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Chapter 4 · Determinants

Example 4: A determinant of sines and cosines equals 0

ExamplesDeterminant of a matrix of order 3 × 3 · Section 4.2.3
The question

Evaluate

Δ=∣0sin⁡α−cos⁡α−sin⁡α0sin⁡βcos⁡α−sin⁡β0∣.\Delta = \begin{vmatrix} 0 & \sin\alpha & -\cos\alpha \\ -\sin\alpha & 0 & \sin\beta \\ \cos\alpha & -\sin\beta & 0 \end{vmatrix}.
Exam answer

Expanding along R1R_1⁠:

Δ=0∣0sin⁡β−sin⁡β0∣−sin⁡α∣−sin⁡αsin⁡βcos⁡α0∣−cos⁡α∣−sin⁡α0cos⁡α−sin⁡β∣=0−sin⁡α(0−sin⁡βcos⁡α)−cos⁡α(sin⁡αsin⁡β−0)=sin⁡αsin⁡βcos⁡α−cos⁡αsin⁡αsin⁡β\begin{aligned} \Delta &= 0\begin{vmatrix} 0 & \sin\beta \\ -\sin\beta & 0 \end{vmatrix} \\ &\quad - \sin\alpha\begin{vmatrix} -\sin\alpha & \sin\beta \\ \cos\alpha & 0 \end{vmatrix} \\ &\quad - \cos\alpha\begin{vmatrix} -\sin\alpha & 0 \\ \cos\alpha & -\sin\beta \end{vmatrix} \\ &= 0 - \sin\alpha(0 - \sin\beta\cos\alpha) \\ &\quad - \cos\alpha(\sin\alpha\sin\beta - 0) \\ &= \sin\alpha\sin\beta\cos\alpha \\ &\quad - \cos\alpha\sin\alpha\sin\beta \end{aligned}

∴ Δ=0\therefore\ \Delta = 0⁠.

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

The one idea this problem is built on

Expand along a row that contains a zero, exactly as with numbers; sin⁡α\sin\alpha⁠, cos⁡α\cos\alpha and sin⁡β\sin\beta are just numbers whose values we do not need. Two terms survive, and they are the same product sin⁡α sin⁡β cos⁡α\sin\alpha\,\sin\beta\,\cos\alpha with opposite signs, so they cancel. The reason is structural: every entry below the diagonal is the negative of its mirror image above it (the matrix is skew symmetric), and for order 3 that forces the determinant to be 0.

Given, to find, and the plan
Given
A 3×33\times3 determinant whose diagonal entries are all 0 and whose other entries are ±sin⁡α\pm\sin\alpha⁠, ±cos⁡α\pm\cos\alpha⁠, ±sin⁡β\pm\sin\beta⁠, for arbitrary angles α\alpha⁠, β\beta⁠.
To find
The value of Δ\Delta⁠.
Plan
Expand along R1R_1 (its first entry is 0), compute the three 2×22\times2 determinants with brackets around every negative entry, combine, and simplify. Then see why the answer had to be 0.

Basics you need

1Expansion along the first row

From Section 4.2.3,

∣A∣=a11∣a22a23a32a33∣−a12∣a21a23a31a33∣+a13∣a21a22a31a32∣.\begin{aligned} |A| &= a_{11}\begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix}\\ &\quad - a_{12}\begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix}\\ &\quad + a_{13}\begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix}. \end{aligned}

The signs +,−,++, -, + come from (−1)1+j(-1)^{1+j}⁠. Lesson: expanding a 3 × 3 determinant

2The 2×22\times2 rule

Main-diagonal product minus other-diagonal product. Basics: where the determinant comes from

3Signs and order in products

(−p)(−q)=pq(-p)(-q) = pq⁠, (−p)q=−pq(-p)q = -pq⁠, and the order of factors does not matter: sin⁡α sin⁡β cos⁡α\sin\alpha\,\sin\beta\,\cos\alpha is the same as cos⁡α sin⁡α sin⁡β\cos\alpha\,\sin\alpha\,\sin\beta⁠. Basics: algebra the chapter leans on

4The sign pattern

Along R1R_1⁠: +,−,++, -, +⁠. Basics: the + − + sign pattern

Solution

  1. Choose the row and name the entries

    Every row and every column contains exactly one 0, so no line is better than another. Following the book, we expand along R1R_1⁠, whose entries are

    a11=0,a12=sin⁡α,a13=−cos⁡α.a_{11} = 0,\quad a_{12} = \sin\alpha,\quad a_{13} = -\cos\alpha.
    With one zero in every line, any choice costs the same: two 2×22\times2 determinants. Taking R1R_1 matches the book, so the working can be compared line by line. The signs along R1R_1 are +,−,++, -, +⁠.
    Remark (i), Section 4.2.3: expand along the line with the most zeros; here they tie. Basics: the + − + sign pattern

