Lesson 4.6: Solve Systems of Equations Using Determinants

New Concept

We'll explore determinants, unique real numbers associated with square matrices. You'll learn to calculate these for $2 \times 2$ and $3 \times 3$ matrices and apply this skill using Cramer's Rule to efficiently solve systems of linear equations.

What’s next

First, we'll master calculating determinants for $2 \times 2$ matrices. You’ll then practice with interactive examples and short videos before tackling larger systems.

Property

The determinant of any square matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, where $a, b, c, d$ are real numbers, is

Examples

• To evaluate the determinant of $\begin{bmatrix} 5 & 2 \\ 4 & 3 \end{bmatrix}$, we calculate $\left| \begin{matrix} 5 & 2 \\ 4 & 3 \end{matrix} \right| = (5)(3) - (2)(4) = 15 - 8 = 7$.

• To evaluate the determinant of $\begin{bmatrix} -1 & -3 \\ 2 & 5 \end{bmatrix}$, we calculate $\left| \begin{matrix} -1 & -3 \\ 2 & 5 \end{matrix} \right| = (-1)(5) - (-3)(2) = -5 - (-6) = 1$.

Property

The minor of an entry in a $3 \times 3$ determinant is the $2 \times 2$ determinant found by eliminating the row and column in the $3 \times 3$ determinant that contains the entry.

Examples

For the determinant $\left| \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{matrix} \right|$:

• The minor of entry $a_1$ (the 1) is found by removing the first row and first column, leaving $\left| \begin{matrix} 5 & 6 \\ 8 & 9 \end{matrix} \right| = 45 - 48 = -3$.

Property

To evaluate a $3 \times 3$ determinant by expanding by minors along the first row, use the pattern:

When expanding by any row or column, use the sign pattern chart:

Examples

• Evaluate $\left| \begin{matrix} 1 & 0 & 2 \\ 3 & 4 & 5 \\ 1 & 2 & 1 \end{matrix} \right|$ by expanding along the first row: $1(4 - 10) - 0(...) + 2(6 - 4) = -6 + 4 = -2$.

• Evaluate $\left| \begin{matrix} 2 & 1 & 3 \\ -1 & 2 & 1 \\ 0 & 4 & -2 \end{matrix} \right|$ by expanding along the third row: $0(...) - 4(-2 - (-3)) + (-2)(4 - (-1)) = -4(1) - 2(5) = -14$.

Property

For the system of equations $\begin{cases} a_1 x + b_1 y = k_1 \\ a_2 x + b_2 y = k_2 \end{cases}$, the solution $(x, y)$ can be determined by $x = \frac{D_x}{D}$ and $y = \frac{D_y}{D}$. The determinants are:
$D = \left| \begin{matrix} a_1 & b_1 \\ a_2 & b_2 \end{matrix} \right|$ (coefficients of variables)
$D_x = \left| \begin{matrix} k_1 & b_1 \\ k_2 & b_2 \end{matrix} \right|$ (replace x-coefficients with constants)
$D_y = \left| \begin{matrix} a_1 & k_1 \\ a_2 & k_2 \end{matrix} \right|$ (replace y-coefficients with constants)

Examples

• Solve $\begin{cases} x + 2y = 5 \\ 3x - y = 1 \end{cases}$. We find $D = -7$, $D_x = -7$, and $D_y = -14$. So, $x = \frac{-7}{-7} = 1$ and $y = \frac{-14}{-7} = 2$. The solution is $(1, 2)$.

• Solve $\begin{cases} 2x + 3y = 7 \\ x + y = 3 \end{cases}$. We find $D = -1$, $D_x = -2$, and $D_y = -1$. So, $x = \frac{-2}{-1} = 2$ and $y = \frac{-1}{-1} = 1$. The solution is $(2, 1)$.

Property

For a system of three equations, the solution $(x, y, z)$ can be determined by $x = \frac{D_x}{D}$, $y = \frac{D_y}{D}$, and $z = \frac{D_z}{D}$. The determinant $D$ is formed from the variable coefficients. The determinants $D_x, D_y,$ and $D_z$ are formed by replacing the corresponding variable's column with the constants.

Examples

• For the system $\begin{cases} x+y+z=6 \\ 2x-y+z=3 \\ x+2y-z=2 \end{cases}$, we find $D = -9, D_x = -9, D_y = -18, D_z = -27$. The solution is $(1, 2, 3)$.

• For the system $\begin{cases} x+y=1 \\ y+z=1 \\ x+z=2 \end{cases}$, we find $D = 2, D_x = 2, D_y = 0, D_z = 2$. The solution is $(1, 0, 1)$.

Property

For a system of equations, if the determinant $D = 0$:

• If $D_x, D_y,$ and $D_z$ are all zero, the system is consistent and dependent and there are infinitely many solutions.
• If $D = 0$ and at least one of $D_x, D_y,$ or $D_z$ is not zero, the system is inconsistent and there is no solution.

Examples

• For $\begin{cases} x - 2y = 3 \\ -2x + 4y = 1 \end{cases}$, $D=0$ and $D_x = 14$. Since $D_x \neq 0$, the system is inconsistent and has no solution.

• For $\begin{cases} 3x + y = 2 \\ 6x + 2y = 4 \end{cases}$, $D=0$, $D_x=0$, and $D_y=0$. Since all determinants are zero, the system is dependent and has infinite solutions.

Property

Three points $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ are collinear if and only if

Examples

• Are the points $(1, 2)$, $(3, 6)$, and $(5, 10)$ collinear? The determinant $\left| \begin{matrix} 1 & 2 & 1 \\ 3 & 6 & 1 \\ 5 & 10 & 1 \end{matrix} \right| = 0$. Yes, they are collinear.

• Are the points $(0, 0)$, $(1, 3)$, and $(2, 5)$ collinear? The determinant $\left| \begin{matrix} 0 & 0 & 1 \\ 1 & 3 & 1 \\ 2 & 5 & 1 \end{matrix} \right| = -1$. No, they are not collinear.