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Chapter 4 · Determinants

Miscellaneous · Q9 (MCQ): Where does det⁡A\det A lie?

Miscellaneous ExerciseExpanding a 3 × 3 determinant · Section 4.2.3
The question

Let

A=[1sin⁡θ1−sin⁡θ1sin⁡θ−1−sin⁡θ1],A = \begin{bmatrix} 1 & \sin\theta & 1 \\ -\sin\theta & 1 & \sin\theta \\ -1 & -\sin\theta & 1 \end{bmatrix},

where 0≤θ≤2π0 \le \theta \le 2\pi⁠. Then

(A) Det⁡(A)=0\operatorname{Det}(A) = 0
(B) Det⁡(A)∈(2,∞)\operatorname{Det}(A) \in (2, \infty)
(C) Det⁡(A)∈(2,4)\operatorname{Det}(A) \in (2, 4)
(D) Det⁡(A)∈[2,4]\operatorname{Det}(A) \in [2, 4]

(This question comes under the book's line "Choose the correct answer in Exercise 17 to 19." Those numbers are left over from an older edition; the line refers to Questions 8 and 9.)

Exam answer

Expanding along R1R_1⁠:

∣A∣=1∣1sin⁡θ−sin⁡θ1∣−sin⁡θ∣−sin⁡θsin⁡θ−11∣+1∣−sin⁡θ1−1−sin⁡θ∣=(1+sin⁡2θ)−sin⁡θ(0)+(sin⁡2θ+1)=2+2sin⁡2θ\begin{aligned} |A| &= 1\begin{vmatrix} 1 & \sin\theta \\ -\sin\theta & 1 \end{vmatrix} \\ &\quad - \sin\theta\begin{vmatrix} -\sin\theta & \sin\theta \\ -1 & 1 \end{vmatrix} \\ &\quad + 1\begin{vmatrix} -\sin\theta & 1 \\ -1 & -\sin\theta \end{vmatrix} \\ &= (1 + \sin^2\theta) - \sin\theta(0) \\ &\quad + (\sin^2\theta + 1) \\ &= 2 + 2\sin^2\theta \end{aligned}

For 0≤θ≤2π0 \le \theta \le 2\pi⁠, sin⁡2θ\sin^2\theta takes every value from 0 to 1, so ∣A∣|A| takes every value from 2 to 4.

Answer: (D) Det⁡(A)∈[2,4]\operatorname{Det}(A) \in [2, 4]

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

The one idea this problem is built on

Write s=sin⁡θs = \sin\theta and expand: the determinant is 2+2s22 + 2s^2⁠. So the question becomes "which values can sin⁡2θ\sin^2\theta take?" On 0≤θ≤2π0 \le \theta \le 2\pi⁠, sin⁡θ\sin\theta runs through every number from −1-1 to 11⁠, so sin⁡2θ\sin^2\theta runs through every number from 0 to 1, including both ends. Hence the determinant takes every value from 2 to 4, ends included: the closed interval [2,4][2, 4]⁠.

Given, to find, and the plan
Given
The 3×33\times3 matrix A above, for 0≤θ≤2π0 \le \theta \le 2\pi⁠.
To find
Exactly which values ∣A∣|A| takes, and the matching option.
Plan
(1) Put s=sin⁡θs = \sin\theta and expand along R1R_1⁠. (2) Simplify to 2+2s22 + 2s^2⁠. (3) Find the exact set of values of s2s^2⁠, then of ∣A∣|A|⁠. (4) Examine every option.

Basics you need

1Expansion of a 3×33\times3 determinant

Along R1R_1⁠: a11M11−a12M12+a13M13a_{11}M_{11} - a_{12}M_{12} + a_{13}M_{13}⁠. Lesson: expanding a 3 × 3 determinant

2A 2×22\times2 determinant

Rows (a,b)(a, b) and (c,d)(c, d) give ad−bcad - bc⁠. Basics: where the determinant comes from

3Values of sin⁡θ\sin\theta

−1≤sin⁡θ≤1-1 \le \sin\theta \le 1 for every θ\theta⁠. On 0≤θ≤π20 \le \theta \le \frac{\pi}{2}⁠, sin⁡θ\sin\theta increases from 0 to 1 and takes every value in between. Basics: algebra the chapter leans on

4Interval notation

[2,4][2, 4] includes 2 and 4; (2,4)(2, 4) excludes both; (2,∞)(2, \infty) means all numbers greater than 2.

