 0$ , then $A$ is not invertible and a nonzero solution exists

Inverting Matrices

The Adjugate Matrix Method

If $|A| \neq 0$ , then the inverse of $A$ is:

$A^{-1} = \frac{1}{|A|} \text{adj}(A)$

where $\text{adj}(A)$ is the adjugate matrix of $A$ .

The adjugate matrix: $\text{adj}(A) = (A_{ji})$ , where $A_{ji}$ is the algebraic cofactor of the entry $a_{ji}$ .

Example: Find the inverse of the matrix $A = \begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix}$ .

Solution:

$|A| = 2 \times 3 - 1 \times 1 = 5$
$\text{adj}(A) = \begin{pmatrix} 3 & -1 \\ -1 & 2 \end{pmatrix}$
$A^{-1} = \frac{1}{5} \begin{pmatrix} 3 & -1 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} \frac{3}{5} & -\frac{1}{5} \\ -\frac{1}{5} & \frac{2}{5} \end{pmatrix}$

Inverting a Block Matrix

For a block matrix $\begin{pmatrix} A & B \\ C & D \end{pmatrix}$ , if $A$ is invertible then:

$\begin{pmatrix} A & B \\ C & D \end{pmatrix}^{-1} = \begin{pmatrix} A^{-1} + A^{-1}B(D-CA^{-1}B)^{-1}CA^{-1} & -A^{-1}B(D-CA^{-1}B)^{-1} \\ -(D-CA^{-1}B)^{-1}CA^{-1} & (D-CA^{-1}B)^{-1} \end{pmatrix}$

Geometric Applications

Areas and Volumes

Two dimensions: the area of the parallelogram whose adjacent sides are the vectors $\vec{a} = (a_1, a_2)$ and $\vec{b} = (b_1, b_2)$ is:

$S = \left| \begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \end{vmatrix} \right|$

Three dimensions: the volume of the parallelepiped whose adjacent sides are the vectors $\vec{a} = (a_1, a_2, a_3)$ , $\vec{b} = (b_1, b_2, b_3)$ and $\vec{c} = (c_1, c_2, c_3)$ is:

$V = \left| \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} \right|$

Example: Compute the area of the parallelogram whose adjacent sides are the vectors $(1, 2)$ and $(3, 4)$ .

Solution: $S = \left| \begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} \right| = |1 \times 4 - 2 \times 3| = |4 - 6| = 2$

Linear Transformations

The Geometric Meaning of the Jacobian Determinant

The Jacobian determinant of a linear transformation $T: \mathbb{R}^n \to \mathbb{R}^n$ records the scaling factor of volume under the transformation.

If the matrix of $T$ is $A$ , then:

$\text{vol}(T(S)) = |\det(A)| \cdot \text{vol}(S)$

where $S$ is a measurable set in $\mathbb{R}^n$ .

Example: The Jacobian determinant of the linear transformation $T(x, y) = (2x + y, x + 3y)$ .

Solution: $A = \begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix}$

$\det(A) = 2 \times 3 - 1 \times 1 = 5$

so the transformation magnifies area by a factor of 5.

Eigenvalues and Eigenvectors

The Characteristic Polynomial

The characteristic polynomial of the matrix $A$ is:

$p(\lambda) = \det(A - \lambda I)$

Eigenvalue: the roots of the characteristic polynomial are the eigenvalues of the matrix.

Example: Find the eigenvalues of the matrix $A = \begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix}$ .

Solution: $p(\lambda) = \det(A - \lambda I) = \begin{vmatrix} 2-\lambda & 1 \\ 1 & 3-\lambda \end{vmatrix}$

$= (2-\lambda)(3-\lambda) - 1 \times 1 = \lambda^2 - 5\lambda + 5$

Solving $\lambda^2 - 5\lambda + 5 = 0$ gives: $\lambda = \frac{5 \pm \sqrt{5}}{2}$

Numerical Computation

The Condition Number

The condition number of the matrix $A$ is defined as:

$\kappa(A) = \|A\| \cdot \|A^{-1}\|$

where $\|A\|$ is the norm of the matrix.

Property: if $A$ is invertible, then:

$\kappa(A) = \frac{\sigma_{\max}(A)}{\sigma_{\min}(A)}$

where $\sigma_{\max}(A)$ and $\sigma_{\min}(A)$ are the largest and smallest singular values of $A$ respectively.

Stability Analysis

The Condition Number and the Stability of the Solution

The sensitivity of the solution of the linear system $Ax = b$ to perturbations of the coefficients is governed by the condition number of the matrix.

Conclusion: the larger the condition number, the more unstable the system.

Exercises

Exercise 1

Use Cramer’s rule to solve the system: $\begin{cases} 3x + 2y = 8 \\ 2x + 5y = 12 \end{cases}$

Reference Answer

determinantsystem of linear equationsapplications of determinantsCramer's rule

Approach: Use the formula of Cramer’s rule.

