The exersice requested to find a QR factorization of $A$
$A=\begin{bmatrix} 1&1&0\\1&1&1\\0&1&-1 \end{bmatrix}$
Wich we did and obtainted
$Q=\begin{bmatrix} \frac{-\sqrt{2}}{2}&0&\frac{-\sqrt{2}}{2}\\\frac{-\sqrt{2}}{2}&0&\frac{\sqrt{2}}{2}\\0&-1&0 \end{bmatrix}$ $R=\begin{bmatrix} -\sqrt{2}&-\sqrt{2}&\frac{-\sqrt{2}}{2}\\0&-1&1\\0&0&\frac{\sqrt{2}}{2} \end{bmatrix}$
Then we had to use the factorization to solve (in the sense of least squares) the overdetermined problem:
$x+y=0, x+y=1, y=-1$ and (multiplying by $Q^T$ and obtaining a triangular system) we get that the solution is
$x=\frac{3}{2}, y=-1$
But then we have to use the $QR$ factorization to find the minimum distance between the vectors $(e^{x+y}, x+y, e^{y+1})$ and $(1,1,1)$.
At first it could seem like the solution is the same, because $e^{x+y}=1\Longleftrightarrow x+y=0$ and $e^{y+1}=1 \Longleftrightarrow y=-1$ but that is not true because the distance from $e^t$ to $1$ is not the same as from $t$ to $0$.
Besides that, which we can't use I don't know was else can be done to solve the problem using que factorization we already have. Any ideas to do so? Thanks