Solution

  1. Expand along the first row

    Write s=sin⁡θs = \sin\theta⁠. The rows are (1,s,1)(1, s, 1)⁠, (−s,1,s)(-s, 1, s)⁠, (−1,−s,1)(-1, -s, 1)⁠. Expand along R1R_1 with signs +,−,++, -, +⁠:

    M11=∣1s−s1∣=1−(−s2)=1+s2M_{11} = \begin{vmatrix} 1 & s \\ -s & 1 \end{vmatrix} = 1 - (-s^2) = 1 + s^2 M12=∣−ss−11∣=−s−(−s)=0M_{12} = \begin{vmatrix} -s & s \\ -1 & 1 \end{vmatrix} = -s - (-s) = 0 M13=∣−s1−1−s∣=s2−(−1)=s2+1M_{13} = \begin{vmatrix} -s & 1 \\ -1 & -s \end{vmatrix} = s^2 - (-1) = s^2 + 1 ∣A∣=1⋅(1+s2)−s⋅0+1⋅(s2+1)=2+2s2=2+2sin⁡2θ.\begin{aligned} |A| &= 1\cdot(1 + s^2) - s\cdot0 + 1\cdot(s^2 + 1) \\ &= 2 + 2s^2 = 2 + 2\sin^2\theta. \end{aligned}
    No entry is zero, but R1R_1 has two entries equal to 1, which keeps the products simple, and its middle minor turns out to be 0. In M11M_{11}⁠, 1⋅1=11\cdot1 = 1 and s⋅(−s)=−s2s\cdot(-s) = -s^2⁠. In M12M_{12}⁠, (−s)⋅1=−s(-s)\cdot1 = -s and s⋅(−1)=−ss\cdot(-1) = -s⁠, so the difference is 0. In M13M_{13}⁠, (−s)(−s)=s2(-s)(-s) = s^2 and 1⋅(−1)=−11\cdot(-1) = -1⁠.
    Section 4.2.3 (expansion along R1R_1⁠). Basics: the + − + sign pattern

  2. Bounds: 2≤∣A∣≤42 \le |A| \le 4

    For every real θ\theta⁠, −1≤sin⁡θ≤1-1 \le \sin\theta \le 1⁠, so

    0≤sin⁡2θ≤1.0 \le \sin^2\theta \le 1.

    Multiply by 2 and add 2:

    2≤2+2sin⁡2θ≤4,that is,2≤∣A∣≤4.2 \le 2 + 2\sin^2\theta \le 4, \quad\text{that is,}\quad 2 \le |A| \le 4.
    Squaring a number between −1-1 and 11 gives a number between 0 and 1 (a square is never negative, and its size is at most 1). Multiplying an inequality by the positive number 2 and adding 2 keeps its direction.
    Range of the sine function. Basics: algebra the chapter leans on

  3. Every value in [2,4][2, 4] actually occurs

    The ends are reached:

    θ=0,π,2π:sin⁡2θ=0, ∣A∣=2θ=π2,3π2:sin⁡2θ=1, ∣A∣=4\begin{aligned} \theta = 0, \pi, 2\pi &: \sin^2\theta = 0, \ |A| = 2 \\ \theta = \tfrac{\pi}{2}, \tfrac{3\pi}{2} &: \sin^2\theta = 1, \ |A| = 4 \end{aligned}

    Values in between are reached too. Take any dd with 2≤d≤42 \le d \le 4⁠. We need sin⁡2θ=d−22\sin^2\theta = \frac{d - 2}{2}⁠, a number tt with 0≤t≤10 \le t \le 1⁠. Then t\sqrt{t} lies in [0,1][0, 1]⁠, and as θ\theta goes from 0 to π2\frac{\pi}{2}⁠, sin⁡θ\sin\theta increases continuously from 0 to 1, so it passes through t\sqrt{t} at some θ\theta in [0,π2][0, \frac{\pi}{2}]⁠. At that θ\theta⁠, sin⁡2θ=t\sin^2\theta = t and ∣A∣=2+2t=d|A| = 2 + 2t = d⁠.

    So the set of values of ∣A∣|A| for 0≤θ≤2π0 \le \theta \le 2\pi is exactly [2,4][2, 4].

    43210 π/2π3π/22π largest value 4 (θ = π/2, 3π/2) smallest value 2 (θ = 0, π, 2π) det A θ
    The graph of ∣A∣=2+2sin⁡2θ|A| = 2 + 2\sin^2\theta for 0≤θ≤2π0 \le \theta \le 2\pi (horizontal and vertical scales differ). It never leaves the shaded band from 2 to 4, it touches both edges, and it passes through every height in between: the set of values is the closed interval [2,4][2, 4]⁠.
    Step 2 only shows that the values lie inside [2,4][2, 4]⁠. To say the values form exactly [2,4][2, 4] we also need each number in it to occur; this matters for the options, which differ only in whether 2 and 4 are included. The continuity of sin⁡θ\sin\theta on [0,π2][0, \frac{\pi}{2}] guarantees that no value in between is skipped.
    sin⁡0=sin⁡π=sin⁡2π=0\sin 0 = \sin\pi = \sin 2\pi = 0⁠; sin⁡π2=1\sin\frac{\pi}{2} = 1⁠, sin⁡3π2=−1\sin\frac{3\pi}{2} = -1⁠. Basics: algebra the chapter leans on

