Chapter 4 · Determinants

Exercise 4.1 · Q7: Find $x$ from equal determinants

Exercise 4.1Determinant of a matrix of order two · Section 4.2.2

The question

Find values of $x$ ⁠, if

\text{(i)}\ \begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix} = \begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}

\text{(ii)}\ \begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = \begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}

Exam answer

(i)

\begin{aligned} \begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix} &= \begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix} \\ \Rightarrow\ 2 - 20 &= 2x^2 - 24 \\ \Rightarrow\ 2x^2 &= 6 \\ \Rightarrow\ x^2 &= 3 \end{aligned}

$\therefore\ x = \pm\sqrt3$ ⁠.

(ii)

\begin{aligned} \begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} &= \begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix} \\ \Rightarrow\ 10 - 12 &= 5x - 6x \\ \Rightarrow\ -2 &= -x \end{aligned}

$\therefore\ x = 2$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

A determinant is a number (or, with $x$ in it, an expression in $x$ ⁠). So "determinant = determinant" is simply an equation between two numbers. Evaluate each side with $ad - bc$ ⁠, then solve the resulting equation for $x$ ⁠, keeping every solution. In (i) the $x$ appears twice on a diagonal, giving $x^2$ and two answers; in (ii) the $x$ 's are in one column, giving a linear equation and one answer.

Given, to find, and the plan

Given: Two equations, each between a numerical $2\times2$ determinant and a $2\times2$ determinant containing $x$ ⁠.
To find: All values of $x$ that make each equation true.
Plan: For each part: (1) evaluate the numerical determinant; (2) evaluate the determinant in $x$ ⁠; (3) set them equal and solve, finding all real solutions; (4) substitute each solution back into the original determinants.

Basics you need

1The $2\times2$ determinant

$|A| = a_{11}a_{22} - a_{21}a_{12}$ (Section 4.2.2). With $x$ in the entries, the result is an expression in $x$ ⁠. Lesson: determinant of a square matrix

2Solving $x^2 = c$

If $c > 0$ ⁠, then $x^2 = c$ has exactly two real solutions, $x = \sqrt{c}$ and $x = -\sqrt{c}$ ⁠, because $(-\sqrt c)^2 = c$ too. Writing only $\sqrt c$ loses a solution. Basics: algebra the chapter leans on

3Solving a linear equation

Collect the $x$ terms on one side and divide by the coefficient. Basics: algebra the chapter leans on

Part (i): an equation in $x^2$

Evaluate the left side
$\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix} = 2\cdot1 - 5\cdot4 = 2 - 20 = -18.$
Main diagonal $2\cdot1 = 2$ ⁠, other diagonal $5\cdot4 = 20$ ⁠, subtract.

Section 4.2.2. Lesson: determinant of a square matrix
Evaluate the right side
$\begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix} = (2x)(x) - 6\cdot4 = 2x^2 - 24.$
The main diagonal holds $2x$ and $x$ ⁠, whose product is $2x^2$ ⁠. The other diagonal holds $6$ and $4$ ⁠. Both $x$ 's sit on the same diagonal, which is why $x^2$ appears.
Set equal and solve
$\begin{aligned} 2x^2 - 24 &= -18\\ 2x^2 &= 6\\ x^2 &= 3\\ x &= \sqrt3 \ \text{ or } \ x = -\sqrt3. \end{aligned}$
Add 24 to both sides, then divide by 2. Since $3 > 0$ ⁠, the equation $x^2 = 3$ has two real solutions, $\pm\sqrt3$ (Basic 2). Both must be kept: the question asks for the values of $x$ ⁠.

Same method as the book's Example 5, where $x^2 = 8$ gives $x = \pm2\sqrt2$ ⁠.
Check both solutions

For $x = \sqrt3$ ⁠: the right side is $2\sqrt3\cdot\sqrt3 - 24$ ⁠, which is $2\cdot3 - 24 = -18$ ✓.

