Let \(A = \left( {\begin{aligned}{{}{}}{\bf{3}}&{\bf{4}}\\{ - {\bf{2}}}&{\bf{1}}\\{\bf{3}}&{\bf{4}}\end{aligned}} \right)\), \({\bf{b}} = \left( {\begin{aligned}{{}{}}{{\bf{11}}}\\{ - {\bf{9}}}\\{\bf{5}}\end{aligned}} \right)\), \({\bf{u}} = \left( {\begin{aligned}{{}{}}{\bf{5}}\\{ - {\bf{1}}}\end{aligned}} \right)\), and \({\bf{v}} = \left( {\begin{aligned}{{}{}}{\bf{5}}\\{ - {\bf{2}}}\end{aligned}} \right)\). Compute \(A{\bf{u}}\) and \(A{\bf{v}}\), and compare them with b. Could u possibly be a least-squares solution of \(A{\bf{x}} = {\bf{b}}\)?

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

\(\begin{aligned}{}{\bf{b}} - A{\bf{u}} &= \left[ {\begin{aligned}{{}{}}{11}\\{ - 9}\\5\end{aligned}} \right] - \left[ {\begin{aligned}{{}{}}3&4\\{ - 2}&1\\3&4\end{aligned}} \right]\left[ {\begin{aligned}{{}{}}5\\{ - 1}\end{aligned}} \right]\\ & = \left[ {\begin{aligned}{{}{}}{11}\\{ - 9}\\5\end{aligned}} \right] - \left[ {\begin{aligned}{{}{}}{11}\\{ - 11}\\{11}\end{aligned}} \right]\\ & = \left[ {\begin{aligned}{{}{}}0\\2\\{ - 6}\end{aligned}} \right]\end{aligned}\)

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

\(\begin{aligned}{}\left\| {{\bf{b}} - A{\bf{u}}} \right\| & = \sqrt {{{\left( 0 \right)}^2} + {{\left( 2 \right)}^2} + {{\left( { - 6} \right)}^2}} \\ & = \sqrt {0 + 4 + 36} \\ & = \sqrt {40} \\ & = 2\sqrt {10} \end{aligned}\)

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

\(\begin{aligned}{}{\bf{b}} - A{\bf{v}} & = \left( {\begin{aligned}{{}{}}{11}\\{ - 9}\\5\end{aligned}} \right) - \left( {\begin{aligned}{{}{}}3&4\\{ - 2}&1\\3&4\end{aligned}} \right)\left( {\begin{aligned}{{}{}}5\\{ - 2}\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}{11}\\{ - 9}\\5\end{aligned}} \right) - \left( {\begin{aligned}{{}{}}7\\{ - 12}\\7\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}4\\3\\{ - 2}\end{aligned}} \right)\end{aligned}\)

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

\(\begin{aligned}{}\left\| {{\bf{b}} - A{\bf{v}}} \right\| & = \sqrt {{{\left( 4 \right)}^2} + {{\left( 3 \right)}^2} + {{\left( { - 2} \right)}^2}} \\ & = \sqrt {16 + 9 + 4} \\ & = \sqrt {29} \end{aligned}\)

It can be observed that \(A{\bf{v}}\) is closer to bas compared to \(A{\bf{u}}\).

Thus, u cannot be a least-squares solution for the equation \(A{\bf{x}} = {\bf{b}}\).