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Chapter 3 · Matrices

Exercise 3.2 · Q6: Simplify a trigonometric matrix sum

Exercise 3.2Scalar multiplication and addition of matrices · Section 3.4.2
The question

Simplify

cos⁡θ[cos⁡θsin⁡θ−sin⁡θcos⁡θ]+sin⁡θ[sin⁡θ−cos⁡θcos⁡θsin⁡θ]\begin{aligned}&\cos\theta\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}\\&+\sin\theta\begin{bmatrix}\sin\theta&-\cos\theta\\\cos\theta&\sin\theta\end{bmatrix}\end{aligned}
Exam answer

Let EE be the given expression. Then

E=[cos⁡2θcos⁡θsin⁡θ−sin⁡θcos⁡θcos⁡2θ]+[sin⁡2θ−sin⁡θcos⁡θsin⁡θcos⁡θsin⁡2θ]\begin{aligned} &E \\ &= \begin{bmatrix} \cos^2\theta & \cos\theta\sin\theta \\ -\sin\theta\cos\theta & \cos^2\theta \end{bmatrix} \\ &\quad + \begin{bmatrix} \sin^2\theta & -\sin\theta\cos\theta \\ \sin\theta\cos\theta & \sin^2\theta \end{bmatrix} \end{aligned}

Adding corresponding entries, with E=[cij]E = [c_{ij}]⁠,

c11=cos⁡2θ+sin⁡2θ=1c12=cos⁡θsin⁡θ−sin⁡θcos⁡θ=0c21=−sin⁡θcos⁡θ+sin⁡θcos⁡θ=0c22=cos⁡2θ+sin⁡2θ=1\begin{aligned} c_{11} &= \cos^2\theta + \sin^2\theta = 1 \\ c_{12} &= \cos\theta\sin\theta - \sin\theta\cos\theta \\ &= 0 \\ c_{21} &= -\sin\theta\cos\theta + \sin\theta\cos\theta \\ &= 0 \\ c_{22} &= \cos^2\theta + \sin^2\theta = 1 \end{aligned}

∴ E=[1001]=I\therefore\ E = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I

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

The one idea this problem is built on

Here cos⁡θ\cos\theta and sin⁡θ\sin\theta in front of the matrices are just numbers (scalars), so each multiplies every entry of its matrix. After that, the two matrices are added position by position. On the diagonal each position collects cos⁡2θ+sin⁡2θ=1\cos^2\theta+\sin^2\theta=1⁠; off the diagonal each position collects a term and its negative, which cancel. The result is the identity matrix II⁠, whatever θ\theta is.

Given, to find, and the plan
Given
A real number θ\theta and two 2×22\times2 matrices, each multiplied by a scalar.
To find
The simplest form of the sum.
Plan
(1) Check orders. (2) Multiply the first matrix by cos⁡θ\cos\theta⁠. (3) Multiply the second by sin⁡θ\sin\theta⁠. (4) Add position by position. (5) Simplify each entry.

Basics you need

1A number times a matrix

kA=[k aij]kA=[k\,a_{ij}]⁠: every entry is multiplied by kk⁠. Here kk is cos⁡θ\cos\theta or sin⁡θ\sin\theta⁠, which are real numbers for each value of θ\theta⁠. Basics: adding and scaling matrices

2Adding matrices

Same order; add the entries in matching positions.

3sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1

True for every real θ\theta⁠: the point (cos⁡θ,sin⁡θ)(\cos\theta,\sin\theta) lies on the unit circle x2+y2=1x^2+y^2=1⁠. (Class XI, Trigonometric Functions.)

4The identity matrix

The square matrix with 1 on the diagonal and 0 elsewhere; the 2×22\times2 one is written II (or I2I_2⁠). Basics: identity

Solution

  1. Orders

    Both matrices are 2×22\times2⁠. Multiplying by a scalar keeps the order, so both terms are 2×22\times2⁠, the sum is defined, and the answer is 2×22\times2⁠.

