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Chapter 4 · Determinants

Exercise 4.1 · Q8 (MCQ): Find xx from equal determinants

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

If

∣x218x∣=∣62186∣,\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = \begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix},

then xx is equal to

(A) 66
(B) ±6\pm 6
(C) −6-6
(D) 00

Exam answer ∣x218x∣=∣62186∣⇒ x2−36=36−36=0⇒ x=±6\begin{aligned} \begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} &= \begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix} \\ \Rightarrow\ x^2 - 36 &= 36 - 36 = 0 \\ \Rightarrow\ x &= \pm 6 \end{aligned}

Answer: (B) ±6\pm 6

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

The one idea this problem is built on

Evaluate both determinants with ad−bcad - bc⁠. The left side is x2−36x^2 - 36⁠, because both xx's lie on the main diagonal, and the right side is 36−36=036 - 36 = 0⁠. The equation x2=36x^2 = 36 has two solutions, 66 and −6-6⁠, so the complete answer is ±6\pm6⁠.

Given, to find, and the plan
Given
An equation between a 2×22\times2 determinant containing xx and a numerical one.
To find
Which option gives the values of xx⁠.
Plan
(1) Evaluate the right side. (2) Evaluate the left side. (3) Solve, finding all real solutions. (4) Check each solution. (5) Examine every option.

Basics you need

1The 2×22\times2 determinant

∣A∣=a11a22−a21a12|A| = a_{11}a_{22} - a_{21}a_{12}⁠: main diagonal product minus the other diagonal product (Section 4.2.2). Lesson: determinant of a square matrix

2Solving x2=cx^2 = c

For c>0c > 0 there are exactly two real solutions, c\sqrt c and −c-\sqrt c⁠. Equivalently, x2−c=(x−c)(x+c)=0x^2 - c = (x - \sqrt c)(x + \sqrt c) = 0⁠, and a product is 0 only when a factor is 0. Basics: algebra the chapter leans on

3What "xx is equal to" asks for

The correct option must describe all values of xx that satisfy the equation, and nothing else.

Solution

  1. Evaluate the right side

    ∣62186∣=6⋅6−18⋅2=36−36=0.\begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix} = 6\cdot6 - 18\cdot2 = 36 - 36 = 0.
    Main diagonal 6⋅6=366\cdot6 = 36⁠; other diagonal 18⋅2=3618\cdot2 = 36⁠. They are equal, so the determinant is 0. (The first row 6,26, 2 is one third of the second row 18,618, 6⁠, which is why: with rows tc,tdtc, td and c,dc, d the rule gives tcd−ctd=0tcd - ctd = 0 for any tt⁠.)
    Section 4.2.2. Lesson: determinant of a square matrix

  2. Evaluate the left side

    ∣x218x∣=x⋅x−18⋅2=x2−36.\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = x\cdot x - 18\cdot2 = x^2 - 36.
    Both xx's are on the main diagonal, so they multiply to x2x^2⁠. The other diagonal is 18⋅2=3618\cdot2 = 36⁠.

  3. Solve x2−36=0x^2 - 36 = 0

    x2−36=0(x−6)(x+6)=0x=6  or  x=−6.\begin{aligned} x^2 - 36 &= 0\\ (x - 6)(x + 6) &= 0\\ x = 6 \ &\text{ or } \ x = -6. \end{aligned}
    Factorise the difference of squares x2−62x^2 - 6^2⁠. A product is 0 exactly when one of its factors is 0, so x−6=0x - 6 = 0 or x+6=0x + 6 = 0⁠. There are no other solutions.
    a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b)⁠. Basics: algebra the chapter leans on

  4. Check both solutions

    For x=6x = 6 the left-hand array becomes rows 6,26, 2 and 18,618, 6⁠, the same as the right-hand one, so the two sides are equal ✓.

    For x=−6x = -6⁠: (−6)(−6)−18⋅2=36−36=0(-6)(-6) - 18\cdot2 = 36 - 36 = 0 ✓.

    So x=±6x = \pm6.

    Substituting into the original determinant confirms the evaluation in Step 2 and the algebra in Step 3 together. The value x=−6x = -6 works because (−6)2=36(-6)^2 = 36 as well.

  5. Examine every option

    OptionVerdict
    (A) 66Incomplete. x=6x = 6 satisfies the equation, but so does x=−6x = -6⁠, which this option leaves out.
    (B) ±6\pm6Correct. Exactly the two solutions found in Step 3, both checked in Step 4.
    (C) −6-6Incomplete, for the same reason: it leaves out x=6x = 6⁠.
    (D) 00Wrong. At x=0x = 0 the left side is 0−36=−36≠00 - 36 = -36 \ne 0⁠.
    (A) and (C) each name a genuine solution, but "xx is equal to" asks for the values of xx⁠, and each of these options misses one of them. Only (B) lists the full solution set. Option (D) probably comes from noticing that the right side is 0 and wrongly concluding that xx must be 0.
Result

The correct answer is (B): x=±6x = \pm6⁠, because x2−36=0x^2 - 36 = 0⁠. (Textbook answer: (B).)

Check it with numbers

Try values around the solutions: at x=5x = 5 the left side is 25−36=−1125 - 36 = -11⁠; at x=7x = 7 it is 49−36=1349 - 36 = 13⁠. It changes sign between 5 and 7 and is 0 at 6 ✓. The same happens between −7-7 and −5-5⁠, at −6-6 ✓, since only x2x^2 appears.

Common mistakes
  1. Choosing (A) by solving x2=36x^2 = 36 as x=6x = 6 only. The negative square root is also a solution.
  2. Choosing (D) because the right side is 0. The equation is x2−36=0x^2 - 36 = 0⁠, not x=0x = 0⁠.
  3. Comparing entries instead of determinants. Matching the arrays entry by entry gives only x=6x = 6⁠; but the question equates determinants (numbers), and different arrays can have the same determinant.
What to take away
  1. Equal determinants means equal numbers, not equal arrays: evaluate, then solve.
  2. When xx appears on one diagonal twice, expect x2x^2 and two solutions.

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, 2, 4, 5Section 4.2.2 (Lesson: determinant of a square matrix)
a2−b2=(a−b)(a+b)a^2 - b^2 = (a-b)(a+b)⁠; a product is 0 only if a factor is 0Step 3Algebra (Basics: algebra)
Equating two determinants to find xxSteps 1–4Example 5, Exercise 4.1 Q7
Values and option testsSteps 4, 5, CheckRecomputed in the site's verification script
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