Question

If \(a,b,\) and \(c\) are distinct numbers, then the following system is inconsistent because the graphs of the equations are parallel planes. Show that the set of all least-squares solutions of the system is precisely the plane whose equation is \(x - 2y + 5z = \frac{{\left( {a + b + c} \right)}}{3}\).

\(\begin{array}{c}x - 2y + 5z = a\\x - 2y + 5z = b\\x - 2y + 5z = c\end{array}\)

Short answer

It is proved that the set of all least-square of the system is precisely the plane whose equation is \(x - 2y + 5z = \frac{{a + b + c}}{3}\).

Step-by-step solution

Least-Square Solution

When \(A\) is an \(m \times n\) matrix and \({\bf{b}}\) in \({\mathbb{R}^m}\), then \(\widehat {\bf{x}}\) in \({\mathbb{R}^n}\) is aleast-squares solutionof \(A{\bf{x}} = {\bf{b}}\) such that;

\(\left\| {{\bf{b}} - A\widehat {\bf{x}}} \right\| \le \left\| {{\bf{b}} - A{\bf{x}}} \right\|\) for every \({\bf{x}}\) in \({\mathbb{R}^n}\).

Show that the set of least-squares solutions of the system is precisely thee planes whose equation  \(x - 2y + 5z = \frac{{\left( {a + b + c} \right)}}{3}\)

Consider that \({\bf{x}} = \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right],{\bf{b}} = \left[ {\begin{array}{*{20}{c}}a\\b\\c\end{array}} \right],{\bf{v}} = \left[ {\begin{array}{*{20}{c}}1\\{ - 2}\\5\end{array}} \right]\), and \(A = \left[ {\begin{array}{*{20}{c}}{{{\bf{v}}^T}}\\{{{\bf{v}}^T}}\\{{{\bf{v}}^T}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{ - 2}&5\\1&{ - 2}&5\\1&{ - 2}&5\end{array}} \right]\).

The given set of equations is of the form \(A{\bf{x}} = {\bf{b}}\). The set of all least-square solutions corresponds to the set of solutions of the normal equations.

Obtain the column-row expansion of \({A^T}A\) and \({A^T}{\bf{b}}\) as shown below:

\(\begin{array}{c}{A^T}A = {\bf{v}}{{\bf{v}}^T} + {\bf{v}}{{\bf{v}}^T} + {\bf{v}}{{\bf{v}}^T}\\ = 3{\bf{v}}{{\bf{v}}^T}\\{A^T}{\bf{b}} = a{\bf{v}} + b{\bf{v}} + c{\bf{v}}\\ = {\bf{v}}\left( {a + b + c} \right)\end{array}\)

Therefore, \({A^T}A{\bf{x}} = 3\left( {{\bf{v}}{{\bf{v}}^T}} \right){\bf{x}} = 3{\bf{v}}\left( {{{\bf{v}}^T}{\bf{x}}} \right) = 3\left( {{{\bf{v}}^T}{\bf{x}}} \right){\bf{v}}\) because \({{\bf{v}}^T}{\bf{x}}\) is scalar, and now, the normal equations are \(3\left( {{{\bf{v}}^T}{\bf{x}}} \right){\bf{v}} = \left( {a + b + c} \right){\bf{v}}\).

Therefore, \(3\left( {{{\bf{v}}^T}{\bf{x}}} \right) = \left( {a + b + c} \right)\), or \({{\bf{v}}^T}{\bf{x}} = \frac{{\left( {a + b + c} \right)}}{3}\).

When computing \({{\bf{v}}^T}{\bf{b}}\), it produces the equation \(x - 2y + 5z = \frac{{a + b + c}}{3}\), that should be satisfied by all least-squares solutions to \(A{\bf{x}} = {\bf{b}}\).

Thus, it is proved that the set of all least-square of the system is precisely the plane whose equation is \(x - 2y + 5z = \frac{{a + b + c}}{3}\).