Applications of determinants – Matrices and Determinants of Order 2 (Senior 4 Mathematics) | Ganzaa

Matrices and Determinants of Order 2

Lessons

Senior 4 · Mathematics

Matrices and Determinants of Order 2

5 lessons

Lesson 5 of 5

Applications of determinants

25 min

Two shopping lists, two unknown prices

At a Kigali market, 3 kg of beans and 2 kg of rice cost 4900 Frw, while 1 kg of beans and 4 kg of rice cost 5300 Frw (illustrative prices). With $x$ and $y$ the prices per kilogram, in Frw,

\begin{cases} 3x + 2y = 4900 \\ x + 4y = 5300 \end{cases} \iff \begin{pmatrix} 3 & 2 \\ 1 & 4 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 4900 \\ 5300 \end{pmatrix}.

The system is now one matrix equation $A X = B$ . The determinant $\det A = 12 - 2 = 10$ is not zero, so $A$ has an inverse and there is exactly one pair of prices. This lesson turns that observation into two methods, and then uses determinants for areas and for points on a line.

Definition

Matrix form of a system: the system $a x + b y = e$ , $c x + d y = f$ is $A X = B$ with the coefficient matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ , the unknown $X = \begin{pmatrix} x \\ y \end{pmatrix}$ and the constants $B = \begin{pmatrix} e \\ f \end{pmatrix}$ . Its determinant of the system is $D = \det A = ad - bc$ .

Example: $2 x - 3 y = 7$ , $5 x + 4 y = 6$ has $A = \begin{pmatrix} 2 & -3 \\ 5 & 4 \end{pmatrix}$ and $D = 8 + 15 = 23$ .

Important

Theorem (Cramer's rule). If $D = ad - bc \ne 0$ , the system $a x + b y = e$ , $c x + d y = f$ has exactly one solution,

x = \frac{D_x}{D} = \frac{\begin{vmatrix} e & b \\ f & d \end{vmatrix}}{\begin{vmatrix} a & b \\ c & d \end{vmatrix}}, \qquad y = \frac{D_y}{D} = \frac{\begin{vmatrix} a & e \\ c & f \end{vmatrix}}{\begin{vmatrix} a & b \\ c & d \end{vmatrix}}.

$D_{x}$ is $D$ with the $x$ -column replaced by the constants; $D_{y}$ is $D$ with the $y$ -column replaced.

Proof. Multiply the first equation by $d$ , the second by $b$ , and subtract: $(a d - b c) x = e d - b f$ , that is $D x = D_{x}$ . Multiply the first by $c$ , the second by $a$ , and subtract: $(b c - a d) y = e c - a f$ , that is $D y = D_{y}$ . Divide by $D \ne 0$ . (Equivalently, $X = A^{-1}B$ written out entry by entry.) $\blacksquare$

Worked example 1: the market prices

Solve $3 x + 2 y = 4900$ , $x + 4 y = 5300$ by Cramer's rule.

Solution. $D = 3 (4) - 2 (1) = 10$ .

D_x = \begin{vmatrix} 4900 & 2 \\ 5300 & 4 \end{vmatrix} = 19\,600 - 10\,600 = 9000, \qquad D_y = \begin{vmatrix} 3 & 4900 \\ 1 & 5300 \end{vmatrix} = 15\,900 - 4900 = 11\,000.

So $x = \frac{9000}{10} = 900$ and $y = \frac{11\,000}{10} = 1100$ : beans cost 900 Frw and rice 1100 Frw per kilogram. Check: $3 (900) + 2 (1100) = 4900$ and $900 + 4 (1100) = 5300$ . ✓

Worked example 2: the inverse-matrix method

Solve $2 x - 3 y = 7$ , $5 x + 4 y = 6$ using $X = A^{-1}B$ .

Solution. $D = 23$ , so $A^{-1} = \frac{1}{23}\begin{pmatrix} 4 & 3 \\ -5 & 2 \end{pmatrix}$ and

X = \frac{1}{23}\begin{pmatrix} 4 & 3 \\ -5 & 2 \end{pmatrix}\begin{pmatrix} 7 \\ 6 \end{pmatrix} = \frac{1}{23}\begin{pmatrix} 28 + 18 \\ -35 + 12 \end{pmatrix} = \frac{1}{23}\begin{pmatrix} 46 \\ -23 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}.

