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Chapter 4 · Determinants

Example 5: Find xx from two equal determinants

ExamplesDeterminant of a matrix of order two · Section 4.2.2
The question

Find values of xx for which

∣3xx1∣=∣3241∣.\begin{vmatrix} 3 & x \\ x & 1 \end{vmatrix} = \begin{vmatrix} 3 & 2 \\ 4 & 1 \end{vmatrix}.
Exam answer ∣3xx1∣=∣3241∣⇒ 3−x2=3−8⇒ x2=8\begin{aligned} \begin{vmatrix} 3 & x \\ x & 1 \end{vmatrix} &= \begin{vmatrix} 3 & 2 \\ 4 & 1 \end{vmatrix} \\ \Rightarrow\ 3 - x^2 &= 3 - 8 \\ \Rightarrow\ x^2 &= 8 \end{aligned}

∴ x=±22\therefore\ x = \pm 2\sqrt2⁠.

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

The one idea this problem is built on

Each side is a determinant, which is just a number (or an expression in xx⁠). Evaluating both sides turns the "determinant equation" into an ordinary equation in xx⁠. Here it becomes x2=8x^2 = 8⁠, which has two real solutions, x=22x = 2\sqrt2 and x=−22x = -2\sqrt2⁠. Both must be given.

Given, to find, and the plan
Given
Two 2×22\times2 determinants, the left one containing xx twice, the right one all numbers; they are equal.
To find
Every real xx that makes them equal.
Plan
Evaluate each side with the 2×22\times2 rule, solve the resulting equation, simplify the surd, and check each root.

Basics you need

1The 2×22\times2 rule

(main-diagonal product) −- (other-diagonal product), Section 4.2.2. Basics: where the determinant comes from

2Solving x2=cx^2 = c with c>0c > 0

There are exactly two real solutions, x=cx = \sqrt c and x=−cx = -\sqrt c⁠, because (c)2=c(\sqrt c)^2 = c and (−c)2=c(-\sqrt c)^2 = c too. Taking only the positive root loses half the answer. Basics: algebra the chapter leans on

3Simplifying 8\sqrt8

ab=a b\sqrt{ab} = \sqrt a\,\sqrt b for a,b≥0a, b \ge 0⁠. Since 8=4×28 = 4\times 2⁠, 8=4 2=22\sqrt 8 = \sqrt4\,\sqrt2 = 2\sqrt2⁠. Basics: algebra the chapter leans on

Solution

  1. Evaluate the left-hand determinant

    ∣3xx1∣=3(1)−x(x)=3−x2.\begin{vmatrix} 3 & x \\ x & 1 \end{vmatrix} = 3(1) - x(x) = 3 - x^2.
    The main diagonal is 33 and 11⁠; the other diagonal is xx and xx⁠. The rule subtracts the second product from the first. The result still contains xx⁠, because the entries do.
    ∣A∣=a11a22−a21a12|A| = a_{11}a_{22} - a_{21}a_{12}⁠, Section 4.2.2. Lesson: determinant of a square matrix

  2. Evaluate the right-hand determinant

    ∣3241∣=3(1)−4(2)=3−8=−5.\begin{vmatrix} 3 & 2 \\ 4 & 1 \end{vmatrix} = 3(1) - 4(2) = 3 - 8 = -5.
    The same rule: 3×13\times1 on the main diagonal, 4×24\times2 on the other. This side is a fixed number, −5-5⁠.

  3. Set the two values equal and solve

    3−x2=3−83−x2=−5−x2=−8x2=8.\begin{aligned} 3 - x^2 &= 3 - 8\\ 3 - x^2 &= -5\\ -x^2 &= -8\\ x^2 &= 8. \end{aligned}
    The question says the two determinants are equal, so their values are equal. Subtracting 3 from both sides gives −x2=−8-x^2 = -8⁠, and multiplying both sides by −1-1 gives x2=8x^2 = 8⁠. These steps are reversible, so no solution is lost or gained.
    Doing the same operation to both sides keeps an equation true. Basics: algebra the chapter leans on

  4. Take both square roots and simplify

    x=±8=±4⋅2=±22.\begin{aligned} x &= \pm\sqrt8\\ &= \pm\sqrt{4\cdot 2}\\ &= \pm 2\sqrt2. \end{aligned}
    Since 8>08 > 0⁠, the equation x2=8x^2 = 8 has two real solutions, one positive and one negative (Basic 2). Writing 8\sqrt8 as 222\sqrt2 takes out the largest perfect-square factor, 4, which is the simplest form of the surd.
    x2=c, c>0  ⟺  x=cx^2 = c,\ c > 0 \iff x = \sqrt c or x=−cx = -\sqrt c⁠. ab=ab\sqrt{ab} = \sqrt a\sqrt b⁠. Basics: algebra the chapter leans on

  5. Check both roots in the original equation

    For either root, x2=(±22)2=4×2=8x^2 = (\pm 2\sqrt2)^2 = 4\times 2 = 8⁠. So the left-hand determinant is

    3−x2=3−8=−5,3 - x^2 = 3 - 8 = -5,

    which equals the right-hand side. Both x=22x = 2\sqrt2 and x=−22x = -2\sqrt2 satisfy the equation.

    Substituting back into the original equation (not into a later line) confirms nothing went wrong in the algebra. The left side depends on xx only through x2x^2⁠, which is why the positive and negative roots behave identically.
Result x=22  or  x=−22,x = 2\sqrt2 \ \text{ or } \ x = -2\sqrt2,

that is, x=±22x = \pm 2\sqrt2⁠.

This agrees with the textbook's solution.

Check it with numbers

22≈2.82842712\sqrt2 \approx 2.8284271⁠, and 2.82842712≈8.00000002.8284271^2 \approx 8.0000000⁠. Then 3−8=−53 - 8 = -5⁠, matching 3×1−4×2=−53\times1 - 4\times2 = -5⁠. A value that is not a root, say x=3x = 3⁠, gives 3−9=−6≠−53 - 9 = -6 \ne -5⁠, so the equation really does pick out only ±22\pm2\sqrt2⁠.

Common mistakes
  1. Giving only x=22x = 2\sqrt2⁠. x2=8x^2 = 8 has a negative solution as well.
  2. Subtracting the diagonals the wrong way round. The right side is 3×1−4×2=−53\times1 - 4\times2 = -5⁠. Computing 4×2−3×1=54\times2 - 3\times1 = 5 instead gives 3−x2=53 - x^2 = 5⁠, so x2=−2x^2 = -2⁠, which has no real solution at all.
  3. Writing 8=42\sqrt8 = 4\sqrt2 or 242\sqrt4⁠. 8=42=22\sqrt8 = \sqrt4\sqrt2 = 2\sqrt2⁠.
What to take away
  1. An equation between determinants is solved by evaluating each determinant and solving the ordinary equation that results.
  2. When xx appears on both diagonals, a quadratic often appears; give every real root and check each one.

Every fact used, and where it comes from

FactUsed inSource
∣A∣=a11a22−a21a12|A| = a_{11}a_{22} - a_{21}a_{12} for order 2Steps 1, 2Section 4.2.2 (Lesson: determinant of a square matrix)
Equal determinants have equal valuesStep 3A determinant is a number, Section 4.2
x2=cx^2 = c⁠, c>0c>0⁠, has roots ±c\pm\sqrt cStep 4Algebra (earlier classes)
8=22\sqrt8 = 2\sqrt2Step 4ab=ab\sqrt{ab} = \sqrt a\sqrt b
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