Exercise 4.4 · Q18 (MCQ): $\det(A^{-1})$ for an invertible matrix of order 2

Basics you need

Solution

1. Take the determinant of $AA^{-1} = I$

   $A$ and $A^{-1}$ are both $2\times2$⁠, so Theorem 3 applies:

   $$\begin{aligned} \det(AA^{-1}) &= \det I\\ \det(A)\,\det(A^{-1}) &= 1 \end{aligned}$$
   $\det I = 1\cdot1 - 0\cdot0 = 1$ for the $2\times2$ identity. Theorem 3 turns the determinant of the product into a product of two numbers, and the unknown $\det(A^{-1})$ is one of them.
   $|XY| = |X||Y|$⁠. Basics: matrix facts the chapter uses

2. Conclude $\det(A^{-1}) = \dfrac{1}{\det(A)}$

   Two numbers with product $1$ are both non-zero, so $\det(A) \neq 0$ (this is also Theorem 4: invertible matrices are non-singular). Dividing by $\det(A)$⁠:

   $$\det(A^{-1}) = \frac{1}{\det(A)}$$

   This holds for every invertible $2\times2$ matrix, so the answer is (B).

   We never needed the entries of $A$⁠: only $AA^{-1} = I$⁠, Theorem 3 and $\det I = 1$⁠. So the argument works equally for invertible matrices of any order $n$⁠, with the $n\times n$ identity, whose determinant is also $1$⁠.

3. A second proof, straight from the entries

   Let $A$ have rows $p, q$ and $r, s$⁠, and put $D = \det A = ps - qr$⁠. Since $A$ is invertible, $D \neq 0$ (Theorem 4), and

   $$A^{-1} = \frac{1}{D}\begin{bmatrix} s & -q \\ -r & p \end{bmatrix}.$$

   Its entries are $\frac{s}{D}, \frac{-q}{D}$ (row 1) and $\frac{-r}{D}, \frac{p}{D}$ (row 2), so

   $$\begin{aligned} \det(A^{-1}) &= \frac{s}{D}\cdot\frac{p}{D} - \frac{-q}{D}\cdot\frac{-r}{D}\\[1mm] &= \frac{ps - qr}{D^2} = \frac{D}{D^2} = \frac{1}{D} \end{aligned}$$
   This proof does not use Theorem 3, which the book states without proof. Only $D \neq 0$ was quoted from Theorem 4 (whose proof in the book uses Theorem 3), and for order 2 that can be seen directly too. Multiplying out, $A(\operatorname{adj}A)$ has entries $ps - qr$⁠, $-pq + qp$⁠, $rs - sr$⁠, $-rq + sp$⁠, so it is $DI$⁠. If $D$ were $0$⁠, this would be $O$⁠; multiplying on the left by $A^{-1}$ would give $\operatorname{adj}A = O$⁠, so $p = q = r = s = 0$ and $A = O$⁠, which has no inverse. The proof also shows where the answer comes from: the factor $\frac1D$ is in both rows, which contributes $\frac{1}{D^2}$ (Basic 4), and $\operatorname{adj}A$ has the same determinant $sp - (-q)(-r) = D$ as $A$⁠.
   $2\times2$ adjoint and inverse. Lesson: the adjoint · Lesson: inverse

4. Examine every option

   Test matrix: the diagonal matrix with entries $2$ and $1$⁠.

   $$A = \begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix}, \quad A^{-1} = \begin{bmatrix} \frac12 & 0 \\ 0 & 1 \end{bmatrix}$$

   Here $\det A = 2\cdot1 - 0 = 2$ and $\det(A^{-1}) = \frac12\cdot1 - 0 = \frac12$⁠. (Check: $AA^{-1}$ has diagonal $2\cdot\frac12 = 1$ and $1\cdot1 = 1$ and zeros elsewhere, so it is $I$⁠.)

   Option	Value	Verdict
   (A) $\det(A)$	$2$	Wrong in general (it holds only when $\det A$ is $1$ or $-1$⁠).
   (B) $\frac{1}{\det(A)}$	$\frac12$	Correct, proved in Steps 1–3 for every invertible $A$⁠.
   (C) $1$	$1$	Wrong. $1$ is the determinant of the product $AA^{-1}$⁠.
   (D) $0$	$0$	Wrong for every invertible $A$⁠: a product equal to $1$ cannot have a factor $0$⁠.

   Each option claims a formula for every invertible $2\times2$ matrix, so a single matrix where it fails rejects it. One test matrix with $\det A = 2$ separates all four options, because $2$⁠, $\frac12$⁠, $1$ and $0$ are all different.
   One counterexample defeats a "for all" claim. Basics: algebra the chapter leans on

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