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Chapter 3 · Matrices

Exercise 3.3 · Q12 (MCQ): Find α\alpha when A+A′=IA+A'=I for a rotation matrix

Exercise 3.3Transpose and equality of matrices · Section 3.5
The question

If

A=[cos⁡α−sin⁡αsin⁡αcos⁡α],A = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix},

and A+A′=IA + A' = I⁠, then the value of α\alpha is

(A) π6\dfrac{\pi}{6}    (B) π3\dfrac{\pi}{3}    (C) π\pi    (D) 3π2\dfrac{3\pi}{2}

Exam answer A′=[cos⁡αsin⁡α−sin⁡αcos⁡α]A' = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix} A+A′=[2cos⁡α002cos⁡α]A + A' = \begin{bmatrix} 2\cos\alpha & 0 \\ 0 & 2\cos\alpha \end{bmatrix}

A+A′=IA + A' = I gives 2cos⁡α=12\cos\alpha = 1⁠, so cos⁡α=12\cos\alpha = \tfrac12 and α=π3\alpha = \tfrac{\pi}{3}⁠.

Answer: (B) π3\dfrac{\pi}{3}

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

The one idea this problem is built on

Transposing AA swaps −sin⁡α-\sin\alpha and sin⁡α\sin\alpha⁠. So when we add AA and A′A'⁠, the sines cancel off the diagonal and the cosines double on the diagonal: A+A′A + A' has 2cos⁡α2\cos\alpha on the diagonal and 0 elsewhere. Setting this equal to II and comparing entries gives one equation, 2cos⁡α=12\cos\alpha = 1⁠, so cos⁡α=12\cos\alpha = \tfrac12⁠.

Given, to find, and the plan
Given
A 2×22\times2 matrix AA in α\alpha⁠, and the condition A+A′=IA + A' = I (here II is the 2×22\times2 identity).
To find
Which of the four values of α\alpha satisfies the condition.
Plan
(1) Write A′A'⁠. (2) Compute A+A′A + A' entry by entry. (3) Equate entries with II and solve. (4) Test every option.

Basics you need

1The transpose

(A′)ij=aji(A')_{ij} = a_{ji}⁠; for a 2×22\times2 matrix the off-diagonal entries swap. Basics: reading a table sideways

2Equal matrices

Two matrices of the same order are equal exactly when every pair of corresponding entries is equal. Each position gives one equation. Basics: solving one unknown at a time

3Standard cosine values

cos⁡π6=32\cos\tfrac{\pi}{6} = \tfrac{\sqrt3}{2}⁠, cos⁡π3=12\cos\tfrac{\pi}{3} = \tfrac12⁠, cos⁡π=−1\cos\pi = -1⁠, cos⁡3π2=0\cos\tfrac{3\pi}{2} = 0⁠. On the unit circle, cos⁡α\cos\alpha is the xx-coordinate of the point at angle α\alpha⁠. Basics: trigonometric tools

Solution

  1. Write A′A'

    Row 1 of AA⁠, cos⁡α,−sin⁡α\cos\alpha, -\sin\alpha⁠, becomes column 1; row 2, sin⁡α,cos⁡α\sin\alpha, \cos\alpha⁠, becomes column 2:

    A′=[cos⁡αsin⁡α−sin⁡αcos⁡α].A' = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}.

    AA and A′A' are 2×22\times2⁠, so A+A′A + A' is defined and has the same order as II⁠.

    Transpose, Definition 3 (Basic 1). Lesson: transpose

  2. Compute A+A′A + A'

    Entry by entry:

    (1,1): cos⁡α+cos⁡α=2cos⁡α,(1,2): −sin⁡α+sin⁡α=0,(2,1): sin⁡α+(−sin⁡α)=0,(2,2): cos⁡α+cos⁡α=2cos⁡α.\begin{aligned} (1,1):&\ \cos\alpha + \cos\alpha = 2\cos\alpha, \\ (1,2):&\ -\sin\alpha + \sin\alpha = 0, \\ (2,1):&\ \sin\alpha + (-\sin\alpha) = 0, \\ (2,2):&\ \cos\alpha + \cos\alpha = 2\cos\alpha. \end{aligned} A+A′=[2cos⁡α002cos⁡α].A + A' = \begin{bmatrix} 2\cos\alpha & 0 \\ 0 & 2\cos\alpha \end{bmatrix}.
    The off-diagonal entries of AA are opposites, and transposing swaps them, so in the sum they cancel for every α\alpha⁠. This is a special case of Theorem 1: A+A′A + A' is always symmetric.
    Sum entry by entry, Section 3.4.1. Basics: adding like with like

  3. Equate with II and solve

    [2cos⁡α002cos⁡α]=[1001].\begin{bmatrix} 2\cos\alpha & 0 \\ 0 & 2\cos\alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.

    Compare the four entries:

    • (1,1) and (2,2): 2cos⁡α=12\cos\alpha = 1⁠, so cos⁡α=12\cos\alpha = \tfrac12⁠;
    • (1,2) and (2,1): 0=00 = 0⁠, true for every α\alpha⁠, so no further condition.

