Inverse matrix norm under simple conditions Let $A$ be a real $2\times 2$ matrix such that $\det A=1$, show that $\|A\|=\left\|A^{-1}\right\|$.
Any hint would be appreciated, thanks.
EDIT: $\|\cdot\|$ is the operator norm $\|A\|=\max_{\|x\|=1}\|Ax\|$, all vector norms are Euclidean norms.
 A: Hint: Use the definition of the inverse matrix

If A is invertible, then $$  \det(\mathbf{A}) \mathbf{A}^{-1}=\mathrm{adj}(\mathbf{A}) , $$
  where $\mathrm{adj}(\mathbf{A})$ is the adjugate matrix. 

The adjugate of the 2 × 2 matrix
$$\mathbf{A} = \begin{pmatrix} {{a}} & {{b}}\\ {{c}} & {{d}} \end{pmatrix}$$
is
$$\operatorname{adj}(\mathbf{A}) = \begin{pmatrix} \,\,\,{{d}} & \!\!{{-b}}\\ {{-c}} & {{a}} \end{pmatrix}.$$ 
A: Recall the singular value decomposition of $A$ as $A_{2 \times 2} = U \begin{bmatrix}\Sigma_{11} & 0\\ 0 & \Sigma_{22} \end{bmatrix}V^*$, then $A^{-1} = V^{*^{-1}} \begin{bmatrix}\dfrac1{\Sigma_{11}} & 0\\ 0 & \dfrac1{\Sigma_{22}} \end{bmatrix} U^{-1}$. Since $\det(A) = 1$, we have that $\Sigma_{11} \Sigma_{22} = 1$. Also, recall that $$\Vert B \Vert_2 = \max \{\sigma_i: \sigma_i \text{ is a singular value of }B\}$$
Hence, $\Vert A \Vert_2 = \Sigma_{11}$ and $\Vert A^{-1} \Vert_2 = \dfrac1{\Sigma_{22}} = \Sigma_{11}$. Hence, we get that $$\Vert A \Vert_2 = \Vert A^{-1} \Vert_2$$
The equality also hold good for the Frobenius norm since $$\left \Vert A \right \Vert_{F} = \sqrt{\Sigma_{11}^2 + \dfrac1{\Sigma_{11}^2}} = \left \Vert A^{-1} \right \Vert_F$$
