Class 12 Maths, Worked in Full Past papersGet updates
Chapter 4 · Determinants

Miscellaneous · Q2: Evaluate a determinant in α\alpha⁠, β\beta

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

Evaluate

∣cos⁡αcos⁡βcos⁡αsin⁡β−sin⁡α−sin⁡βcos⁡β0sin⁡αcos⁡βsin⁡αsin⁡βcos⁡α∣.\begin{vmatrix} \cos\alpha\cos\beta & \cos\alpha\sin\beta & -\sin\alpha \\ -\sin\beta & \cos\beta & 0 \\ \sin\alpha\cos\beta & \sin\alpha\sin\beta & \cos\alpha \end{vmatrix}.
Exam answer

Let Δ\Delta be the given determinant.

Expanding along R2R_2⁠:

Δ=−(−sin⁡β)M21+cos⁡β M22−0⋅M23\begin{aligned} \Delta &= -(-\sin\beta)M_{21} + \cos\beta\,M_{22} \\ &\quad - 0\cdot M_{23} \end{aligned}

where

M21=∣cos⁡αsin⁡β−sin⁡αsin⁡αsin⁡βcos⁡α∣=cos⁡2αsin⁡β+sin⁡2αsin⁡β=sin⁡β(cos⁡2α+sin⁡2α)=sin⁡β\begin{aligned} M_{21} &= \begin{vmatrix} \cos\alpha\sin\beta & -\sin\alpha \\ \sin\alpha\sin\beta & \cos\alpha \end{vmatrix} \\ &= \cos^2\alpha\sin\beta + \sin^2\alpha\sin\beta \\ &= \sin\beta(\cos^2\alpha + \sin^2\alpha) \\ &= \sin\beta \end{aligned} M22=∣cos⁡αcos⁡β−sin⁡αsin⁡αcos⁡βcos⁡α∣=cos⁡2αcos⁡β+sin⁡2αcos⁡β=cos⁡β(cos⁡2α+sin⁡2α)=cos⁡β\begin{aligned} M_{22} &= \begin{vmatrix} \cos\alpha\cos\beta & -\sin\alpha \\ \sin\alpha\cos\beta & \cos\alpha \end{vmatrix} \\ &= \cos^2\alpha\cos\beta + \sin^2\alpha\cos\beta \\ &= \cos\beta(\cos^2\alpha + \sin^2\alpha) \\ &= \cos\beta \end{aligned} ∴ Δ=sin⁡β⋅sin⁡β+cos⁡β⋅cos⁡β=sin⁡2β+cos⁡2β\begin{aligned} \therefore\ \Delta &= \sin\beta\cdot\sin\beta + \cos\beta\cdot\cos\beta \\ &= \sin^2\beta + \cos^2\beta \end{aligned}

∴ Δ=1\therefore\ \Delta = 1⁠.

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

The one idea this problem is built on

Expand along the line that contains the zero, the second row. Each of its two remaining minors has a common factor that comes out as cos⁡2α+sin⁡2α=1\cos^2\alpha + \sin^2\alpha = 1⁠, leaving just sin⁡β\sin\beta and cos⁡β\cos\beta⁠. The determinant then becomes sin⁡2β+cos⁡2β\sin^2\beta + \cos^2\beta⁠, which is 1. The Pythagorean identity is used twice, once for each angle.

Given, to find, and the plan
Given
A 3×33\times3 determinant Δ\Delta whose entries are products of sines and cosines of two angles α\alpha and β\beta⁠.
To find
The value of Δ\Delta⁠.
Plan
(1) Expand along R2R_2⁠, which contains a 0. (2) Evaluate the minors M21M_{21} and M22M_{22} and simplify each with cos⁡2α+sin⁡2α=1\cos^2\alpha + \sin^2\alpha = 1⁠. (3) Form the signed sum and use sin⁡2β+cos⁡2β=1\sin^2\beta + \cos^2\beta = 1⁠. (4) Check by expanding along another line, and with numbers.

Basics you need

1A 2×22\times2 determinant

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

2Expanding along any row

A 3×33\times3 determinant can be expanded along any row or column; a zero entry makes its term vanish, so a row with a zero means less work. Lesson: expanding a 3 × 3 determinant

3The sign pattern

Position (i,j)(i, j) has sign (−1)i+j(-1)^{i+j}⁠. Along the second row the signs are −,+,−-, +, -⁠. Basics: the + − + sign pattern

4The Pythagorean identity

sin⁡2t+cos⁡2t=1\sin^2 t + \cos^2 t = 1 for every angle tt⁠; here it is used with t=αt = \alpha and with t=βt = \beta⁠. Basics: algebra the chapter leans on

Solution

  1. Expand along the second row

    Row 2 is (−sin⁡β, cos⁡β, 0)(-\sin\beta,\ \cos\beta,\ 0)⁠. Its signs are (−1)2+1=−(-1)^{2+1} = -⁠, (−1)2+2=+(-1)^{2+2} = +⁠, (−1)2+3=−(-1)^{2+3} = -⁠. So

