An interesting way of using matrices to solve linear systems

Cramer's rule is used to solve systems of equations for which we have $n$ linearly-independent equations and $n$ unknowns. We'll show the rule here, use it in a couple of examples, then show how the rule is derived so you'll know where it comes from.

The general idea of Cramer's rule is that if we have an $n \times n$ system represented by the matrix-vector equation

$$A \vec x = \vec b$$

Example 1

The simplest example is a $2 \times 2$ system, so let's solve one of those. Solve for the intersection of the lines

$$ \begin{align} 2x - y &= 3 \\[5pt] -3x + 2y &= -5 \end{align}$$

Solution: The matrix-vector representation of this system is

$$\begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 3 \\ -5 \end{bmatrix}$$

Using Cramer's rule, the x-component of our solution is

$$x = \frac{\text{det}(A_1)}{\text{det}(A)} = \frac{\text{det} \begin{bmatrix} \color{red}{3} & -1 \\ \color{red}{-5} & 2 \end{bmatrix}}{\text{det}\begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix}}$$

The red is the $\vec b$ vector substituted for the first column of matrix $A$. Then we can calculate $x$.

$$x = \frac{3(2)-(-5)(-1)}{2(2) - (-3)(-1)} = \frac{6-5}{4-3} = 1$$

Example 2

Use Cramer's rule to solve this $3 \times 3$ system:

$$ \begin{align} x_1 - 2x_2 - x_3 &= -4 \\[5pt] -x_1 + x_2 + 4x_3 &= -1 \\[5pt] 2x_1 + 2x_2 - x_3 &= 6 \end{align}$$

Solution: The matrix-vector representation of this system is

$$\begin{bmatrix} 1 & -2 & -1 \\ -1 & 1 & 4 \\ 2 & 2 & -1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -4 \\ -1 \\ 6 \end{bmatrix}$$

Using Cramer's rule, the x₁-component of our solution is

$$x_1 = \frac{\text{det}(A_1)}{\text{det}(A)} = \frac{\text{det} \begin{bmatrix} \color{red}{-4} & -2 & -1 \\ \color{red}{-1} & 1 & 4 \\ \color{red}{6} & 2 & -1 \end{bmatrix}}{\text{det}\begin{bmatrix} 1 & -2 & -1 \\ -1 & 1 & 4 \\ 2 & 2 & -1 \end{bmatrix}}$$

The red is the $\vec b$ vector substituted for the first column of matrix $A$. Then we can calculate $x_1$.

$$ \begin{align} x_1 &= \frac{-4(1)(-1) -2(4)(6) + (-1)(-1)(2) - \\ 6(1)(-1) - 2(4)(-4) -(-1)(-1)(-2)}{1(1)(-1) -2(4)(2) + (-1)(-1)(2) - \\ 2(1)(-1) - 2(4)(1) - (-1)(-1)(-2)} \\[5pt] &= \frac{4-48+2+6+32+2}{-1-16+2+2-8+2} \\[5pt] &= \frac{-2}{-19} = \frac{2}{19} \end{align}$$

The $x_2$ component is

$$x_2 = \frac{\text{det}(A_2)}{\text{det}(A)} = \frac{\text{det} \begin{bmatrix} 1 & \color{red}{-4} & -1 \\ -1 & \color{red}{-1} & 4 \\ 2 & \color{red}{6} & -1 \end{bmatrix}}{\text{det}\begin{bmatrix} 1 & -2 & -1 \\ -1 & 1 & 4 \\ 2 & 2 & -1 \end{bmatrix}}$$

The red is the $\vec b$ vector substituted for the second column of matrix $A$. Then we can calculate $x_2$.

$$ \begin{align} x_2 &= \frac{1(-1)(-1) -4(4)(2) + (-1)(-1)(6) - \\ 2(-1)(-1) - 6(4)(1) -(-1)(-1)(-4)}{1(1)(-1) -2(4)(2) + (-1)(-1)(2) - \\ 2(1)(-1) - 2(4)(1) - (-1)(-1)(-2)} \\[5pt] &= \frac{1-32+6-2-24+4}{-1-16+2+2-8+2} \\[5pt] &= \frac{-47}{-19} = \frac{47}{19} \end{align}$$

A derivation (sort of)

A full derivation or proof of Cramer's rule is a little beyond the scope of what we can do here, but we can make a hand-waving demonstration of how Cramer's rule is connected to a $2 \times 2$ system. Take the system

$$ \begin{align} ax + by &= u \\[5pt] cx + dy &= v \end{align}$$

Now let's solve that system by multiplying the top equation by $c$ and the bottom equation by $a$ to get

$$ \begin{align} cax + cby &= cu \\[5pt] acx + ady &= av \end{align}$$

Now let's subtract the first equation from the second:

$$\require{cancel} \cancel{acx} - \cancel{acx} + ady - bcy = av - cu$$

Practice problems

Solve for the intersection of these pairs of lines using Cramer's rule.

$$1. \phantom{00000} \begin{align} 7x - y &= 15 \\[5pt] x + 4y &= 10 \end{align}$$

$$2. \phantom{00000} \begin{align} 3x - 3y &= 11 \\[5pt] 2x - 3y &= 4 \end{align}$$