The solution $(2, - 1)$ is the point where the two lines cross in the diagram below. The inverse method pays off when you must solve several systems with the same left-hand sides: compute $A^{- 1}$ once.

When the determinant is zero

Each equation is a line, and the solution is where the lines meet. The determinant decides which picture you get:

Determinant	The two lines	Solutions
$D \ne 0$	cross at one point	exactly one: $x = \frac{D_x}{D}$ , $y = \frac{D_y}{D}$
$D = 0$ , and the equations are not multiples of each other	parallel and distinct	none (the system is inconsistent)
$D = 0$ , and one equation is a multiple of the other	the same line	infinitely many

$D = 0$ says the rows $(a, b)$ and $(c, d)$ are proportional, so the two lines have the same slope. The grey dashed line $2 x - 3 y = - 3$ is parallel to $2 x - 3 y = 7$ : together they have $D = 2 (- 3) - (- 3) (2) = 0$ and no common point.

Worked example 3: a system with a parameter

Discuss, according to the value of $k$ , the number of solutions of $x + k y = 2$ , $k x + 4 y = 4$ .

Solution. $D = 4 - k^{2} = (2 - k) (2 + k)$ .

If $k \ne 2$ and $k \ne -2$ : $D \ne 0$ , exactly one solution. (Cramer: $D_{x} = 8 - 4 k$ , $D_{y} = 4 - 2 k$ , so $x = \frac{4(2 - k)}{(2 - k)(2 + k)} = \frac{4}{2 + k}$ and $y = \frac{2}{2 + k}$ .)
If $k = 2$ : $x + 2 y = 2$ and $2 x + 4 y = 4$ . The second is twice the first: the same line, infinitely many solutions $(2 - 2t,\ t)$ , $t \in \mathbb{R}$ .
If $k = - 2$ : $x - 2 y = 2$ and $- 2 x + 4 y = 4$ , that is $x - 2 y = - 2$ . Parallel lines: no solution.

Areas of triangles and parallelograms

In the previous lesson, $|\det|$ was an area scale factor. Applied to two vectors it gives an area directly: the parallelogram built on $\overrightarrow{AB} = (p, q)$ and $\overrightarrow{AC} = (r, s)$ is the image of the unit square under $\begin{pmatrix} p & r \\ q & s \end{pmatrix}$ , so

\text{area of parallelogram} = \left| \begin{vmatrix} p & r \\ q & s \end{vmatrix} \right|, \qquad \text{area of triangle } ABC = \frac{1}{2}\left| \begin{vmatrix} p & r \\ q & s \end{vmatrix} \right| .

Three points $A, B, C$ are collinear exactly when this determinant is 0: the "triangle" has no area because $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are parallel.

Worked example 4: area and collinearity

(a) Find the area of the triangle $A (1, 1)$ , $B (6, 3)$ , $C (2, 5)$ . (b) Find $k$ so that $P (1, 2)$ , $Q (3, 5)$ and $R (k, 11)$ are collinear.

Solution. (a) $\overrightarrow{AB} = (5, 2)$ , $\overrightarrow{AC} = (1, 4)$ : $\begin{vmatrix} 5 & 1 \\ 2 & 4 \end{vmatrix} = 20 - 2 = 18$ , so the area is $\frac{1}{2}(18) = 9$ square units.

(b) $\overrightarrow{PQ} = (2, 3)$ , $\overrightarrow{PR} = (k - 1, 9)$ : $\begin{vmatrix} 2 & k - 1 \\ 3 & 9 \end{vmatrix} = 18 - 3(k - 1) = 0$ , so $k - 1 = 6$ and $k = 7$ . Check: $\overrightarrow{PR} = (6, 9) = 3(2, 3)$ . ✓

Gabriel Cramer (1704–1752), who taught mathematics in Geneva, in a portrait from around 1750. The rule that carries his name appears in his 1750 book on algebraic curves, where he needed to find a curve through given points. Portrait: unknown artist, public domain, via Wikimedia Commons.

Warning

$D = 0$ is a fork, not an answer. Cramer's rule cannot be used when $D = 0$ , and $D = 0$ alone does not tell you whether there are no solutions or infinitely many: compare the two equations, as in worked example 3. And keep the columns straight: $D_{x}$ replaces the column of $x$ -coefficients, not the first row.

