Lesson 11.8: Solving Systems with Cramer's Rule

New Concept

Cramer's Rule offers a formula to solve systems of equations using determinants. You'll find each variable by computing a ratio of determinants, like $x = \frac{D_x}{D}$, providing a direct path to the system's unique solution.

What’s next

Next, you’ll master calculating determinants through guided practice. Then, you'll apply this skill in interactive examples to solve systems with Cramer's Rule.

Property

The determinant of a $2 \times 2$ matrix, given

is defined as

There are several ways to indicate the determinant, including $\operatorname{det}(A)$ and replacing the brackets in a matrix with straight lines, $|A|$.

Examples

• For the matrix $A = \begin{bmatrix} 4 & 1 \\ -2 & 5 \end{bmatrix}$, the determinant is $\operatorname{det}(A) = (4)(5) - (1)(-2) = 20 + 2 = 22$.
• For the matrix $B = \begin{bmatrix} 0 & 7 \\ 3 & -1 \end{bmatrix}$, the determinant is $\operatorname{det}(B) = (0)(-1) - (7)(3) = 0 - 21 = -21$.
• For the matrix $C = \begin{bmatrix} -5 & -3 \\ -1 & -2 \end{bmatrix}$, the determinant is $\operatorname{det}(C) = (-5)(-2) - (-3)(-1) = 10 - 3 = 7$.

Explanation

A determinant is a single number calculated from a square matrix. For a $2 \times 2$ matrix, it's found by multiplying the main diagonal entries and subtracting the product of the other diagonal entries. This value is key to solving systems.

Property

Cramer's Rule is a method that uses determinants to solve systems of equations that have the same number of equations as variables. Consider a system of two linear equations in two variables.

The solution using Cramer's Rule is given as

If we are solving for $x$, the $x$ column is replaced with the constant column. If we are solving for $y$, the $y$ column is replaced with the constant column.

Examples

• Solve the system $x + y = 3$ and $2x - y = 0$. The determinants are $D = -3$, $D_x = -3$, and $D_y = -6$. The solution is $x = \frac{-3}{-3} = 1$ and $y = \frac{-6}{-3} = 2$, or $(1, 2)$.
• Solve the system $3x + 2y = 7$ and $x + y = 3$. The determinants are $D = 1$, $D_x = 1$, and $D_y = 2$. The solution is $x = \frac{1}{1} = 1$ and $y = \frac{2}{1} = 2$, or $(1, 2)$.
• For the system $2x - y = 5$ and $4x - 2y = 10$, the determinant of the coefficient matrix is $D = (2)(-2) - (-1)(4) = 0$. Since $D=0$, Cramer's Rule cannot be used to find a unique solution.

Explanation

Cramer's Rule offers a quick way to solve for variables using ratios of determinants. To find $x$, divide the determinant of the matrix with its x-column replaced by the constants, by the determinant of the original coefficient matrix.

Property

To find the determinant of a $3 \times 3$ matrix, augment the matrix with the first two columns. Then calculate the sum of the products of entries down each of the three diagonals (upper left to lower right), and subtract the products of entries up each of the three diagonals (lower left to upper right).
For $A = \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix}$, the determinant is:

Examples

• For $A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{bmatrix}$, $\operatorname{det}(A) = (1(1)(0) + 2(4)(5) + 3(0)(6)) - (5(1)(3) + 6(4)(1) + 0(0)(2)) = 40 - 39 = 1$.
• For $B = \begin{bmatrix} 1 & 0 & 2 \\ 3 & 4 & 0 \\ 0 & 5 & 6 \end{bmatrix}$, $\operatorname{det}(B) = (1(4)(6) + 0(0)(0) + 2(3)(5)) - (0(4)(2) + 5(0)(1) + 6(3)(0)) = 54 - 0 = 54$.
• For $C = \begin{bmatrix} 2 & -1 & 0 \\ 1 & 2 & -1 \\ 3 & 0 & -2 \end{bmatrix}$, $\operatorname{det}(C) = (2(2)(-2) + (-1)(-1)(3) + 0(1)(0)) - (3(2)(0) + 0(-1)(2) + (-2)(1)(-1)) = -5 - 2 = -7$.

Explanation

To find a $3 \times 3$ determinant, use the diagonal method. Copy the first two columns to the right side. Add the products of the three downward diagonals and subtract the products of the three upward diagonals for your answer.

Property

For a $3 \times 3$ system of equations where

the solutions are given by $x = \frac{D_x}{D}$, $y = \frac{D_y}{D}$, and $z = \frac{D_z}{D}$, provided $D \neq 0$. The determinants are:

Examples

• For the system $x+y+z=6$, $2y+5z=-4$, $2x+5y-z=27$, the coefficient determinant is $D = -21$. The determinant for $x$ is $D_x = -105$, so $x = \frac{-105}{-21} = 5$.
• For the system $x+y+z=2$, $2x-y-z=1$, $x+2y-z=4$, the coefficient determinant is $D=9$. The determinant for $y$ is $D_y = -9$, so $y = \frac{-9}{9} = -1$.
• For the system $x-y=1$, $y+z=4$, $x+z=3$, the coefficient determinant is $D=-2$. The determinant for $z$ is $D_z = -6$, so $z = \frac{-6}{-2} = 3$.

Explanation

This rule extends to $3 \times 3$ systems. To find any variable, calculate the determinant of the coefficient matrix where that variable's column is replaced by the constant column. Then, divide this by the determinant of the original coefficient matrix.

Property

1. If the matrix is in upper triangular form, the determinant equals the product of entries down the main diagonal.
2. When two rows are interchanged, the determinant changes sign.
3. If either two rows or two columns are identical, the determinant equals zero.
4. If a matrix contains either a row of zeros or a column of zeros, the determinant equals zero.
5. The determinant of an inverse matrix $A^{-1}$ is the reciprocal of the determinant of the matrix $A$.
6. If any row or column is multiplied by a constant, the determinant is multiplied by the same factor.

Examples

• The matrix $A = \begin{bmatrix} 5 & 9 & 1 \\ 0 & 2 & 7 \\ 0 & 0 & 3 \end{bmatrix}$ is in upper triangular form. By Property 1, its determinant is the product of the diagonal entries: $\operatorname{det}(A) = 5 \cdot 2 \cdot 3 = 30$.
• The matrix $B = \begin{bmatrix} 1 & 4 & 5 \\ 1 & 4 & 5 \\ 9 & 8 & 7 \end{bmatrix}$ has two identical rows. By Property 3, its determinant is $\operatorname{det}(B) = 0$.
• The matrix $C = \begin{bmatrix} 2 & 3 & 8 \\ 0 & 0 & 0 \\ 1 & 5 & 9 \end{bmatrix}$ has a row of zeros. By Property 4, its determinant is $\operatorname{det}(C) = 0$.

Explanation

Determinants have special properties that act like calculation shortcuts. For example, if two rows are identical or an entire row is zeros, the determinant is automatically zero. Knowing these rules can save you a lot of time and effort.