  2. The first term: a11=0a_{11} = 0

    Delete R1R_1 and C1C_1⁠:

    ∣0sin⁡β−sin⁡β0∣\begin{vmatrix} 0 & \sin\beta \\ -\sin\beta & 0 \end{vmatrix} =0⋅0−(−sin⁡β)(sin⁡β)=sin⁡2β.\begin{aligned} &= 0\cdot 0 - (-\sin\beta)(\sin\beta)\\ &= \sin^2\beta. \end{aligned}

    The term is + 0×sin⁡2β=0+\,0\times\sin^2\beta = 0⁠.

    The 2×22\times2 value is worked out only to show that it is an ordinary finite number. Whatever it is, multiplying by a11=0a_{11} = 0 gives 0, so this term drops out.

  3. The second term: a12=sin⁡αa_{12} = \sin\alpha⁠, sign −-

    Delete R1R_1 and C2C_2⁠:

    ∣−sin⁡αsin⁡βcos⁡α0∣\begin{vmatrix} -\sin\alpha & \sin\beta \\ \cos\alpha & 0 \end{vmatrix} =(−sin⁡α)(0)−(cos⁡α)(sin⁡β)=0−sin⁡βcos⁡α.\begin{aligned} &= (-\sin\alpha)(0) - (\cos\alpha)(\sin\beta)\\ &= 0 - \sin\beta\cos\alpha. \end{aligned}

    With its sign and entry, the term is

    −sin⁡α (0−sin⁡βcos⁡α)=+sin⁡α sin⁡βcos⁡α.\begin{aligned} &-\sin\alpha\,(0 - \sin\beta\cos\alpha)\\ &= +\sin\alpha\,\sin\beta\cos\alpha. \end{aligned}
    The position (1,2)(1,2) has sign (−1)1+2=−1(-1)^{1+2} = -1⁠. The 2×22\times2 value is itself negative, −sin⁡βcos⁡α-\sin\beta\cos\alpha⁠, and minus times minus is plus. This is the step where a sign is most easily lost, so the bracket (0−sin⁡βcos⁡α)(0 - \sin\beta\cos\alpha) is kept until the last moment.
    Sign of a12a_{12}⁠: (−1)1+2=−1(-1)^{1+2} = -1⁠. (−p)(−q)=pq(-p)(-q) = pq⁠. Basics: algebra the chapter leans on

  4. The third term: a13=−cos⁡αa_{13} = -\cos\alpha⁠, sign ++

    Delete R1R_1 and C3C_3⁠:

    ∣−sin⁡α0cos⁡α−sin⁡β∣\begin{vmatrix} -\sin\alpha & 0 \\ \cos\alpha & -\sin\beta \end{vmatrix} =(−sin⁡α)(−sin⁡β)−(cos⁡α)(0)=sin⁡αsin⁡β−0.\begin{aligned} &= (-\sin\alpha)(-\sin\beta) - (\cos\alpha)(0)\\ &= \sin\alpha\sin\beta - 0. \end{aligned}

    With its sign and entry, the term is

    +(−cos⁡α)(sin⁡αsin⁡β−0)=−cos⁡α sin⁡α sin⁡β.\begin{aligned} &+(-\cos\alpha)(\sin\alpha\sin\beta - 0)\\ &= -\cos\alpha\,\sin\alpha\,\sin\beta. \end{aligned}
    The position sign is (−1)1+3=+1(-1)^{1+3} = +1⁠, but the entry itself is −cos⁡α-\cos\alpha⁠, so the term is negative. The book writes this as "−cos⁡α (sin⁡αsin⁡β−0)-\cos\alpha\,(\sin\alpha\sin\beta - 0)": the ++ of the pattern and the −- of the entry combine into one minus.
    Sign of a13a_{13}⁠: (−1)1+3=+1(-1)^{1+3} = +1⁠; the entry's own minus is separate. Basics: the + − + sign pattern

  5. Add the three terms

    Δ=0+sin⁡α sin⁡βcos⁡α−cos⁡α sin⁡α sin⁡β=sin⁡αsin⁡βcos⁡α−sin⁡αsin⁡βcos⁡α=0.\begin{aligned} \Delta &= 0 + \sin\alpha\,\sin\beta\cos\alpha\\ &\quad - \cos\alpha\,\sin\alpha\,\sin\beta\\ &= \sin\alpha\sin\beta\cos\alpha - \sin\alpha\sin\beta\cos\alpha\\ &= 0. \end{aligned}
    Multiplication of numbers is commutative, so cos⁡α sin⁡α sin⁡β\cos\alpha\,\sin\alpha\,\sin\beta is the same product as sin⁡α sin⁡β cos⁡α\sin\alpha\,\sin\beta\,\cos\alpha⁠, just written in another order. A number minus itself is 0. No trigonometric identity is needed, and nothing was assumed about α\alpha or β\beta⁠, so Δ=0\Delta = 0 for all angles.
    Commutativity: pqr=rpqpqr = rpq⁠. Basics: algebra the chapter leans on