    Δ=−(−sin⁡β) M21+cos⁡β M22−0⋅M23=sin⁡β M21+cos⁡β M22.\begin{aligned} \Delta &= -(-\sin\beta)\,M_{21} + \cos\beta\,M_{22} - 0\cdot M_{23} \\ &= \sin\beta\,M_{21} + \cos\beta\,M_{22}. \end{aligned}
    The third term is 0⋅M23=00\cdot M_{23} = 0⁠, so we never need M23M_{23}⁠. This is the book's advice (Remark (i) of Section 4.2.3): expand along the row or column with the most zeros. Column 3 also has a zero; it would work equally well and is used as a check below. Note the double sign in the first term: the pattern gives −- and the entry is −sin⁡β-\sin\beta⁠, so the product is +sin⁡β+\sin\beta⁠.
    Sign of position (2,j)(2, j) is (−1)2+j(-1)^{2+j}⁠. Basics: the + − + sign pattern

  2. The minor M21M_{21}

    Delete row 2 and column 1, keeping rows 1 and 3, columns 2 and 3:

    M21=∣cos⁡αsin⁡β−sin⁡αsin⁡αsin⁡βcos⁡α∣=cos⁡2αsin⁡β−(−sin⁡α)(sin⁡αsin⁡β)=cos⁡2αsin⁡β+sin⁡2αsin⁡β=sin⁡β (cos⁡2α+sin⁡2α)=sin⁡β.\begin{aligned} M_{21} &= \begin{vmatrix} \cos\alpha\sin\beta & -\sin\alpha \\ \sin\alpha\sin\beta & \cos\alpha \end{vmatrix} \\ &= \cos^2\alpha\sin\beta - (-\sin\alpha)(\sin\alpha\sin\beta) \\ &= \cos^2\alpha\sin\beta + \sin^2\alpha\sin\beta \\ &= \sin\beta\,(\cos^2\alpha + \sin^2\alpha) \\ &= \sin\beta. \end{aligned}
    The down-right product is (cos⁡αsin⁡β)(cos⁡α)=cos⁡2αsin⁡β(\cos\alpha\sin\beta)(\cos\alpha) = \cos^2\alpha\sin\beta⁠. The up-right product is (−sin⁡α)(sin⁡αsin⁡β)=−sin⁡2αsin⁡β(-\sin\alpha)(\sin\alpha\sin\beta) = -\sin^2\alpha\sin\beta⁠, and subtracting it adds sin⁡2αsin⁡β\sin^2\alpha\sin\beta⁠. Both terms contain sin⁡β\sin\beta⁠; taking it out leaves cos⁡2α+sin⁡2α=1\cos^2\alpha + \sin^2\alpha = 1⁠.
    Minor MijM_{ij}⁠: delete row ii and column jj (Section 4.4, Definition 1). Lesson: minors and cofactors

  3. The minor M22M_{22}

    Delete row 2 and column 2, keeping rows 1 and 3, columns 1 and 3:

    M22=∣cos⁡αcos⁡β−sin⁡αsin⁡αcos⁡βcos⁡α∣=cos⁡2αcos⁡β−(−sin⁡α)(sin⁡αcos⁡β)=cos⁡2αcos⁡β+sin⁡2αcos⁡β=cos⁡β (cos⁡2α+sin⁡2α)=cos⁡β.\begin{aligned} M_{22} &= \begin{vmatrix} \cos\alpha\cos\beta & -\sin\alpha \\ \sin\alpha\cos\beta & \cos\alpha \end{vmatrix} \\ &= \cos^2\alpha\cos\beta - (-\sin\alpha)(\sin\alpha\cos\beta) \\ &= \cos^2\alpha\cos\beta + \sin^2\alpha\cos\beta \\ &= \cos\beta\,(\cos^2\alpha + \sin^2\alpha) \\ &= \cos\beta. \end{aligned}
    This is the same calculation as Step 2 with sin⁡β\sin\beta replaced by cos⁡β\cos\beta⁠, because columns 1 and 2 of the determinant differ only in that factor (in rows 1 and 3).
    cos⁡2α+sin⁡2α=1\cos^2\alpha + \sin^2\alpha = 1⁠. Basics: algebra the chapter leans on

  4. Put the pieces together

    Substitute the two minors into Step 1:

    Δ=sin⁡β⋅sin⁡β+cos⁡β⋅cos⁡β=sin⁡2β+cos⁡2β=1.\begin{aligned} \Delta &= \sin\beta\cdot\sin\beta + \cos\beta\cdot\cos\beta \\ &= \sin^2\beta + \cos^2\beta \\ &= 1. \end{aligned}

    So Δ=1\Delta = 1, for every pair of angles α\alpha⁠, β\beta⁠.

