Solving systems of three linear equations – Matrices and Determinants of Order 3 (Senior 5 Mathematics) | Ganzaa

Matrices and Determinants of Order 3

Lessons

Senior 5 · Mathematics

Matrices and Determinants of Order 3

5 lessons

Lesson 5 of 5

Solving systems of three linear equations

45 min

Three orders, three prices

A school canteen buys rice, beans and maize flour from one supplier. Three receipts show only totals: 2 kg of rice, 3 kg of beans and 1 kg of maize flour cost 5,800 FRW; 1 kg, 2 kg and 4 kg cost 5,800 FRW; 3 kg, 1 kg and 2 kg cost 5,900 FRW (illustrative figures). What is the price of each? With prices $x, y, z$ in hundreds of francs per kilogram, the receipts are three linear equations in three unknowns. This lesson solves such systems with the inverse matrix of lesson 4 and with determinants alone, by Cramer's rule, and explains what happens when the determinant is 0.

Definition

A system of three linear equations in $x, y, z$

\begin{cases} a_{11}x + a_{12}y + a_{13}z = b_1 \\ a_{21}x + a_{22}y + a_{23}z = b_2 \\ a_{31}x + a_{32}y + a_{33}z = b_3 \end{cases} \quad \text{is the matrix equation} \quad AX = B,

where $A = (a_{ij})$ is the coefficient matrix, $X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$ and $B = \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}$ . Its determinant is $\Delta = \det A$ . The system is consistent if it has at least one solution and inconsistent if it has none.

Important

The inverse-matrix method. If $\Delta \ne 0$ , multiply $A X = B$ on the left by $A^{- 1}$ : $X = A^{-1}B$ . The system has exactly one solution.

Worked example 1: by the inverse matrix

Solve $\; x + 2y - z = 1, \;\; 2x + y = 5, \;\; -x + y + 3z = 8$ .

Solution. The coefficient matrix is the matrix $A$ of lesson 4, worked example 1, with $\det A = -12$ and $\operatorname{adj} A = \begin{pmatrix} 3 & -7 & 1 \\ -6 & 2 & -2 \\ 3 & -3 & -3 \end{pmatrix}$ . So

X = A^{-1}B = -\frac{1}{12}\begin{pmatrix} 3 & -7 & 1 \\ -6 & 2 & -2 \\ 3 & -3 & -3 \end{pmatrix}\begin{pmatrix} 1 \\ 5 \\ 8 \end{pmatrix} = -\frac{1}{12}\begin{pmatrix} -24 \\ -12 \\ -36 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix} .

Check in the third equation: $- 2 + 1 + 9 = 8$ . ✓ The solution is $(x, y, z) = (2, 1, 3)$ .

Important

Cramer's rule. If $\Delta \ne 0$ , the unique solution is

x = \frac{\Delta_x}{\Delta}, \qquad y = \frac{\Delta_y}{\Delta}, \qquad z = \frac{\Delta_z}{\Delta},

where $\Delta_x$ , $\Delta_y$ , $\Delta_z$ are the determinants obtained from $\Delta$ by replacing the column of $x$ , of $y$ or of $z$ by the column $B$ .

Reason (for $x$ ). Let the columns of $A$ be $c_{1}, c_{2}, c_{3}$ , so the system says $x c_{1} + y c_{2} + z c_{3} = B$ . By property 4 of determinants, $x\Delta = \det(xc_1, c_2, c_3)$ . By property 5, adding $y c_{2} + z c_{3}$ to the first column changes nothing: $x\Delta = \det(xc_1 + yc_2 + zc_3, c_2, c_3) = \det(B, c_2, c_3) = \Delta_x$ .

Worked example 2: by Cramer's rule

Solve $\; 2x + y - z = 5, \;\; x - y + 2z = -3, \;\; 3x + 2y + z = 6$ .

Solution.

\Delta = \begin{vmatrix} 2 & 1 & -1 \\ 1 & -1 & 2 \\ 3 & 2 & 1 \end{vmatrix} = 2(-1 - 4) - 1(1 - 6) - 1(2 + 3) = -10 + 5 - 5 = -10 .

\Delta_x = \begin{vmatrix} 5 & 1 & -1 \\ -3 & -1 & 2 \\ 6 & 2 & 1 \end{vmatrix} = -10, \quad \Delta_y = \begin{vmatrix} 2 & 5 & -1 \\ 1 & -3 & 2 \\ 3 & 6 & 1 \end{vmatrix} = -20, \quad \Delta_z = \begin{vmatrix} 2 & 1 & 5 \\ 1 & -1 & -3 \\ 3 & 2 & 6 \end{vmatrix} = 10 .

So $x = \frac{-10}{-10} = 1$ , $y = \frac{-20}{-10} = 2$ , $z = \frac{10}{-10} = -1$ . Check in the first equation: $2 + 2 + 1 = 5$ . ✓

Worked example 3: the canteen's prices

Solve the opening problem: $\; 2x + 3y + z = 58, \;\; x + 2y + 4z = 58, \;\; 3x + y + 2z = 59$ , with prices in hundreds of francs per kilogram.

Solution. $\Delta = 2(4 - 4) - 3(2 - 12) + 1(1 - 6) = 0 + 30 - 5 = 25$ . Replacing columns in turn, $\Delta_x = 300$ , $\Delta_y = 225$ , $\Delta_z = 175$ . So

x = \frac{300}{25} = 12, \qquad y = \frac{225}{25} = 9, \qquad z = \frac{175}{25} = 7 .

Rice costs 1,200 FRW, beans 900 FRW and maize flour 700 FRW per kilogram. Check the third receipt: $3 (12) + 9 + 2 (7) = 59$ . ✓

When $\Delta = 0$ : the geometry of three planes

Each equation $a x + b y + c z = d$ describes a plane in space (unit 8), and a solution is a point on all three planes. If $\Delta \ne 0$ , the three planes meet in exactly one point. If $\Delta = 0$ , there are only two possibilities: the planes share a whole line (or a plane), giving infinitely many solutions, or they have no point in common, giving none. Two of the planes may also be parallel, which is another way to have no solution. Neither the inverse method nor Cramer's rule can be used; eliminate instead.

Worked example 4: a system with $\Delta = 0$

Discuss, according to the value of $k$ , the system $\; x + y + z = 6, \;\; x + 2y + 3z = 14, \;\; 2x + 3y + 4z = k$ .

Solution. $\Delta = 1(8 - 9) - 1(4 - 6) + 1(3 - 4) = -1 + 2 - 1 = 0$ . The left side of the third equation is the sum of the first two, so the third equation must say $2 x + 3 y + 4 z = 6 + 14 = 20$ .

If $k \ne 20$ , the equations contradict each other: no solution (the planes form a tube).
If $k = 20$ , the third equation repeats the others. Subtracting the first from the second gives $y + 2 z = 8$ . Let $z = t$ : then $y = 8 - 2 t$ and $x = 6 - y - z = - 2 + t$ . The solutions are $(x, y, z) = (-2 + t,\ 8 - 2t,\ t)$ , $t \in \mathbb{R}$ : infinitely many, all on one line.

Warning

Check $\Delta \ne 0$ before using Cramer's rule or $A^{- 1}$ . $\Delta = 0$ does not say which of the two cases you are in: you must look at the equations.
Replace the right column. $\Delta_y$ replaces the column of the $y$ -coefficients, the second column, by $B$ .
Write every equation with $x, y, z$ in the same order, and put a 0 where an unknown is missing ( $2 x + y = 5$ has $z$ -coefficient 0).
$X = A^{-1}B$ , not $BA^{-1}$ : $BA^{-1}$ does not even exist, since $B$ is $3 \times 1$ .

Tip

Always check a solution by substituting it in the equation you did not use last. One substitution catches almost every arithmetic slip.

3Blue1Brown: Cramer's rule, explained geometrically (about 12 min). Watch why each unknown is a ratio of two determinants, the same idea as this lesson's proof with columns.

✦

Practice

Questions 1 to 3 are routine; 4 to 6 need more care; 7 asks for a reason.

Use the inverse of $A = \begin{pmatrix} 1 & 2 & -1 \\ 2 & 1 & 0 \\ -1 & 1 & 3 \end{pmatrix}$ to solve $x + 2 y - z = - 3$ , $2 x + y = 1$ , $- x + y + 3 z = 4$ .
Solve by Cramer's rule: $x + y + z = 2$ , $2 x - y + 3 z = 9$ , $x + 3 y - z = - 4$ .
At a school concert, day 1 sold 20 adult, 50 student and 30 child tickets for 130,000 FRW; day 2 sold 15, 60 and 40 for 140,000 FRW; day 3 sold 25, 40 and 10 for 115,000 FRW (illustrative). Find the price of each ticket.
For which values of $k$ does $x + 2 y + z = 3$ , $2 x + y + k z = 3$ , $x - y + 2 z = 0$ have exactly one solution?
Show that $x + y - z = 1$ , $2 x + 3 y + z = 5$ , $3 x + 4 y = 6$ has $\Delta = 0$ , and find all its solutions.
The third equation of question 5 is changed to $3 x + 4 y = 7$ . Show that the new system has no solution.
Explain why a system of three linear equations cannot have exactly two solutions. (Hint: if $X_{1}$ and $X_{2}$ are solutions, test $X_{1} + t (X_{2} - X_{1})$ .)

Show answersHide answers

$X = -\frac{1}{12}\operatorname{adj}A \begin{pmatrix} -3 \\ 1 \\ 4 \end{pmatrix} = -\frac{1}{12}\begin{pmatrix} -12 \\ 12 \\ -24 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}$ .
$\Delta = 4$ , $\Delta_x = 4$ , $\Delta_y = -4$ , $\Delta_z = 8$ : $(x, y, z) = (1, - 1, 2)$ .
$\Delta = -4500$ ; adult 2,000 FRW, student 1,500 FRW, child 500 FRW.
$\Delta = 3k - 9$ , so exactly one solution when $k \ne 3$ .
Row 3 = row 1 + row 2 and $6 = 1 + 5$ , so $\Delta = 0$ and the third equation repeats the others. Solutions: $(x, y, z) = (-2 + 4t,\ 3 - 3t,\ t)$ , $t \in \mathbb{R}$ .
Adding the first two equations gives $3 x + 4 y = 6$ , which contradicts $3 x + 4 y = 7$ .
$A\left(X_1 + t(X_2 - X_1)\right) = B + t(B - B) = B$ for every real $t$ . If $X_1 \ne X_2$ , these are infinitely many different solutions, so two solutions always bring infinitely many.

1Three orders, three prices

2When Δ=0 \Delta = 0 Δ=0: the geometry of three planes

Three orders, three prices

When $\Delta = 0$ : the geometry of three planes