Worked examples · Sheet 3

Linear systems with Cramer's rule

Theory used: Cramer's rule Linear system in matrix form Systems of two equations in two unknowns Sarrus's rule Determinant of a 2 × 2 matrix

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

Solve with Cramer’s rule the system

\begin{cases} 3x - 2y = 4 \\[4pt] x + 4y = 6 \end{cases}

\det A = \begin{vmatrix} 3 & -2 \\ 1 & 4 \end{vmatrix} = 12 + 2 = 14 \neq 0

x = \frac{\begin{vmatrix} 4 & -2 \\ 6 & 4 \end{vmatrix}}{14} = \frac{16 + 12}{14} = 2, \qquad y = \frac{\begin{vmatrix} 3 & 4 \\ 1 & 6 \end{vmatrix}}{14} = \frac{18 - 4}{14} = 1

y = \dfrac{3 x - 4}{2}

y = \dfrac{6 - x}{4}

The lines

3 x - 2 y = 4

and

x + 4 y = 6

meet at the point

(2, 1)

The system has two equations and two unknowns, and the determinant of the coefficient matrix is $14 \neq 0$ ; therefore Cramer’s rule applies and the system has one and only one solution. For $x$ we replace the first column, that of the coefficients of $x$ , with the column of the constant terms $(4, 6)$ , and divide the resulting determinant by $14$ . For $y$ we replace the second column. We obtain the solution $(2, 1)$ . As a check, $3 \cdot 2 - 2 \cdot 1 = 4$ and $2 + 4 \cdot 1 = 6$ . In the plane the solution is the point where the two lines meet.

Exercise 2

Solve with Cramer’s rule the system

\begin{cases} x + y + z = 6 \\[4pt] 2x - y + z = 3 \\[4pt] x + 2y - z = 2 \end{cases}

\begin{aligned} \det A = \begin{vmatrix} 1 & 1 & 1 \\ 2 & -1 & 1 \\ 1 & 2 & -1 \end{vmatrix} &= (1 + 1 + 4) - (-1 + 2 - 2) \\[6pt] &= 6 + 1 = 7 \end{aligned}

\det A_1 = \begin{vmatrix} 6 & 1 & 1 \\ 3 & -1 & 1 \\ 2 & 2 & -1 \end{vmatrix} = (6 + 2 + 6) - (-2 + 12 - 3) = 7

\det A_2 = \begin{vmatrix} 1 & 6 & 1 \\ 2 & 3 & 1 \\ 1 & 2 & -1 \end{vmatrix} = (-3 + 6 + 4) - (3 + 2 - 12) = 14

\det A_3 = \begin{vmatrix} 1 & 1 & 6 \\ 2 & -1 & 3 \\ 1 & 2 & 2 \end{vmatrix} = (-2 + 3 + 24) - (-6 + 6 + 4) = 21

x = \frac{7}{7} = 1, \qquad y = \frac{14}{7} = 2, \qquad z = \frac{21}{7} = 3

The determinant of the coefficient matrix is computed with Sarrus’s rule. The products along the descending diagonals are $1 \cdot (-1) \cdot (-1) = 1$ , $1 \cdot 1 \cdot 1 = 1$ and $1 \cdot 2 \cdot 2 = 4$ ; those along the ascending diagonals are $1 \cdot (-1) \cdot 1 = -1$ , $2 \cdot 1 \cdot 1 = 2$ and $(-1) \cdot 2 \cdot 1 = -2$ . Therefore $\det A = 6 - (-1) = 7 \neq 0$ , and by Cramer’s rule the system has only one solution.

The matrix $A_j$ is obtained from $A$ by replacing column $j$ with the column of the constant terms $(6, 3, 2)$ . The three determinants, again computed with Sarrus, equal $7$ , $14$ and $21$ , and dividing each by $\det A = 7$ we obtain the solution $(1, 2, 3)$ . Substituting into the equations we verify that $1 + 2 + 3 = 6$ , $2 - 2 + 3 = 3$ and $1 + 4 - 3 = 2$ .

Exercise 3

A firm produces three items, which pass through three departments. One unit of the first item requires $1$ hour of cutting, $2$ of assembly and $1$ of finishing; one unit of the second requires $2$ , $1$ and $1$ hours; one unit of the third requires $1$ , $1$ and $2$ hours. The hours available each week are $80$ for cutting, $70$ for assembly and $90$ for finishing. Find the quantities of the three items that use up all the hours of every department.

\begin{cases} x_1 + 2x_2 + x_3 = 80 \\[4pt] 2x_1 + x_2 + x_3 = 70 \\[4pt] x_1 + x_2 + 2x_3 = 90 \end{cases}

\det A = \begin{vmatrix} 1 & 2 & 1 \\ 2 & 1 & 1 \\ 1 & 1 & 2 \end{vmatrix} = (2 + 2 + 2) - (1 + 1 + 8) = -4

\det A_1 = \begin{vmatrix} 80 & 2 & 1 \\ 70 & 1 & 1 \\ 90 & 1 & 2 \end{vmatrix} = (160 + 180 + 70) - (90 + 80 + 280) = -40

\det A_2 = \begin{vmatrix} 1 & 80 & 1 \\ 2 & 70 & 1 \\ 1 & 90 & 2 \end{vmatrix} = (140 + 80 + 180) - (70 + 90 + 320) = -80

\det A_3 = \begin{vmatrix} 1 & 2 & 80 \\ 2 & 1 & 70 \\ 1 & 1 & 90 \end{vmatrix} = (90 + 140 + 160) - (80 + 70 + 360) = -120

x_1 = \frac{-40}{-4} = 10, \qquad x_2 = \frac{-80}{-4} = 20, \qquad x_3 = \frac{-120}{-4} = 30

Denote by $x_1$ , $x_2$ , $x_3$ the weekly quantities of the three items. The cutting hours used are $x_1 + 2x_2 + x_3$ , and they must equal the $80$ available; in the same way we write the equations for assembly and finishing. We obtain a linear system of three equations in three unknowns, in which each row of the coefficient matrix corresponds to a department and each column to an item.

The determinant of the coefficient matrix is $-4 \neq 0$ , so by Cramer’s rule the production plan that uses all the hours exists and is unique. The determinants $\det A_1$ , $\det A_2$ , $\det A_3$ equal $-40$ , $-80$ and $-120$ , hence the firm must produce $10$ , $20$ and $30$ units of the three items. The quantities are positive, so the solution makes economic sense. As a check, the cutting hours are $10 + 40 + 30 = 80$ .