Cramer's rule

#### 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

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

– remember that $det(A) \ne 0$ is required for the system to have a solution – The components $(x_i)$ of vector $\vec x$ are

$$x_i = \frac{\text{det}(A_i)}{\text{det}(A)}, \phantom{00} i = 1, \dots , n$$

where the matrix $A_i$ is formed by replacing the $i^{th}$ column of $A$ by the column vector $\vec b$.

### 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 $x$.

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

Likewise, the y-component of the solution is

$$y = \frac{\text{det}(A_2)}{\text{det}(A)} = \frac{\text{det} \begin{bmatrix} 2 & \color{red}{3} \\ -3 & \color{red}{-5} \end{bmatrix}}{\text{det}\begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix}}$$

$$y = \frac{2(-5)-3(-3)}{2(2) - (-3(-1))} = \frac{-10+9}{4-3} = -1$$

Now let's check that the solution $(x, y) = (1, -1)$ is valid:

\begin{align} 2(1) - (-1) &= 2 + 1 = 3 \phantom{000} \checkmark \\[5pt] -3(1) + 2(-1) &= -3 - 2 = -5 \phantom{000} \checkmark \end{align}

That's it!

### 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}

Finally, the $x_3$ component is

$$x_3 = \frac{\text{det}(A_3)}{\text{det}(A)} = \frac{\text{det} \begin{bmatrix} 1 & -2 & \color{red}{-4} \\ -1 & 1 & \color{red}{-1} \\ 2 & 2 & \color{red}{6} \end{bmatrix}}{\text{det}\begin{bmatrix} 1 & -2 & -1 \\ -1 & 1 & 4 \\ 2 & 2 & -1 \end{bmatrix}}$$

The red is the $\vec b$ vector substituted for the second column of matrix $A$. Then we can calculate $x_3$.

\begin{align} x_3 &= \frac{1(1)(6) + (-2)(-1)(2) + (-4)(-1)(2) - \\ 2(1)(-4) - 2(-1)(1) - 6(-1)(-2)}{1(1)(-1) -2(4)(2) + (-1)(-1)(2) - \\ 2(1)(-1) - 2(4)(1) - (-1)(-1)(-2)} \\[5pt] &= \frac{6+4+8+8+2-12}{-1-16+2+2-8+2} \\[5pt] &= \frac{16}{-19} = -\frac{16}{19} \end{align}

So our solution is

$$(x_1, x_2, x_3) = \left( \frac{2}{19}, \frac{47}{19}, \frac{-16}{19} \right)$$

... but we should check our solution:

\begin{align} \frac{4}{19} - 2 \left( \frac{47}{19} \right) - \frac{-16}{19} &= -4 \\[5pt] 2 - 94 + 16 &= -4(19) \\[5pt] -76 &= -76 \phantom{000} \checkmark \end{align}

\begin{align} -\frac{4}{19} + \frac{47}{19} + 4 \left( \frac{-16}{19} \right) &= -1 \\[5pt] -2 + 47 -64 &= -1(19) \\[5pt] -19 = -19 \phantom{000} \checkmark \end{align}

\begin{align} 2 \left( \frac{4}{19} \right) - 2 \left( \frac{47}{19} \right) - \frac{-16}{19} &= 6 \\[5pt] 4 + 94 + 16 &= 6(19) \\[5pt] 114 &= 114 \phantom{000} \checkmark \end{align}

Yep. That's it!

#### 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:

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

Now solve for y in two steps:

\begin{align} y(ad - bc) &= av - cu \\[5pt] y &= \frac{av - cu}{ad - bc} \end{align}

Now this is just the Cramer's rule equation for $y$,

$$y = \frac{\text{det}\begin{bmatrix} a & \color{red}{u} \\ c & \color{red}{v} \end{bmatrix}}{\text{det}\begin{bmatrix} a & b \\ c & d \end{bmatrix}},$$

where the red column was substituted for the second column of the coefficient matrix, just as Cramer's rule prescribes. You can convince yourself that solvind for $x$ yields the correct Cramer's rule expression in that case.

