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I have a problem that says to find the least squares solution to

[tex]\newcommand{\colv}[2] {\left(\begin{array}{c} #1 \\ #2 \end{array}\right)}

K x = \colv{2}{2}[/tex] for

[tex] K = \left(

\begin{array}{cc}

1 & 2\\

2 & 4

\end{array} \right)[/tex]. Then express the solution in the form x = w + z, where w is in the corange, and z is in the kernel.

Since the determinate of the matrix K will be zero, then there should be infinite least squares solutions since the kernel is non-zero, correct? When I try to find a solution by multiplying each side by K^T, I get no solution rather than infinite solutions. Specifically the sytem

[tex] \newcommand{\colv}[2] {\left(\begin{array}{c} #1 \\ #2

\left(

\begin{array}{cc}

5 & 10\\

0 & 0

\end{array} \right) x = \colv{0}{10}[/tex]

How can I find the least square of a matrix with determinate zero? Everything I know is based off the fact that the matrix will be invertible.

*My last matrix is messed up somehow, it should be the augmented matrix {{5,10 | 5},{0,0 | 10}}.

[tex]\newcommand{\colv}[2] {\left(\begin{array}{c} #1 \\ #2 \end{array}\right)}

K x = \colv{2}{2}[/tex] for

[tex] K = \left(

\begin{array}{cc}

1 & 2\\

2 & 4

\end{array} \right)[/tex]. Then express the solution in the form x = w + z, where w is in the corange, and z is in the kernel.

Since the determinate of the matrix K will be zero, then there should be infinite least squares solutions since the kernel is non-zero, correct? When I try to find a solution by multiplying each side by K^T, I get no solution rather than infinite solutions. Specifically the sytem

[tex] \newcommand{\colv}[2] {\left(\begin{array}{c} #1 \\ #2

\left(

\begin{array}{cc}

5 & 10\\

0 & 0

\end{array} \right) x = \colv{0}{10}[/tex]

How can I find the least square of a matrix with determinate zero? Everything I know is based off the fact that the matrix will be invertible.

*My last matrix is messed up somehow, it should be the augmented matrix {{5,10 | 5},{0,0 | 10}}.

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