Miscellaneous · Q9 (MCQ): Condition from $A^2=I$

Basics you need

Solution

1. Compute $A^2$⁠, entry by entry

   A is $2\times2$⁠, so $AA$ is defined and $2\times2$⁠. The rows of A are $(\alpha, \beta)$ and $(\gamma, -\alpha)$⁠; the columns are $(\alpha, \gamma)$ and $(\beta, -\alpha)$⁠.

   $$\begin{aligned} c_{11} &= \alpha\cdot\alpha + \beta\cdot\gamma = \alpha^2 + \beta\gamma \\ c_{12} &= \alpha\cdot\beta + \beta\cdot(-\alpha) = 0 \\ c_{21} &= \gamma\cdot\alpha + (-\alpha)\cdot\gamma = 0 \\ c_{22} &= \gamma\cdot\beta + (-\alpha)(-\alpha) = \beta\gamma + \alpha^2 \end{aligned}$$ $$A^2 = \begin{bmatrix} \alpha^2 + \beta\gamma & 0 \\ 0 & \alpha^2 + \beta\gamma \end{bmatrix}$$
   The off-diagonal entries cancel because the two diagonal entries of A are $\alpha$ and $-\alpha$⁠: $\alpha\beta - \alpha\beta = 0$ and $\gamma\alpha - \alpha\gamma = 0$ (numbers commute, so $\gamma\alpha = \alpha\gamma$⁠). In $c_{22}$⁠, $(-\alpha)(-\alpha) = +\alpha^2$⁠. So $A^2$ is the scalar matrix $(\alpha^2 + \beta\gamma)I$⁠.
   Entry formula of Section 3.4.5. Basics: pair up, multiply, add

2. Equate with I

   $A^2 = I$ means the four entries match those of $I_2$⁠. The off-diagonal ones give $0 = 0$⁠, always true. Both diagonal ones give the same equation:

   $$\alpha^2 + \beta\gamma = 1.$$

   Subtracting $\alpha^2 + \beta\gamma$ from both sides,

   $$1 - \alpha^2 - \beta\gamma = 0.$$

   This is option (C). The steps are reversible, so it is in fact equivalent to $A^2 = I$⁠: (C) holds exactly when $A^2 = I$⁠.

   Rearranging $\alpha^2 + \beta\gamma = 1$ into the form "$\dots = 0$" lets us match it against the options, which are all written that way. Notice that both signs in front of $\alpha^2$ and $\beta\gamma$ are minus: they came from moving $+\alpha^2$ and $+\beta\gamma$ across.
   Definition 2 (Section 3.3.1). Lesson: equality of matrices

3. Examine every option

   Under the condition $\beta\gamma = 1 - \alpha^2$ from Step 2, each option's left side becomes:

   Option	Left side becomes	Verdict
   (A)	$1 + \alpha^2 + (1 - \alpha^2) = 2$	Wrong. It is never 0 when $A^2 = I$⁠.
   (B)	$1 - \alpha^2 + (1 - \alpha^2)$⁠, that is $2 - 2\alpha^2$	Wrong. Not forced. Take $\alpha = 0$⁠, $\beta = \gamma = 1$⁠: A has rows $(0, 1)$⁠, $(1, 0)$ and $A^2 = I$⁠, but (B) gives $1 - 0 + 1 = 2 \ne 0$⁠.
   (C)	$1 - \alpha^2 - (1 - \alpha^2) = 0$	Correct. It always holds when $A^2 = I$ (Step 2).
   (D)	$1 + \alpha^2 - (1 - \alpha^2)$⁠, that is $2\alpha^2$	Wrong. Not forced. Take $\alpha = 1$⁠, $\beta = \gamma = 0$⁠: A has rows $(1, 0)$⁠, $(0, -1)$ and $A^2 = I$⁠, but (D) gives $1 + 1 - 0 = 2 \ne 0$⁠.

   Options (B) and (D) are true for some matrices with $A^2 = I$ (for instance (B) holds when $\alpha = \pm1$⁠, and (D) when $\alpha = 0$⁠), so it is not enough to say they "look wrong". A single matrix with $A^2 = I$ that breaks the option shows it does not follow from $A^2 = I$⁠. Option (A) fails for every such matrix.
   A counterexample disproves a general claim. Lesson: matrix multiplication

Every fact used, and where it comes from