Find values of $a$ and $b$ that make matrix orthogonal 
Given the matrix 
$$A=\begin{bmatrix}1/2&a\\b&1/2\\ \end{bmatrix}$$ 
find the values of $a$ and $b$ that make it orthogonal. 

So far I have tried using dot product $$(1/2)a+(1/2)b=0$$ and we can conclude that $a=-b$ and $b=-a$. I also tried the following theorem 
$$A^T=A^{-1}$$
so
$$\begin{bmatrix}1/2&b\\a&1/2\\ \end{bmatrix}=
\begin{bmatrix}
(2+\frac{4ab}{1-4ab})&\frac{-4a}{1-4ab}\\
\frac{-4b}{1-4ab}&\frac{2}{1-4ab}\\ \end{bmatrix}$$
Can someone tell if I'am on the right track and point me in the right direction? Thanks!
 A: Orthogonal matrix means $$AA^T=I$$
Hence,
$$\begin{bmatrix}
\frac{1}{2} & a\\ 
 b& \frac{1}{2}
\end{bmatrix}\begin{bmatrix}
\frac{1}{2} & b\\ 
 a& \frac{1}{2}
\end{bmatrix}=\begin{bmatrix}
\frac{1}{4}+a^2 & \frac{1}{2}(a+b)\\ 
 \frac{1}{2}(a+b)& \frac{1}{4}+b^2
\end{bmatrix}=\begin{bmatrix}
1 & 0\\ 
 0& 1\end{bmatrix}$$
which implies $$a=-b=\pm\frac{\sqrt{3}}{2}$$
A: A square matrix $A$ is orthogonal iff $A^TA=AA^T=I$. Computing $A^TA$, we have
$$ 
A^TA = \begin{bmatrix} \frac12& b\\ a &\frac12
\end{bmatrix}\begin{bmatrix} \frac12& a\\ b &\frac12
\end{bmatrix} = \begin{bmatrix} \frac14 + b^2& \frac12(a+b)\\ \frac12(a+b) &\frac14 + a^2.
\end{bmatrix}
$$
From $A^TA$ we obtain the system of equations
\begin{align}
\frac14 + b^2 &= 1\\
\frac14 + a^2 &= 1\\
\frac12 (a+b) &= 0,
\end{align}
which has solution set $$(a,b) = \pm\left(\frac{\sqrt3}2, -\frac{\sqrt3}2 \right).$$
A: We know $A^TA=I$ where $I$ is the identity matrix.
So applying this, we have $a^2+0.25=1$ giving us $a= \pm \frac{\sqrt 3}{2}$.  Next, $\frac{b} {2}+\frac{a}{2}=0$ giving us $b=-a$.
