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