  4. Examine every option

    OptionClaimVerdict
    (A)∣A∣=0|A| = 0Wrong: ∣A∣≥2|A| \ge 2 always, so it is never 0.
    (B)∣A∣∈(2,∞)|A| \in (2, \infty)Wrong: at θ=0\theta = 0⁠, ∣A∣=2|A| = 2⁠, which is not in (2,∞)(2, \infty)⁠. (Also, values above 4 never occur.)
    (C)∣A∣∈(2,4)|A| \in (2, 4)Wrong: ∣A∣=2|A| = 2 at θ=0\theta = 0 and ∣A∣=4|A| = 4 at θ=π2\theta = \frac{\pi}{2}⁠, and the open interval excludes both.
    (D)∣A∣∈[2,4]|A| \in [2, 4]Correct: true for every θ\theta (Step 2), and the values fill the whole interval (Step 3).
    Each wrong option fails at a specific θ\theta in the given range, and naming that θ\theta is a complete disproof. The difference between (C) and (D) is only the endpoints, which is why Step 3 checked that 2 and 4 are actually attained.
    A single counterexample disproves a claim that is supposed to hold for every θ\theta⁠.
Result

∣A∣=2+2sin⁡2θ|A| = 2 + 2\sin^2\theta⁠, whose values for 0≤θ≤2π0 \le \theta \le 2\pi are exactly the numbers from 2 to 4. The correct answer is (D): Det⁡(A)∈[2,4]\operatorname{Det}(A) \in [2, 4]⁠. (Textbook answer: D.)

Check it with numbers

At θ=π6\theta = \frac{\pi}{6}⁠, s=12s = \frac12⁠: the rows are (1,12,1)(1, \frac12, 1)⁠, (−12,1,12)(-\frac12, 1, \frac12)⁠, (−1,−12,1)(-1, -\frac12, 1)⁠, and along R1R_1 the determinant is 1⋅54−12⋅0+1⋅54=521\cdot\frac54 - \frac12\cdot0 + 1\cdot\frac54 = \frac52⁠, which equals 2+2⋅142 + 2\cdot\frac14 ✓. At θ=2\theta = 2 (radians), sin⁡2≈0.909297\sin 2 \approx 0.909297⁠, sin⁡22≈0.826822\sin^2 2 \approx 0.826822⁠, and the determinant is about 3.653644=2+2(0.826822)3.653644 = 2 + 2(0.826822) ✓. The verification script evaluates the determinant at 2001 equally spaced θ\theta in [0,2π][0, 2\pi]⁠: the smallest value found is 2 and the largest is 4.

Common mistakes
  1. Choosing (C) by forgetting that sin⁡2θ\sin^2\theta actually reaches 0 and 1 in the given range.
  2. Choosing (B) from 2+2sin⁡2θ≥22 + 2\sin^2\theta \ge 2 alone, without the upper bound and without noticing the value 2 itself.
  3. Sign slips in the minors, for example writing M11=1−s2M_{11} = 1 - s^2 (forgetting that subtracting s⋅(−s)s\cdot(-s) adds s2s^2⁠). The s2s^2 terms would then cancel, and the determinant would wrongly appear to be the constant 2.
What to take away
  1. To find where a trigonometric expression lies, reduce it to one function such as sin⁡2θ\sin^2\theta and use that function's exact set of values.
  2. Open versus closed intervals matter: check whether each end value is actually reached in the given range of θ\theta⁠.

Every fact used, and where it comes from

FactUsed inSource
Expansion of a 3×33\times3 determinant along R1R_1⁠; 2×22\times2 determinantsStep 1, CheckSections 4.2.2, 4.2.3 (Lesson: expanding a 3 × 3 determinant)
−1≤sin⁡θ≤1-1 \le \sin\theta \le 1⁠; sin⁡θ\sin\theta increases continuously from 0 to 1 on [0,π2][0, \frac{\pi}{2}]Steps 2, 3Class XI trigonometry (Basics: algebra the chapter leans on)
Inequalities keep their direction when multiplied by a positive numberStep 2Algebra of inequalities
∣A∣=2+2sin⁡2θ|A| = 2 + 2\sin^2\theta symbolically, the values at sample angles, and the figure's coordinatesSteps 1–4, Check, figureRecomputed with sympy and numerically in the site's verification script
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