  6. Why the answer had to be 0

    Compare each entry below the diagonal with its mirror image above it: a21=−sin⁡α=−a12a_{21} = -\sin\alpha = -a_{12}⁠, a31=cos⁡α=−a13a_{31} = \cos\alpha = -a_{13}⁠, a32=−sin⁡β=−a23a_{32} = -\sin\beta = -a_{23}⁠, and the diagonal is 0. So the matrix has the form below, with p=sin⁡αp = \sin\alpha⁠, q=−cos⁡αq = -\cos\alpha⁠, r=sin⁡βr = \sin\beta⁠:

    ∣0pq−p0r−q−r0∣.\begin{vmatrix} 0 & p & q \\ -p & 0 & r \\ -q & -r & 0 \end{vmatrix}.

    Expanding along R1R_1 in exactly the same way, for arbitrary numbers p,q,rp, q, r⁠:

    0(0+r2)−p(0+qr)+q(pr−0)=−pqr+pqr=0.\begin{aligned} &0(0 + r^2) - p(0 + qr) + q(pr - 0)\\ &= -pqr + pqr = 0. \end{aligned}
    This shows the cancellation in Step 5 was not luck. A matrix with aji=−aija_{ji} = -a_{ij} is called skew symmetric (Chapter 3), and the general calculation shows that every 3×33\times3 skew symmetric determinant is 0, whatever its entries. Example 4 is the special case p=sin⁡αp = \sin\alpha⁠, q=−cos⁡αq = -\cos\alpha⁠, r=sin⁡βr = \sin\beta⁠.
    Skew symmetric: A′=−AA' = -A⁠, so the diagonal is 0 and mirror entries are negatives of each other. Basics: transpose and identity
Result Δ=0  for all values of α and β.\Delta = 0\ \text{ for all values of } \alpha \text{ and } \beta.

This agrees with the textbook's solution.

Check it with numbers

Take α=0.7\alpha = 0.7 and β=1.3\beta = 1.3 (radians). To 6 decimal places, sin⁡α=0.644218\sin\alpha = 0.644218⁠, cos⁡α=0.764842\cos\alpha = 0.764842⁠, sin⁡β=0.963558\sin\beta = 0.963558⁠. The two surviving terms of Step 5 are

sin⁡αsin⁡βcos⁡α≈0.474769,cos⁡αsin⁡αsin⁡β≈0.474769,\begin{aligned} \sin\alpha\sin\beta\cos\alpha &\approx 0.474769,\\ \cos\alpha\sin\alpha\sin\beta &\approx 0.474769, \end{aligned}

so Δ≈0.474769−0.474769=0\Delta \approx 0.474769 - 0.474769 = 0⁠. With α=2\alpha = 2⁠, β=−0.4\beta = -0.4 the product is ≈0.147356\approx 0.147356⁠, and again the two terms cancel exactly.

Common mistakes
  1. Mixing the position sign with the entry's sign. The third term has pattern sign ++ and entry −cos⁡α-\cos\alpha⁠; together they give one minus, not two.
  2. Dropping the minus inside a minor. In Step 3 the 2×22\times2 value is −sin⁡βcos⁡α-\sin\beta\cos\alpha⁠; losing that sign gives Δ=−2sin⁡αsin⁡βcos⁡α\Delta = -2\sin\alpha\sin\beta\cos\alpha⁠.
  3. Looking for a trig identity. None is needed; the cancellation is pure algebra.
What to take away
  1. Trigonometric entries are handled exactly like numbers: expand, keep every sign in brackets, and simplify.
  2. Every 3×33\times3 skew symmetric determinant (zero diagonal, aji=−aija_{ji} = -a_{ij}⁠) equals 0. Spotting that shape tells you the answer before you start.
  3. Products in a different order are the same product; rearrange them to see cancellations.

Every fact used, and where it comes from

FactUsed inSource
Expansion along R1R_1 with signs +,−,++, -, +Steps 1–6Section 4.2.3 (Lesson: expanding a 3 × 3 determinant)
2×22\times2 ruleSteps 2–4Section 4.2.2
Sign (−1)i+j(-1)^{i+j}Steps 3, 4Section 4.2.3, Remark (ii)
Commutativity of multiplicationStep 5Properties of real numbers
Skew symmetric matrix: A′=−AA' = -AStep 6Chapter 3, Section 3.6, Definition 5
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