    This is a sum of two scalar multiples, so the only size condition is that the two matrices have the same order, which they do.

  2. Multiply the first matrix by cos⁡θ\cos\theta

    Every entry is multiplied by cos⁡θ\cos\theta⁠:

    (1,1): cos⁡θ⋅cos⁡θ=cos⁡2θ(1,2): cos⁡θ⋅sin⁡θ=sin⁡θcos⁡θ(2,1): cos⁡θ⋅(−sin⁡θ)=−sin⁡θcos⁡θ(2,2): cos⁡θ⋅cos⁡θ=cos⁡2θ\begin{aligned} (1,1)&:\ \cos\theta\cdot\cos\theta=\cos^2\theta\\ (1,2)&:\ \cos\theta\cdot\sin\theta=\sin\theta\cos\theta\\ (2,1)&:\ \cos\theta\cdot(-\sin\theta)=-\sin\theta\cos\theta\\ (2,2)&:\ \cos\theta\cdot\cos\theta=\cos^2\theta \end{aligned} cos⁡θ[cos⁡θsin⁡θ−sin⁡θcos⁡θ]=[cos⁡2θsin⁡θcos⁡θ−sin⁡θcos⁡θcos⁡2θ]\begin{aligned}&\cos\theta\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}\\ &=\begin{bmatrix}\cos^2\theta&\sin\theta\cos\theta\\-\sin\theta\cos\theta&\cos^2\theta\end{bmatrix}\end{aligned}
    For a fixed θ\theta⁠, cos⁡θ\cos\theta is a real number, so this is an ordinary scalar multiple. Writing every product as sin⁡θcos⁡θ\sin\theta\cos\theta (sine first) makes it easy to spot cancellations later.
    kA=[k aij]kA=[k\,a_{ij}]⁠. Lesson: scalar multiplication

  3. Multiply the second matrix by sin⁡θ\sin\theta

    (1,1): sin⁡θ⋅sin⁡θ=sin⁡2θ(1,2): sin⁡θ⋅(−cos⁡θ)=−sin⁡θcos⁡θ(2,1): sin⁡θ⋅cos⁡θ=sin⁡θcos⁡θ(2,2): sin⁡θ⋅sin⁡θ=sin⁡2θ\begin{aligned} (1,1)&:\ \sin\theta\cdot\sin\theta=\sin^2\theta\\ (1,2)&:\ \sin\theta\cdot(-\cos\theta)=-\sin\theta\cos\theta\\ (2,1)&:\ \sin\theta\cdot\cos\theta=\sin\theta\cos\theta\\ (2,2)&:\ \sin\theta\cdot\sin\theta=\sin^2\theta \end{aligned} sin⁡θ[sin⁡θ−cos⁡θcos⁡θsin⁡θ]=[sin⁡2θ−sin⁡θcos⁡θsin⁡θcos⁡θsin⁡2θ]\begin{aligned}&\sin\theta\begin{bmatrix}\sin\theta&-\cos\theta\\\cos\theta&\sin\theta\end{bmatrix}\\ &=\begin{bmatrix}\sin^2\theta&-\sin\theta\cos\theta\\\sin\theta\cos\theta&\sin^2\theta\end{bmatrix}\end{aligned}

  4. Add position by position

    (1,1): cos⁡2θ+sin⁡2θ(1,2): sin⁡θcos⁡θ−sin⁡θcos⁡θ(2,1): −sin⁡θcos⁡θ+sin⁡θcos⁡θ(2,2): cos⁡2θ+sin⁡2θ\begin{aligned} (1,1)&:\ \cos^2\theta+\sin^2\theta\\ (1,2)&:\ \sin\theta\cos\theta-\sin\theta\cos\theta\\ (2,1)&:\ -\sin\theta\cos\theta+\sin\theta\cos\theta\\ (2,2)&:\ \cos^2\theta+\sin^2\theta \end{aligned}
    Each position of the sum takes the entry from Step 2 and the entry from Step 3 in that same position. The off-diagonal positions pair a term with its exact negative; the diagonal positions pair cos⁡2θ\cos^2\theta with sin⁡2θ\sin^2\theta⁠.
    Basics: adding like with like