    The second use of the Pythagorean identity, now with the angle β\beta⁠. Both uses are valid for all real angles, so the answer holds with no restriction on α\alpha or β\beta⁠.
    Signed sum along R2R_2 (Section 4.2.3, "Expansion along second row"). Lesson: expanding a 3 × 3 determinant
Result ∣cos⁡αcos⁡βcos⁡αsin⁡β−sin⁡α−sin⁡βcos⁡β0sin⁡αcos⁡βsin⁡αsin⁡βcos⁡α∣=1\begin{vmatrix} \cos\alpha\cos\beta & \cos\alpha\sin\beta & -\sin\alpha \\ -\sin\beta & \cos\beta & 0 \\ \sin\alpha\cos\beta & \sin\alpha\sin\beta & \cos\alpha \end{vmatrix} = 1

for all α\alpha⁠, β\beta⁠. (Textbook answer: 1.)

Check it with numbers

Another line. Column 3 is (−sin⁡α,0,cos⁡α)(-\sin\alpha, 0, \cos\alpha) with signs +,−,++, -, +⁠:

Δ=−sin⁡α M13+cos⁡α M33.\Delta = -\sin\alpha\,M_{13} + \cos\alpha\,M_{33}.

Here

M13=(−sin⁡β)(sin⁡αsin⁡β)−(cos⁡β)(sin⁡αcos⁡β)=−sin⁡α(sin⁡2β+cos⁡2β)=−sin⁡α\begin{aligned} M_{13} &= (-\sin\beta)(\sin\alpha\sin\beta) \\ &\quad - (\cos\beta)(\sin\alpha\cos\beta) \\ &= -\sin\alpha(\sin^2\beta + \cos^2\beta) \\ &= -\sin\alpha \end{aligned} M33=(cos⁡αcos⁡β)(cos⁡β)−(cos⁡αsin⁡β)(−sin⁡β)=cos⁡α(cos⁡2β+sin⁡2β)=cos⁡α\begin{aligned} M_{33} &= (\cos\alpha\cos\beta)(\cos\beta) \\ &\quad - (\cos\alpha\sin\beta)(-\sin\beta) \\ &= \cos\alpha(\cos^2\beta + \sin^2\beta) \\ &= \cos\alpha \end{aligned}

So Δ=(−sin⁡α)(−sin⁡α)+cos⁡α⋅cos⁡α\Delta = (-\sin\alpha)(-\sin\alpha) + \cos\alpha\cdot\cos\alpha =sin⁡2α+cos⁡2α=1= \sin^2\alpha + \cos^2\alpha = 1 ✓.

Numbers. With α=0.3\alpha = 0.3⁠, β=1.1\beta = 1.1 (radians): cos⁡0.3≈0.955336\cos 0.3 \approx 0.955336⁠, sin⁡0.3≈0.295520\sin 0.3 \approx 0.295520⁠, cos⁡1.1≈0.453596\cos 1.1 \approx 0.453596⁠, sin⁡1.1≈0.891207\sin 1.1 \approx 0.891207⁠. Then M21≈0.891207=sin⁡1.1M_{21} \approx 0.891207 = \sin 1.1⁠, M22≈0.453596=cos⁡1.1M_{22} \approx 0.453596 = \cos 1.1⁠, and Δ≈0.794251+0.205749\Delta \approx 0.794251 + 0.205749 =1.000000= 1.000000 ✓. The verification script repeats this at 200 random pairs of angles.

Common mistakes
  1. Using +,−,++, -, + along row 2. The second row starts with −-⁠: the signs are −,+,−-, +, -⁠.
  2. Losing the double minus in the first term: −(−sin⁡β)=+sin⁡β-(-\sin\beta) = +\sin\beta⁠.
  3. Sign slip inside the minors: the up-right product contains −sin⁡α-\sin\alpha⁠, so subtracting it gives +sin⁡2αsin⁡β+\sin^2\alpha\sin\beta (and +sin⁡2αcos⁡β+\sin^2\alpha\cos\beta⁠).
  4. Expanding along R1R_1 without need. It works, but every minor is longer and the chance of a slip is much higher.
What to take away
  1. A zero entry is a gift: expand along its row or column and one whole minor disappears.
  2. In trigonometric determinants, pull out common factors from each minor; the leftover bracket is often sin⁡2+cos⁡2=1\sin^2 + \cos^2 = 1⁠.
  3. Expanding along a second line is a quick, independent check of the answer.

Every fact used, and where it comes from

FactUsed inSource
Value of a 2×22\times2 determinant: ad−bcad - bcSteps 2, 3, CheckSection 4.2.2 (Lesson: determinant of a square matrix)
Expansion along R2R_2 or C3C_3⁠; all lines give the same value; prefer the line with zerosSteps 1, 4, CheckSection 4.2.3 and its Remarks (i), (ii) (Lesson: expanding a 3 × 3 determinant)
Minor MijM_{ij}Steps 2, 3Section 4.4, Definition 1 (Lesson: minors and cofactors)
sin⁡2t+cos⁡2t=1\sin^2 t + \cos^2 t = 1Steps 2–4Class XI trigonometry (Basics: algebra the chapter leans on)
Δ=1\Delta = 1 symbolically, both expansions, and the decimalsSteps 1–4, CheckRecomputed with sympy (symbolic and 200 random angle pairs) in the site's verification script
Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →