Exercise 4.1 · Q2: Evaluate two determinants

Basics you need

Part (i): the cosine and sine determinant

1. Read off the entries

   $$\begin{aligned} a_{11} &= \cos\theta, & a_{12} &= -\sin\theta,\\ a_{21} &= \sin\theta, & a_{22} &= \cos\theta. \end{aligned}$$
   The minus sign in the top right entry belongs to that entry: $a_{12}$ is the number $-\sin\theta$⁠. Keeping it attached is what makes the next step come out right.
   Basics: rows, columns and positions

2. Apply the definition

   $$\begin{aligned} \Delta &= a_{11}a_{22} - a_{21}a_{12}\\ &= (\cos\theta)(\cos\theta) - (\sin\theta)(-\sin\theta)\\ &= \cos^2\theta + \sin^2\theta. \end{aligned}$$
   The main diagonal gives $\cos\theta\cdot\cos\theta = \cos^2\theta$⁠. The other diagonal gives $\sin\theta\cdot(-\sin\theta) = -\sin^2\theta$⁠, and subtracting a negative quantity adds it: $-(-\sin^2\theta) = +\sin^2\theta$⁠.
   $|A| = a_{11}a_{22} - a_{21}a_{12}$⁠, Section 4.2.2. Lesson: determinant of a square matrix

3. Use $\cos^2\theta + \sin^2\theta = 1$

   $$\Delta = \cos^2\theta + \sin^2\theta = 1.$$

   So the determinant equals 1 for every real $\theta$.

   The identity holds for every real angle, so the value does not depend on $\theta$ at all. That is why the answer is a plain number rather than an expression in $\theta$⁠.
   Pythagorean identity (Basic 2), from Class XI trigonometry.

Part (ii): the determinant in $x$

1. Read off the entries

   $$\begin{aligned} a_{11} &= x^2 - x + 1, & a_{12} &= x - 1,\\ a_{21} &= x + 1, & a_{22} &= x + 1. \end{aligned}$$
   Each entry is a whole expression and must stay together as one quantity, so every entry goes inside its own bracket when we multiply.

2. Apply the definition

   $$\begin{aligned} \Delta &= a_{11}a_{22} - a_{21}a_{12}\\ &= (x^2 - x + 1)(x + 1)\\ &\quad - (x + 1)(x - 1). \end{aligned}$$
   Main diagonal: $x^2 - x + 1$ and $x + 1$⁠. Other diagonal: $x + 1$ (bottom left) and $x - 1$ (top right). The minus sign in front of the second product applies to the whole product, which is why that product is kept in brackets.
   $|A| = a_{11}a_{22} - a_{21}a_{12}$⁠, Section 4.2.2, exactly as in Example 2.

3. Expand the first product

   $$\begin{aligned} (x^2 - x + 1)(x + 1) &= x^3 + x^2 - x^2\\ &\quad - x + x + 1\\ &= x^3 + 1. \end{aligned}$$
   Each term of $x^2 - x + 1$ is multiplied by $x$ and by $1$⁠: $x^2\cdot x = x^3$⁠, $x^2\cdot1 = x^2$⁠, $(-x)\cdot x = -x^2$⁠, $(-x)\cdot 1 = -x$⁠, $1\cdot x = x$⁠, $1\cdot1 = 1$⁠. The $x^2$ terms cancel and the $x$ terms cancel, leaving $x^3 + 1$⁠. (This is the sum-of-cubes product $(a+b)(a^2 - ab + b^2) = a^3 + b^3$ with $a = x$⁠, $b = 1$⁠.)
   Basics: algebra the chapter leans on

4. Expand the second product and subtract

   $$(x + 1)(x - 1) = x^2 - 1$$ $$\begin{aligned} \Delta &= (x^3 + 1) - (x^2 - 1)\\ &= x^3 + 1 - x^2 + 1\\ &= x^3 - x^2 + 2. \end{aligned}$$

   So the determinant equals $x^3 - x^2 + 2$.

   $(x+1)(x-1)$ is a difference of squares, $x^2 - 1^2$⁠. When it is subtracted, the bracket's minus sign changes the sign of both terms: $-(x^2 - 1) = -x^2 + 1$⁠. Forgetting the second sign change gives $x^3 - x^2$⁠, the most common wrong answer.
   $(a+b)(a-b) = a^2 - b^2$⁠. Basics: algebra the chapter leans on

5. The same answer by factorising first

   Both products contain the factor $x + 1$⁠, so take it out before expanding:

   $$\begin{aligned} \Delta &= (x + 1)\big[(x^2 - x + 1) - (x - 1)\big]\\ &= (x + 1)(x^2 - 2x + 2)\\ &= x^3 - 2x^2 + 2x + x^2 - 2x + 2\\ &= x^3 - x^2 + 2. \end{aligned}$$
   Getting the same polynomial by a second route is a strong check. The factorised form also tells us something extra: $x^2 - 2x + 2 = (x-1)^2 + 1$ is never 0 for real $x$⁠, so for real $x$ the determinant is 0 only at $x = -1$⁠, where the whole second row becomes 0.

Every fact used, and where it comes from