    So the condition A+A′=IA + A' = I holds exactly when cos⁡α=12\cos\alpha = \tfrac12⁠. The angles with this cosine are α=2nπ±π3\alpha = 2n\pi \pm \tfrac{\pi}{3}⁠, nn an integer; in [0,2π)[0, 2\pi) they are π3\tfrac{\pi}{3} and 5π3\tfrac{5\pi}{3}⁠.

    Matrix equality gives one equation per position (Basic 2). Two of them repeat each other and two are automatically true, so in the end there is a single equation to solve. Listing all solutions shows that, among the options offered, only π3\tfrac{\pi}{3} can work; the other solution 5π3\tfrac{5\pi}{3} is not an option.
    Equality of matrices, Definition 2, Section 3.3.1. Basics: solving equations

  4. Examine every option

    A+A′=(2cos⁡α) IA + A' = (2\cos\alpha)\, I⁠, so we need cos⁡α=12\cos\alpha = \tfrac12⁠:

    • (A) π6\tfrac{\pi}{6}⁠: wrong. cos⁡π6=32\cos\tfrac{\pi}{6} = \tfrac{\sqrt3}{2}⁠, so A+A′=3 IA + A' = \sqrt3\, I (diagonal ≈1.732\approx 1.732⁠), not II⁠.
    • (B) π3\tfrac{\pi}{3}⁠: correct. cos⁡π3=12\cos\tfrac{\pi}{3} = \tfrac12⁠, so the diagonal is 2⋅12=12\cdot\tfrac12 = 1 and A+A′=IA + A' = I⁠.
    • (C) π\pi⁠: wrong. cos⁡π=−1\cos\pi = -1⁠, so A+A′=−2IA + A' = -2I⁠.
    • (D) 3π2\tfrac{3\pi}{2}⁠: wrong. cos⁡3π2=0\cos\tfrac{3\pi}{2} = 0⁠, so A+A′=OA + A' = O⁠, the zero matrix.
    x y (A) π/6 (B) π/3 (C) π (D) 3π/2 5π/3 x = 1/2
    cos⁡α\cos\alpha is the xx-coordinate of the point at angle α\alpha on the unit circle. Only points on the dashed line x=12x = \tfrac12 satisfy cos⁡α=12\cos\alpha = \tfrac12⁠: angle π3\tfrac{\pi}{3} (option B) and 5π3\tfrac{5\pi}{3} (not offered).
    A multiple-choice answer must make the condition true; substituting each option confirms (B) and shows exactly why each other option fails.
    Standard cosine values (Basic 3). Basics: trigonometric tools
Result

(B) α=π3\alpha = \dfrac{\pi}{3}⁠, because A+A′=(2cos⁡α)IA + A' = (2\cos\alpha)I equals II exactly when cos⁡α=12\cos\alpha = \tfrac12⁠. (Textbook answer: B.)

Check it with numbers

At α=π3≈1.0471975512\alpha = \tfrac{\pi}{3} \approx 1.0471975512⁠: cos⁡α=0.5\cos\alpha = 0.5 and sin⁡α≈0.8660254038\sin\alpha \approx 0.8660254038⁠. The (1,1) entry of A+A′A + A' is 0.5+0.5=10.5 + 0.5 = 1⁠, and the (1,2) entry is −0.8660254038+0.8660254038=0-0.8660254038 + 0.8660254038 = 0⁠. ✓ At α=π6\alpha = \tfrac{\pi}{6} the (1,1) entry would be 2(0.8660254038)≈1.7320508076≠12(0.8660254038) \approx 1.7320508076 \ne 1⁠.

Common mistakes
  1. Mixing up sin and cos values. cos⁡π6=32\cos\tfrac{\pi}{6} = \tfrac{\sqrt3}{2} but sin⁡π6=12\sin\tfrac{\pi}{6} = \tfrac12⁠; choosing (A) usually comes from this slip.
  2. Forgetting the factor 2. The diagonal of A+A′A + A' is 2cos⁡α2\cos\alpha⁠, so the condition is cos⁡α=12\cos\alpha = \tfrac12⁠, not cos⁡α=1\cos\alpha = 1⁠.
  3. Thinking the off-diagonal entries give a condition. They cancel to 0 for every α\alpha⁠.
What to take away
  1. For this rotation-type matrix, A+A′=(2cos⁡α)IA + A' = (2\cos\alpha)I⁠: the sines cancel, the cosines double.
  2. A matrix equation is a set of entry equations; solve each one and keep only the real conditions.
  3. Find all solutions, then compare with the options.

Every fact used, and where it comes from

FactUsed inSource
Transpose (A′)ij=aji(A')_{ij} = a_{ji}Step 1Definition 3, Section 3.5 (Lesson: transpose)
Sum entry by entryStep 2Section 3.4.1 (Lesson: addition)
A+A′A + A' is symmetricStep 2 remarkTheorem 1, Section 3.6
Equal matrices: corresponding entries equalStep 3Definition 2, Section 3.3.1 (Lesson: equality)
cos⁡α=12\cos\alpha = \tfrac12 exactly when α=2nπ±π3\alpha = 2n\pi \pm \tfrac{\pi}{3}⁠; standard valuesSteps 3, 4Trigonometry, Class XI (Basics)
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