Let \(A = \left( {\begin{aligned}{{}{}}{\bf{2}}&{\bf{1}}\\{ - {\bf{3}}}&{ - {\bf{4}}}\\{\bf{3}}&{\bf{2}}\end{aligned}} \right)\), \({\bf{b}} = \left( {\begin{aligned}{{}{}}{\bf{5}}\\{\bf{4}}\\{\bf{4}}\end{aligned}} \right)\), \({\bf{u}} = \left( {\begin{aligned}{{}{}}{\bf{4}}\\{ - {\bf{5}}}\end{aligned}} \right)\), and \({\bf{v}} = \left( {\begin{aligned}{{}{}}{\bf{6}}\\{ - {\bf{5}}}\end{aligned}} \right)\). Compute \(A{\bf{u}}\) and \(A{\bf{v}}\), and compare them with b. Could u possibly be a least-squares solution of \(A{\bf{x}} = {\bf{b}}\)?

(Answer this without computing a least-squares solution.)

The value of \({\bf{b}} - A{\bf{u}}\) can be calculated as follows:

\(\begin{aligned}{}{\bf{b}} - A{\bf{u}} &= \left( {\begin{aligned}{{}{}}5\\4\\4\end{aligned}} \right) - \left( {\begin{aligned}{{}{}}2&1\\{ - 3}&{ - 4}\\3&2\end{aligned}} \right)\left( {\begin{aligned}{{}{}}4\\{ - 5}\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}5\\4\\4\end{aligned}} \right) - \left( {\begin{aligned}{{}{}}3\\8\\2\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}2\\{ - 4}\\2\end{aligned}} \right)\end{aligned}\)

Find the value of \(\left\| {{\bf{b}} - A{\bf{u}}} \right\|\).

\(\begin{aligned}{}\left\| {{\bf{b}} - A{\bf{u}}} \right\| &= \sqrt {{{\left( 2 \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( 2 \right)}^2}} \\ &= \sqrt {4 + 16 + 4} \\ &= \sqrt {24} \\ &= 2\sqrt 6 \end{aligned}\)

The value of \({\bf{b}} - A{\bf{v}}\) can be calculated as follows:

\(\begin{aligned}{}{\bf{b}} - A{\bf{u}} &= \left( {\begin{aligned}{{}{}}5\\4\\4\end{aligned}} \right) - \left( {\begin{aligned}{{}{}}2&1\\{ - 3}&{ - 4}\\3&2\end{aligned}} \right)\left( {\begin{aligned}{{}{}}6\\{ - 5}\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}5\\4\\4\end{aligned}} \right) - \left( {\begin{aligned}{{}{}}7\\2\\8\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}{ - 2}\\2\\{ - 4}\end{aligned}} \right)\end{aligned}\)

Find the value of \(\left\| {{\bf{b}} - A{\bf{u}}} \right\|\).

\(\begin{aligned}{}\left\| {{\bf{b}} - A{\bf{v}}} \right\| & = \sqrt {{{\left( { - 2} \right)}^2} + {{\left( 2 \right)}^2} + {{\left( { - 4} \right)}^2}} \\ & = \sqrt {4 + 4 + 16} \\ & = \sqrt {24} \\ & = 2\sqrt 6 \end{aligned}\)

It can be observed that \(A{\bf{v}}\)and \(A{\bf{u}}\) are at an equal distance fromb.So, they are equally close to b.

Thus, none of them can be the least square solution of\(A{\bf{x}} = {\bf{b}}\).