Steps:

$|A| = \begin{vmatrix} 3 & 2 \\ 2 & 5 \end{vmatrix} = 3 \times 5 - 2 \times 2 = 15 - 4 = 11$
$|A_1| = \begin{vmatrix} 8 & 2 \\ 12 & 5 \end{vmatrix} = 8 \times 5 - 2 \times 12 = 40 - 24 = 16$
$|A_2| = \begin{vmatrix} 3 & 8 \\ 2 & 12 \end{vmatrix} = 3 \times 12 - 8 \times 2 = 36 - 16 = 20$
$x = \frac{16}{11}$ , $y = \frac{20}{11}$

Answer: $x = \frac{16}{11}$ , $y = \frac{20}{11}$ .

Exercise 2

Find the inverse of the matrix $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ .

Reference Answer

determinantsystem of linear equationsapplications of determinantsadjugate matrix

Approach: Use the adjugate matrix method.

Steps:

$|A| = 1 \times 4 - 2 \times 3 = 4 - 6 = -2$
$\text{adj}(A) = \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix}$
$A^{-1} = \frac{1}{-2} \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{pmatrix}$

Answer: $A^{-1} = \begin{pmatrix} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{pmatrix}$ .

Exercise 3

Compute the volume of the parallelepiped whose adjacent sides are the vectors $(1, 1, 1)$ , $(1, 2, 3)$ and $(1, 4, 9)$ .

Reference Answer

determinantsystem of linear equationsapplications of determinantsdeterminant calculation

Approach: Use a third-order determinant to compute the volume.

Steps:

$V = \left| \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 9 \end{vmatrix} \right|$
$= \left| 1 \begin{vmatrix} 2 & 3 \\ 4 & 9 \end{vmatrix} - 1 \begin{vmatrix} 1 & 3 \\ 1 & 9 \end{vmatrix} + 1 \begin{vmatrix} 1 & 2 \\ 1 & 4 \end{vmatrix} \right|$
$= \left| 1(18-12) - 1(9-3) + 1(4-2) \right| = |6 - 6 + 2| = 2$

Answer: the volume is 2.

Exercise 4

Find the eigenvalues of the matrix $A = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$ .

Reference Answer

determinantsystem of linear equationsapplications of determinantscharacteristic polynomial

Approach: Find the roots of the characteristic polynomial.

Steps:

$p(\lambda) = \det(A - \lambda I) = \begin{vmatrix} 1-\lambda & 1 \\ 1 & 2-\lambda \end{vmatrix}$
$= (1-\lambda)(2-\lambda) - 1 \times 1 = \lambda^2 - 3\lambda + 1$
Solving $\lambda^2 - 3\lambda + 1 = 0$ gives: $\lambda = \frac{3 \pm \sqrt{5}}{2}$

Answer: the eigenvalues are $\frac{3 + \sqrt{5}}{2}$ and $\frac{3 - \sqrt{5}}{2}$ .

Exercise 5

Decide whether the homogeneous system $\begin{cases} x + 2y = 0 \\ 2x + 4y = 0 \end{cases}$ has a nonzero solution.

Reference Answer

determinantsystem of linear equationsapplications of determinantscoefficient matrix

Approach: Compute the determinant of the coefficient matrix.

Steps:

$|A| = \begin{vmatrix} 1 & 2 \\ 2 & 4 \end{vmatrix} = 1 \times 4 - 2 \times 2 = 4 - 4 = 0$
Since $|A| = 0$ , the system has a nonzero solution.

Answer: the system has a nonzero solution.

Summary

Symbols Used in This Article

Symbol	Type	Pronunciation / Note	Meaning in This Article
$\det(A)$	Mathematical symbol	determinant of A	The determinant of the coefficient matrix $A$
$Ax = b$	Mathematical symbol	A x equals b	The matrix form of a system of linear equations
$x_i = \det(A_i)/\det(A)$	Mathematical symbol	x i equals A i over A	Cramer’s rule, the solution for the $i$ -th unknown
$A^{-1}$	Mathematical symbol	A inverse	The inverse matrix of the matrix $A$
$\text{adj}(A)$	Mathematical symbol	adjugate of A	The adjugate matrix of the matrix $A$
$p(\lambda) = \det(A - \lambda I)$	Mathematical symbol	characteristic polynomial	The characteristic polynomial of the matrix $A$
$\lambda$	Greek letter	Lambda	An eigenvalue, that is, a root of the characteristic polynomial
$\kappa(A)$	Greek letter	Kappa	The condition number of the matrix $A$

A Zero Determinant and Degeneracy

Comprehensive Exercises on Determinants

Found an issue with this content?

Tell us what could be improved.

Unlock Full Course ·

Your membership covers this course — no need to buy it again.

Login Required to Purchase

Payment Failed

Payment Successful!

Price History

Applications of Determinants