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Chapter 4 · Determinants

Miscellaneous · Q1: A determinant independent of θ\theta

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

Prove that the determinant

∣xsin⁡θcos⁡θ−sin⁡θ−x1cos⁡θ1x∣\begin{vmatrix} x & \sin\theta & \cos\theta \\ -\sin\theta & -x & 1 \\ \cos\theta & 1 & x \end{vmatrix}

is independent of θ\theta⁠.

Exam answer

Let Δ\Delta be the given determinant.

Expanding along R1R_1⁠:

Δ=x∣−x11x∣−sin⁡θ∣−sin⁡θ1cos⁡θx∣+cos⁡θ∣−sin⁡θ−xcos⁡θ1∣=x(−x2−1)−sin⁡θ(−xsin⁡θ−cos⁡θ)+cos⁡θ(−sin⁡θ+xcos⁡θ)=−x3−x+xsin⁡2θ+sin⁡θcos⁡θ−sin⁡θcos⁡θ+xcos⁡2θ=−x3−x+x(sin⁡2θ+cos⁡2θ)\begin{aligned} \Delta &= x\begin{vmatrix} -x & 1 \\ 1 & x \end{vmatrix} \\ &\quad - \sin\theta\begin{vmatrix} -\sin\theta & 1 \\ \cos\theta & x \end{vmatrix} \\ &\quad + \cos\theta\begin{vmatrix} -\sin\theta & -x \\ \cos\theta & 1 \end{vmatrix} \\ &= x(-x^2 - 1) \\ &\quad - \sin\theta(-x\sin\theta - \cos\theta) \\ &\quad + \cos\theta(-\sin\theta + x\cos\theta) \\ &= -x^3 - x + x\sin^2\theta \\ &\quad + \sin\theta\cos\theta - \sin\theta\cos\theta \\ &\quad + x\cos^2\theta \\ &= -x^3 - x \\ &\quad + x(\sin^2\theta + \cos^2\theta) \end{aligned}

Since sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1⁠,

Δ=−x3−x+x\Delta = -x^3 - x + x

∴ Δ=−x3\therefore\ \Delta = -x^3⁠, which does not contain θ\theta⁠. Hence the determinant is independent of θ\theta⁠.

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

The one idea this problem is built on

"Independent of θ\theta" means: once the determinant is worked out, θ\theta is no longer in the answer. So we simply expand the determinant and watch the θ\theta-terms. The two products sin⁡θcos⁡θ\sin\theta\cos\theta cancel each other, and the remaining terms combine into x(sin⁡2θ+cos⁡2θ)x(\sin^2\theta + \cos^2\theta)⁠, which the identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 turns into plain xx⁠. What is left is −x3-x^3⁠.

Given, to show, and the plan
Given
A 3×33\times3 determinant, call it Δ\Delta⁠, whose entries involve a number xx and an angle θ\theta⁠.
To show
The value of Δ\Delta contains no θ\theta⁠: it is the same for every θ\theta⁠.
Plan
(1) Choose a row to expand along. (2) Write the three minors as 2×22\times2 determinants and evaluate them. (3) Form the signed sum. (4) Simplify with sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1⁠. (5) Check with numbers.

Basics you need

1A 2×22\times2 determinant

The determinant with rows (a,b)(a, b) and (c,d)(c, d) is ad−bcad - bc⁠: the down-right diagonal product minus the up-right one. Basics: where the determinant comes from

2Expanding a 3×33\times3 determinant

Pick a row; multiply each entry by its minor (the 2×22\times2 determinant left after deleting that entry's row and column); attach the signs from the pattern, then add. Any row or column gives the same value. Lesson: expanding a 3 × 3 determinant

3The sign pattern

The entry in row ii⁠, column jj gets the sign (−1)i+j(-1)^{i+j}⁠: along the first row this is +,−,++, -, +⁠. Basics: the + − + sign pattern

4The Pythagorean identity

sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 for every angle θ\theta⁠. It is the only trigonometry needed here. Basics: algebra the chapter leans on

Solution

  1. Choose the row to expand along

    No row or column contains a zero, so no choice saves work by killing a term. We expand along the first row R1R_1⁠, whose entries are xx⁠, sin⁡θ\sin\theta⁠, cos⁡θ\cos\theta⁠, with the signs +,−,++, -, +⁠:

    Δ=x M11−sin⁡θ M12+cos⁡θ M13.\Delta = x\,M_{11} - \sin\theta\,M_{12} + \cos\theta\,M_{13}.
    Expanding along any row or column gives the same value (Section 4.2.3), and the book's advice is to choose the line with the most zeros. With none available, R1R_1 is a natural choice: its minors come from rows 2 and 3, and each of them contains only one trigonometric entry per row, so the minors stay short.
    Sign of position (1,j)(1, j) is (−1)1+j(-1)^{1+j}⁠: ++ for j=1j = 1⁠, −- for j=2j = 2⁠, ++ for j=3j = 3⁠. Basics: the + − + sign pattern