  5. Simplify each entry

    Diagonal entries: cos⁡2θ+sin⁡2θ=1\cos^2\theta+\sin^2\theta=1 (Basic 3). Off-diagonal entries: a quantity minus itself is 0. So

    [1001]=I.\begin{bmatrix}1&0\\0&1\end{bmatrix}=I.
    Both simplifications hold for every real θ\theta⁠, so the expression equals the 2×22\times2 identity matrix for all θ\theta⁠. No restriction on θ\theta is needed because sine and cosine are defined for every real number.
    Identity matrix: 1 on the diagonal, 0 elsewhere. Lesson: types of matrices
Result cos⁡θ[cos⁡θsin⁡θ−sin⁡θcos⁡θ]+sin⁡θ[sin⁡θ−cos⁡θcos⁡θsin⁡θ]=[1001] \begin{aligned}&\cos\theta\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}\\&+\sin\theta\begin{bmatrix}\sin\theta&-\cos\theta\\\cos\theta&\sin\theta\end{bmatrix}\\&=\begin{bmatrix}1&0\\0&1\end{bmatrix}\end{aligned}

that is, the identity matrix II⁠, for every real θ\theta⁠. (Textbook answer: the 2×22\times2 identity matrix.)

Check it with numbers

Take θ=30∘\theta=30^\circ⁠: cos⁡θ≈0.8660254038\cos\theta\approx0.8660254038⁠, sin⁡θ=0.5\sin\theta=0.5⁠. Then cos⁡2θ=0.75\cos^2\theta=0.75⁠, sin⁡2θ=0.25\sin^2\theta=0.25⁠, sin⁡θcos⁡θ≈0.4330127019\sin\theta\cos\theta\approx0.4330127019⁠.

  • First term: rows (0.75, 0.4330127019)(0.75,\ 0.4330127019) and (−0.4330127019, 0.75)(-0.4330127019,\ 0.75)⁠.
  • Second term: rows (0.25, −0.4330127019)(0.25,\ -0.4330127019) and (0.4330127019, 0.25)(0.4330127019,\ 0.25)⁠.
  • Sum: rows (1, 0)(1,\ 0) and (0, 1)(0,\ 1)⁠. ✓
Common mistakes
  1. Multiplying only the first row (or first column) by the scalar. A scalar multiplies all four entries.
  2. Treating this as a matrix product. cos⁡θ\cos\theta in front of the matrix is a number, not a matrix; there is no row-times-column work here.
  3. Losing a minus sign in the (1,2)(1,2) entry of the second term: sin⁡θ⋅(−cos⁡θ)=−sin⁡θcos⁡θ\sin\theta\cdot(-\cos\theta)=-\sin\theta\cos\theta⁠.
What to take away
  1. Trigonometric functions in front of a matrix are scalars: they multiply every entry.
  2. Write products in one fixed order (sin⁡θcos⁡θ\sin\theta\cos\theta⁠) so that cancellations are visible.
  3. A simplification that uses only identities true for all θ\theta gives an answer valid for all θ\theta⁠.

Every fact used, and where it comes from

FactUsed inSource
Scalar multiple kA=[k aij]kA=[k\,a_{ij}]Steps 2, 3Section 3.4.2 (Lesson: scalar multiplication)
Sum of matrices of the same orderSteps 1, 4Section 3.4.1
sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1 for all real θ\thetaStep 5Trigonometric identities (Class 11)
Identity matrixStep 5Section 3.3 (vi) (Lesson: types of matrices)
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