For which values of $a,b$ is the matrix invertible? I am trying to figure out the below question:

15. For which values of the constants $a$ and $b$ is the matrix
  $$A = \left[\begin{array}{cc} a & -b \\ b & a \end{array}\right]$$
  invertible? What is the inverse in this case? See Exercise 13.

My understanding is that a matrix is invertible when the determinant is not zero. In this case, when $a^2 - b^2 = 0$ the matrix is not invertible. Thus, for any values $a,b$ such that $a^2$ does not equal $b^2$, the matrix is invertible.
However, the solutions in the back of the book state that the matrix is invertible if $a$ does not equal zero or if $b$ does not equal zero. Can someone explain this to me? 
 A: The determinant of the matrix is $a^2 + b^2$, not $a^2 - b^2$. This is non-zero if and only if at least one of $a$ and $b$ is non-zero.
This assumes that $a$ and $b$ are real; otherwise, the issue is more complicated. For example, the matrix
$$\left[\begin{array}{cc} i & -1 \\ 1 & i\end{array}\right]$$
is not invertible.
A: A matrix $\mathsf{A}$ is said to be invertible, namely it does exist $\mathsf{A}^{-1}$ when it's determinant is not zero.
In your case:
$$\text{Det}\ \mathsf{A} = (a\cdot a) - (b\cdot(-b)) = a^2 + b^2$$
Thence when $$a^2 \neq -b^2$$
So the only case by which the determinant is zero, if $(a, b)\in\mathbb{R}$ is when $a = b = 0$. The trivial solution.
The inverse of a $2x2$ matrix
If a matrix $\mathsf{A}$ is invertible, and $\mathsf{A}$ is a $2x2$ matrix of the form
$$
\mathsf{A} = \left(
\begin{matrix}
a & b \\
c & d
\end{matrix}
\right)
$$
thence its inverse is defined to be
$$
\mathsf{A}^{-1} = \frac{1}{\text{Det}\ \mathsf{A}} \left(
\begin{matrix}
d & -b \\
-c & a
\end{matrix}
\right)
$$
Try to adapt this into your case..