#### 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}

Solution

In matrix form, this system is

$$\begin{bmatrix} 7 & -1 \\ 1 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 15 \\ 10 \end{bmatrix}$$

\begin{align} x &= \frac{\text{det} \begin{bmatrix} 15 & -1 \\ 10 & 4 \end{bmatrix}}{\text{det} \begin{bmatrix} 7 & -1 \\ 1 & 4 \end{bmatrix}} \\[5pt] &= \frac{60 - (-10)}{28 - (-1)} \\[5pt] &= \frac{70}{29} \end{align}

\begin{align} y &= \frac{\text{det} \begin{bmatrix} 7 & 15 \\ 1 & 10 \end{bmatrix}}{29} \\[5pt] &= \frac{70-15}{29} \\[5pt] &= \frac{55}{29} \end{align}

$$(x, y) = \left( \frac{70}{29}, \frac{55}{29} \right)$$

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

Solution

In matrix form, this system is

$$\begin{bmatrix} 3 & -3 \\ 2 & -3 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 11 \\ 4 \end{bmatrix}$$

\begin{align} x &= \frac{\text{det} \begin{bmatrix} 11 & -3 \\ 4 & -3 \end{bmatrix}}{\text{det} \begin{bmatrix} 3 & -3 \\ 2 & -3 \end{bmatrix}} \\[5pt] &= \frac{-33+12}{-9+6} \\[5pt] &= \frac{-21}{-3} = 7 \end{align}

\begin{align} y &= \frac{\text{det} \begin{bmatrix} 3 & 11 \\ 2 & 4 \end{bmatrix}}{-3} \\[5pt] &= \frac{12-22}{-3} \\[5pt] &= \frac{-10}{-3} = \frac{10}{3} \end{align}

$$(x, y) = \left( 7, \frac{10}{3} \right)$$

3. \phantom{00000} \begin{align} 6x + 7y &= 13 \\[5pt] -x + 4y &= -2 \end{align}

Solution

In matrix form, this system is

$$\begin{bmatrix} 6 & 7 \\ -1 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 13 \\ -2 \end{bmatrix}$$

\begin{align} x &= \frac{\text{det} \begin{bmatrix} 13 & 7 \\ -2 & 4 \end{bmatrix}}{\text{det} \begin{bmatrix} 6 & 7 \\ -1 & 4 \end{bmatrix}} \\[5pt] &= \frac{52+14}{24+7} \\[5pt] &= \frac{66}{31} \end{align}

\begin{align} y &= \frac{\text{det} \begin{bmatrix} 6 & 13 \\ -1 & -2 \end{bmatrix}}{31} \\[5pt] &= \frac{-12+13}{31} \\[5pt] &= \frac{1}{31} \end{align}

$$(x, y) = \left( \frac{66}{31}, \frac{1}{31} \right)$$

4. \phantom{00000} \begin{align} 4x - 5y &= -5 \\[5pt] 5x + 2y &= 17 \end{align}

Solution

In matrix form, this system is

$$\begin{bmatrix} 4 & -5 \\ 5 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -5 \\ 17 \end{bmatrix}$$

\begin{align} x &= \frac{\text{det} \begin{bmatrix} -5 & -5 \\ 17 & 2 \end{bmatrix}}{\text{det} \begin{bmatrix} 4 & -5 \\ 5 & 2 \end{bmatrix}} \\[5pt] &= \frac{-10+85}{8+25} \\[5pt] &= \frac{75}{33} \end{align}

\begin{align} y &= \frac{\text{det} \begin{bmatrix} 4 & -5 \\ 5 & 17 \end{bmatrix}}{33} \\[5pt] &= \frac{68+25}{33} \\[5pt] &= \frac{93}{33} \end{align}

$$(x, y) = \left( \frac{75}{33}, \frac{93}{33} \right)$$

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