Hi I'm looking for the least squares solutions of....
$$ \begin{pmatrix} 1&-1 \\ -1&2 \\ -1&0 \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \\ 5\end{pmatrix} $$
So Assuming this goes by $ A\vec{x}=\vec{b}$
Then using.... $A^T A \vec{x} = A^T \vec{b} $
$$ \begin{pmatrix} 1&-1&-1 \\ -1&2&0 \end{pmatrix} \begin{pmatrix} 1&-1 \\ -1&2 \\ -1&0 \end{pmatrix} \vec{x}= \begin{pmatrix} 1&-1&-1 \\ -1&2&0 \end{pmatrix} \begin{pmatrix} 3 \\ 4 \\ 5\end{pmatrix} $$
Which solves to be $$ \begin{pmatrix} 3&-3 \\ -3&5 \end{pmatrix} \vec{x} = \begin{pmatrix} -6 \\ 5 \end{pmatrix} $$
which i solve to get $$ \vec{x} = \begin{pmatrix} -7/3 \\ 3/2 \end{pmatrix} $$
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$\begingroup$Since you are only asking for the solution...
$$ \left( \begin{vmatrix} 1 & -1 \\ -1 & 2 \\ -1 & 0 \end{vmatrix}^\intercal \begin{vmatrix} 1 & -1 \\ -1 & 2 \\ -1 & 0 \end{vmatrix} \right) \begin{pmatrix} x \\ y \end{pmatrix} = \left( \begin{vmatrix} 1 & -1 \\ -1 & 2 \\ -1 & 0 \end{vmatrix}^\intercal \begin{pmatrix} 3 \\ 4 \\ 5 \end{pmatrix} \right) $$
$$ \begin{vmatrix} 3 & -3 \\ -3 & 5 \end{vmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -6 \\ 5 \end{pmatrix} $$
$$ \begin{pmatrix} x = -\frac{5}{2} \\ y = -\frac{1}{2} \end{pmatrix} $$
$\endgroup$ 2 $\begingroup$The normal equations are correct: $$ \mathbf{A}^{*}\mathbf{A} x = \mathbf{A}^{*}b. $$ The error is in the solution which should be $$ x = \left( \mathbf{A}^{*}\mathbf{A} \right)^{-1} \mathbf{A}^{*}b. $$ The inverse is $$ \left( \mathbf{A}^{*}\mathbf{A} \right)^{-1} = \frac{1} {6} \left( \begin{array}{cc} 5 & 3 \\ 3 & 3 \\ \end{array} \right) $$ which provides $$ x = -\frac{1}{2} \left( \begin{array}{r} 5 \\ 1 \end{array} \right). $$
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