Tip

Compute $D$ first, every time. If it is zero, stop and compare the equations; if it is not, Cramer's rule is three $2 \times 2$ determinants and two divisions. Substitute your answer back into both equations before you move on.

The Organic Chemistry Tutor: Cramer's rule for a 2x2 linear system (about 11 min). Watch how each numerator determinant is built by replacing one column with the constants.

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Practice

Work each one, then open the answers.

Solve by Cramer's rule: (a) $4 x + y = 9$ , $3 x - 2 y = 4$ (b) $5 x - 2 y = 1$ , $3 x + 4 y = 11$ .
Solve by the inverse-matrix method: $2 x + 5 y = - 1$ , $3 x + 7 y = 0$ .
How many solutions? (a) $6 x - 4 y = 10$ , $3 x - 2 y = 5$ (b) $6 x - 4 y = 10$ , $3 x - 2 y = 4$ (c) $6 x - 4 y = 10$ , $3 x + 2 y = 5$
For which $m$ does $m x + 3 y = 1$ , $12 x + m y = 2$ fail to have exactly one solution?
Find the area of the triangle $P (- 2, 1)$ , $Q (4, - 1)$ , $R (3, 5)$ .
Are the points $(- 1, - 4)$ , $(2, 2)$ and $(5, 8)$ collinear? Justify with a determinant.
Two exercise books and three pens cost 1300 Frw; five exercise books and one pen cost 1950 Frw (illustrative). Find the price of each.
Explain why the system $A X = B$ has exactly one solution for every $B$ when $\det A \ne 0$ .

Show answersHide answers

(a) $D = - 8 - 3 = - 11$ , $D_{x} = - 18 - 4 = - 22$ , $D_{y} = 16 - 27 = - 11$ : $x = 2$ , $y = 1$ . (b) $D = 20 + 6 = 26$ , $D_{x} = 4 + 22 = 26$ , $D_{y} = 55 - 3 = 52$ : $x = 1$ , $y = 2$ .
$D = 14 - 15 = - 1$ , $A^{-1} = \begin{pmatrix} -7 & 5 \\ 3 & -2 \end{pmatrix}$ , $X = A^{-1}\begin{pmatrix} -1 \\ 0 \end{pmatrix} = \begin{pmatrix} 7 \\ -3 \end{pmatrix}$ .
(a) $D = 0$ and the first equation is twice the second: infinitely many. (b) $D = 0$ , but $2(3x - 2y) = 8 \ne 10$ : none. (c) $D = 12 + 12 = 24 \ne 0$ : exactly one.
$D = m^{2} - 36 = 0$ : $m = 6$ or $m = - 6$ .
$\overrightarrow{PQ} = (6, -2)$ , $\overrightarrow{PR} = (5, 4)$ : $24 + 10 = 34$ , area $17$ square units.
Vectors $(3, 6)$ and $(6, 12)$ from the first point: $3 (12) - 6 (6) = 0$ . Yes, collinear.
$2 b + 3 p = 1300$ , $5 b + p = 1950$ : $D = 2 - 15 = - 13$ , $D_{b} = 1300 - 5850 = - 4550$ , $D_{p} = 3900 - 6500 = - 2600$ . An exercise book costs 350 Frw and a pen 200 Frw.
$A^{- 1}$ exists, so $X = A^{-1}B$ is a solution; and any solution satisfies $X = A^{-1}(AX) = A^{-1}B$ , so it is the only one.

Quick check

Key points

A system $a x + b y = e$ , $c x + d y = f$ is $A X = B$ ; it has exactly one solution when $D = \det A \ne 0$ .
Cramer's rule: $x = \frac{D_x}{D}$ , $y = \frac{D_y}{D}$ ; or $X = A^{-1}B$ .
$D = 0$ : parallel lines (no solution) or the same line (infinitely many).
Triangle $A B C$ has area $\frac{1}{2}|\det(\overrightarrow{AB}, \overrightarrow{AC})|$ ; the points are collinear exactly when the determinant is 0.

1Two shopping lists, two unknown prices

2When the determinant is zero

3Areas of triangles and parallelograms

Two shopping lists, two unknown prices

When the determinant is zero

Areas of triangles and parallelograms