Find the least squares solution to $A\vec{x} = \vec{b}$ using the transpose of $A$ 
Find the least squares solution to $A\vec{x} = \vec{b}$ using the transpose of $A$. Let $A = \begin{bmatrix}1&1&0\\1&0&-1\\0&1&1\\-1&1&-1\end{bmatrix}$ and $\vec{b} = \begin{bmatrix}2\\5\\6\\6\end{bmatrix}$

The formula says $x = (A^TA)^{-1} * A^T b$ so if they want me to use the transpose of $A$ should I do instead $x = (A^{T^T}A)^{-1} * (A^{T^T})b$?

$$A^T A = \left(
\begin{array}{ccc}
 3 & 0 & 0 \\
 0 & 3 & 0 \\
 0 & 0 & 3 \\
\end{array}
\right)$$
$$(A^T A )^{-1} = \left(
\begin{array}{ccc}
 \frac{1}{3} & 0 & 0 \\
 0 & \frac{1}{3} & 0 \\
 0 & 0 & \frac{1}{3} \\
\end{array}
\right)$$
$$A^T b = (1,14,-5)$$
$$(A^T A )^{-1} A^T b = \left(\dfrac{1}{3},\dfrac{14}{3},-\dfrac{5}{3}\right)$$