  2. The three minors

    Delete row 1 and the entry's column each time:

    M11=∣−x11x∣=(−x)(x)−(1)(1)=−x2−1\begin{aligned} M_{11} &= \begin{vmatrix} -x & 1 \\ 1 & x \end{vmatrix} = (-x)(x) - (1)(1) \\ &= -x^2 - 1 \end{aligned} M12=∣−sin⁡θ1cos⁡θx∣=(−sin⁡θ)(x)−(1)(cos⁡θ)=−xsin⁡θ−cos⁡θ\begin{aligned} M_{12} &= \begin{vmatrix} -\sin\theta & 1 \\ \cos\theta & x \end{vmatrix} \\ &= (-\sin\theta)(x) - (1)(\cos\theta) \\ &= -x\sin\theta - \cos\theta \end{aligned} M13=∣−sin⁡θ−xcos⁡θ1∣=(−sin⁡θ)(1)−(−x)(cos⁡θ)=−sin⁡θ+xcos⁡θ\begin{aligned} M_{13} &= \begin{vmatrix} -\sin\theta & -x \\ \cos\theta & 1 \end{vmatrix} \\ &= (-\sin\theta)(1) - (-x)(\cos\theta) \\ &= -\sin\theta + x\cos\theta \end{aligned}
    For M11M_{11} we delete row 1 and column 1, leaving the entries of rows 2 and 3 in columns 2 and 3. For M12M_{12} we keep columns 1 and 3; for M13M_{13}⁠, columns 1 and 2. Each 2×22\times2 determinant is "down-right product minus up-right product". In M13M_{13} the second product is (−x)(cos⁡θ)=−xcos⁡θ(-x)(\cos\theta) = -x\cos\theta⁠, and subtracting it gives +xcos⁡θ+x\cos\theta⁠: this sign is the one most often lost.
    Minor MijM_{ij}⁠: delete row ii and column jj (Section 4.4, Definition 1). Lesson: minors and cofactors

  3. Form the signed sum

    Multiply each entry of R1R_1 by its minor, with the signs +,−,++, -, +⁠:

    Δ=x(−x2−1)−sin⁡θ (−xsin⁡θ−cos⁡θ)+cos⁡θ (−sin⁡θ+xcos⁡θ)\begin{aligned} \Delta &= x(-x^2 - 1) \\ &\quad - \sin\theta\,(-x\sin\theta - \cos\theta) \\ &\quad + \cos\theta\,(-\sin\theta + x\cos\theta) \end{aligned}

    Multiply out each bracket:

    Δ=−x3−x+xsin⁡2θ+sin⁡θcos⁡θ−sin⁡θcos⁡θ+xcos⁡2θ\begin{aligned} \Delta &= -x^3 - x \\ &\quad + x\sin^2\theta + \sin\theta\cos\theta \\ &\quad - \sin\theta\cos\theta + x\cos^2\theta \end{aligned}
    The middle term carries a minus sign from the pattern, and its bracket is itself negative, so both of its products become positive: −sin⁡θ⋅(−xsin⁡θ)=+xsin⁡2θ-\sin\theta \cdot (-x\sin\theta) = +x\sin^2\theta and −sin⁡θ⋅(−cos⁡θ)=+sin⁡θcos⁡θ-\sin\theta\cdot(-\cos\theta) = +\sin\theta\cos\theta⁠. In the last term, cos⁡θ⋅(−sin⁡θ)=−sin⁡θcos⁡θ\cos\theta\cdot(-\sin\theta) = -\sin\theta\cos\theta and cos⁡θ⋅xcos⁡θ=xcos⁡2θ\cos\theta\cdot x\cos\theta = x\cos^2\theta⁠.
    Δ=a11M11−a12M12+a13M13\Delta = a_{11}M_{11} - a_{12}M_{12} + a_{13}M_{13} along R1R_1 (Section 4.2.3). Lesson: expanding a 3 × 3 determinant

  4. Simplify: every θ\theta disappears

    The two sin⁡θcos⁡θ\sin\theta\cos\theta terms are equal and opposite, so they cancel. Group the two remaining θ\theta-terms, which share the factor xx⁠:

    Δ=−x3−x+x(sin⁡2θ+cos⁡2θ)=−x3−x+x⋅1=−x3.\begin{aligned} \Delta &= -x^3 - x + x(\sin^2\theta + \cos^2\theta) \\ &= -x^3 - x + x\cdot 1 \\ &= -x^3. \end{aligned}

    So Δ=−x3\Delta = -x^3 for every angle θ\theta⁠: the determinant is independent of θ\theta.

