Find all matrices where the matrix is its own inverse and the determinant is 1

Question

I need to find all the matrices: $$\begin{bmatrix} a & b \\ c & d \\ \end{bmatrix} $$

such that

$$ad-bc=1$$

and

$$A^{-1}=A$$

How would I go about doing this? I know that $AA=I^2$, but that just gives me gibberish. Any hints?

Answer 1

Hint
The inverse of a $2\times 2$ matrix as you gave it is given by $$\frac1{ad-bc} \pmatrix{d&-b\\-c&a}$$

Bonus:
The answer will be disappointing, but can you see what happens if you allow $\det A = ad-bc$ to be free?

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Solution: Since $1 = \det I = \det (A^2) = \det^2 A = (ad-bc)^2$, we have $ad-bc \in \{\pm 1\}$. By the inversion formula we get: $$\cases{ad-bc = 1, a=d, b=-b, c=-c\\ ad-bc = -1, a=-d, d=-a}$$ The upper one simplifies to $b=c=0, a^2 = 1, a=d$ so $A = \pm I$ (this is the solution to the original question). The lower one simplifies to $a^2 = 1-bc, d=-a$ so chosing $b,c$ such that $bc \le 1$ we get $$A = \pmatrix{\pm \sqrt{1-bc} & b \\ c & \mp\sqrt{1-bc}}$$

Answer 2

Firstly, looking at the $(1, 2)$ and $(2, 1)$ entries of the square of $A$, we have $ab+bd = 0$ and $ac+dc = 0$, so either $b=c=0$ or $a+d=0$.

In the first case $a^2 = 1$ and $d^2 = 1$ and this gives you $4$ solutions, only two of which have the correct determinant.

The other case has $a^2 +bc = 1$ when we examine the result of the multiplication. Compare this to the determinant, and you should be done.

Answer 3

Just another view: the determinant of a matrix is also the product of its eigenvalues. A 2x2-matrix has 2 eigenvalues, so they must satisfy $$\lambda_0 \cdot \lambda_1 = 1 $$ On the other hand, if a matrix $A$ has some eigenvalues $\lambda_0$ and $\lambda_1$ then $A^2$ has the eigenvalues $\lambda_0^2$ and $\lambda_1^2$ ; taking that $A^2 = I$ it means that that squares must both equal $1$. Now there are only two numbers $m$ whose square equal $1$ : $m \in (1,-1)$ Then the $\lambda$s must equal one of the $m$, and since their product must equal 1, they must have the same sign. So any matrix $$A_0 = Q \cdot D_0 \cdot Q^{-1} \qquad \qquad \text{ or } \qquad A_1 = Q \cdot D_1 \cdot Q^{-1} $$ with any invertible matrix $Q$ is a solution, where $\small \operatorname{diag}(D_0) = [1,1]$ and $\small \operatorname{diag}(D_1) = [-1,-1]$

Since $D_0$ is the identity matrix, $A_0 = I$ is also the identity matrix (by the definition of the multiplication of a matrix with its inverse $Q \cdot Q^{-1} = I$). But the same occurs also for $A_1$ only that $A_1 = - 1 \cdot A_0 = - I$ by the possibility to factor out the scalar $-1$ which is common to both entries in $D_1$ .

Remark: The reason why I added this seemingly redundant answer is, that it opens the view also for the more general case, when the size of $A$ is larger than $2x2$ and the explicte expressions for the determinant are intractable.