For $x = -\sqrt3$ ⁠: it is $2(-\sqrt3)(-\sqrt3) - 24$ ⁠, which is $2\cdot3 - 24 = -18$ ✓.

So $x = \pm\sqrt3$ .

Substituting into the determinant itself, not just into $x^2 = 3$ ⁠, checks the evaluation in Step 2 as well as the algebra. The two products $(\sqrt3)(\sqrt3)$ and $(-\sqrt3)(-\sqrt3)$ are both 3, so both values work.

Part (ii): a linear equation

Evaluate the left side
$\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = 2\cdot5 - 4\cdot3 = 10 - 12 = -2.$
Main diagonal $2\cdot5 = 10$ ⁠, other diagonal $4\cdot3 = 12$ ⁠.
Evaluate the right side
$\begin{aligned} \begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix} &= x\cdot5 - (2x)\cdot3\\ &= 5x - 6x = -x. \end{aligned}$
This time the two $x$ 's are in the same column, so they are on different diagonals and never multiply each other. Each diagonal product has a single factor $x$ ⁠, and the result is linear in $x$ ⁠.
Set equal and solve
$\begin{aligned} -x &= -2\\ x &= 2. \end{aligned}$
Multiply both sides by $-1$ ⁠. A linear equation $ax = b$ with $a \ne 0$ has exactly one solution, so there is nothing else to find.
Check

For $x = 2$ the right-hand determinant has rows $2, 3$ and $4, 5$ ⁠, and its value is $10 - 12 = -2$ ✓. In fact, at $x = 2$ the two determinants are identical. So $x = 2$ .

Replacing $x$ by 2 turns $2x$ into 4, and the right-hand array becomes exactly the left-hand one; equal arrays certainly have equal determinants. Step 3 shows this is the only value that works.

Result

$\text{(i)}\ x = \pm\sqrt3 \qquad \text{(ii)}\ x = 2$ (Textbook answer: (i) $x = \pm\sqrt3$ ⁠, (ii) $x = 2$ ⁠.)

Check it with numbers

(i) $\sqrt3 \approx 1.7320508$ ⁠, and $1.7320508^2 \approx 3.0000000$ ⁠, so $2x\cdot x = 2x^2 \approx 6.0000000$ and the right side is $6 - 24 = -18$ ✓. A wrong value such as $x = 2$ gives $8 - 24 = -16 \ne -18$ ⁠.

(ii) Try a non-solution, $x = 3$ ⁠: the right side is $-3 \ne -2$ ⁠, as $-x$ predicts ✓.

Common mistakes

(i): giving only $x = \sqrt3$ ⁠. $x^2 = 3$ has two real roots; the answer is $\pm\sqrt3$ ⁠.
(i): $2x\cdot x = 2x$ ⁠. The product of $2x$ and $x$ is $2x^2$ ⁠.
(ii): $5x - 6x = x$ ⁠. It is $-x$ ⁠; then $-x = -2$ gives $x = 2$ ⁠, while the slip would give $x = -2$ ⁠, which fails the check: $-(-2) = 2 \ne -2$ ⁠.

What to take away

An equation between determinants is an equation between numbers: evaluate both sides, then solve.
Where the unknowns sit decides the type of equation: on one diagonal they multiply (quadratic); in one column they do not (linear).
Find all solutions and check each one in the original determinants.

Every fact used, and where it comes from

Fact	Used in	Source
$\|A\| = a_{11}a_{22} - a_{21}a_{12}$ for order 2	Part (i), Steps 1, 2; Part (ii), Steps 1, 2	Section 4.2.2 (Lesson: determinant of a square matrix)
$x^2 = c$ with $c > 0$ has the two solutions $\pm\sqrt c$	Part (i), Step 3	Algebra (Basics: algebra)
Equating two determinants to find $x$	Both parts	Example 5
All values and checks on this page	Both parts, Check	Recomputed in the site's verification script

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Exercise 4.1 · Q7: Find xxx from equal determinants

Basics you need

Part (i): an equation in x2x^2x2

Evaluate the left side

Evaluate the right side

Set equal and solve

Check both solutions