    The identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 holds for every real θ\theta⁠, with no exceptions, so the simplification is valid for all θ\theta and all xx⁠. The final expression −x3-x^3 contains xx only. It does depend on xx⁠, but the question only asks about θ\theta⁠.
    sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 for all θ\theta⁠. Basics: algebra the chapter leans on
Result ∣xsin⁡θcos⁡θ−sin⁡θ−x1cos⁡θ1x∣=−x3,\begin{vmatrix} x & \sin\theta & \cos\theta \\ -\sin\theta & -x & 1 \\ \cos\theta & 1 & x \end{vmatrix} = -x^3,

which contains no θ\theta⁠, so the determinant is independent of θ\theta. (Proofs have no entry in the answer key.)

Check it with numbers

Take x=2x = 2⁠. Then −x3=−8-x^3 = -8⁠, whatever θ\theta is.

With θ=0\theta = 0⁠: sin⁡θ=0\sin\theta = 0⁠, cos⁡θ=1\cos\theta = 1⁠, and the rows are (2,0,1)(2, 0, 1)⁠, (0,−2,1)(0, -2, 1)⁠, (1,1,2)(1, 1, 2)⁠. Along R1R_1⁠: 2 (−4−1)−0+1 (0+2)2\,(-4 - 1) - 0 + 1\,(0 + 2) =−10+2=−8= -10 + 2 = -8 ✓.

With θ=π2\theta = \frac{\pi}{2}⁠: sin⁡θ=1\sin\theta = 1⁠, cos⁡θ=0\cos\theta = 0⁠, and the rows are (2,1,0)(2, 1, 0)⁠, (−1,−2,1)(-1, -2, 1)⁠, (0,1,2)(0, 1, 2)⁠. Along R1R_1⁠: 2 (−4−1)−1 (−2−0)+02\,(-4 - 1) - 1\,(-2 - 0) + 0 =−10+2=−8= -10 + 2 = -8 ✓.

With θ=0.7\theta = 0.7 (radians): sin⁡0.7≈0.644218\sin 0.7 \approx 0.644218⁠, cos⁡0.7≈0.764842\cos 0.7 \approx 0.764842⁠, and the determinant evaluates to −8.000000-8.000000 to six decimal places ✓. The verification script repeats this for 200 random pairs (x,θ)(x, \theta)⁠.

Common mistakes
  1. Dropping the middle minus sign. The −- in front of sin⁡θ M12\sin\theta\,M_{12} is what turns −xsin⁡2θ-x\sin^2\theta into +xsin⁡2θ+x\sin^2\theta⁠. Without it the θ\theta-terms do not combine into x(sin⁡2θ+cos⁡2θ)x(\sin^2\theta + \cos^2\theta)⁠.
  2. Sign slip in M13M_{13}: (−sin⁡θ)(1)−(−x)(cos⁡θ)(-\sin\theta)(1) - (-x)(\cos\theta) is −sin⁡θ+xcos⁡θ-\sin\theta + x\cos\theta⁠, not −sin⁡θ−xcos⁡θ-\sin\theta - x\cos\theta⁠.
  3. Stopping too early. −x3−x+xsin⁡2θ+xcos⁡2θ-x^3 - x + x\sin^2\theta + x\cos^2\theta still shows θ\theta⁠; the proof is only finished after the identity is applied.
  4. Thinking "independent of θ\theta" means "constant". The value −x3-x^3 changes with xx⁠; it just does not change with θ\theta⁠.
What to take away
  1. To show a determinant is independent of a variable, expand it fully and simplify until that variable is gone.
  2. When sin⁡θ\sin\theta and cos⁡θ\cos\theta both appear, expect the terms to pair up into sin⁡2θ+cos⁡2θ\sin^2\theta + \cos^2\theta and look for it.
  3. When no line has a zero, any row works; keep the signs +,−,++, -, + and the inner minus signs separate and careful.

Every fact used, and where it comes from

FactUsed inSource
Value of a 2×22\times2 determinant: ad−bcad - bcStep 2, CheckSection 4.2.2 (Lesson: determinant of a square matrix)
Expansion of a 3×33\times3 determinant along a row; any row or column gives the same valueSteps 1, 3Section 4.2.3 and its Remarks (i), (ii) (Lesson: expanding a 3 × 3 determinant)
Minor MijM_{ij}Step 2Section 4.4, Definition 1 (Lesson: minors and cofactors)
sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1Step 4Class XI trigonometry (Basics: algebra the chapter leans on)
Δ=−x3\Delta = -x^3 symbolically, and the numerical checksSteps 2–4, CheckRecomputed with sympy (symbolic expansion and 200 random points) in the site's verification script
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