Overdetermined Systems of Equations

The Solve command has been extended to handle overdetermined systems, returning the least-squares solution . Here we give an example of an overdetermined system. Note that (as before) the least-squares solution is the actual solution, when an actual solution exists.

$\blacktriangleright$ Solve + Exact

$\left[\vphantom{
\begin{array}{cc}
1 & 2 \\
3 & 4 \\
5 & 6 \\
7 & 8
\end{array}
}\right.$$\begin{array}{cc}
1 & 2 \\
3 & 4 \\
5 & 6 \\
7 & 8
\end{array}$$\left.\vphantom{
\begin{array}{cc}
1 & 2 \\
3 & 4 \\
5 & 6 \\
7 & 8
\end{array}
}\right]$$\left[\vphantom{
\begin{array}{c}
x \\
y
\end{array}
}\right.$$\begin{array}{c}
x \\
y
\end{array}$$\left.\vphantom{
\begin{array}{c}
x \\
y
\end{array}
}\right]$ = $\left[\vphantom{
\begin{array}{r}
3 \\
7 \\
11 \\
15
\end{array}
}\right.$$\begin{array}{r}
3 \\
7 \\
11 \\
15
\end{array}$$\left.\vphantom{
\begin{array}{r}
3 \\
7 \\
11 \\
15
\end{array}
}\right]$, Solution is : $\left[\vphantom{
\begin{array}{c}
1 \\
1
\end{array}
}\right.$$\begin{array}{c}
1 \\
1
\end{array}$$\left.\vphantom{
\begin{array}{c}
1 \\
1
\end{array}
}\right]$

It is easy to multiply both sides of a matrix equation by AT to check that, when you ``solve'' AX = B, you are actually getting the solution of $\left(\vphantom{ A^TA}\right.$ATA$\left.\vphantom{ A^TA}\right)$X = ATB.

$\left(\vphantom{ A^{T}A}\right.$ATA$\left.\vphantom{ A^{T}A}\right)$X = $\left(\vphantom{ \left[
\begin{array}{cccc}
1 & 3 & 5 & 7 \\
2 & 4 & 6 & ...
...}{cc}
1 & 2 \\
3 & 4 \\
5 & 6 \\
7 & 8
\end{array}
\right] }\right.$$\left[\vphantom{
\begin{array}{cccc}
1 & 3 & 5 & 7 \\
2 & 4 & 6 & 8
\end{array}
}\right.$$\begin{array}{cccc}
1 & 3 & 5 & 7 \\
2 & 4 & 6 & 8
\end{array}$$\left.\vphantom{
\begin{array}{cccc}
1 & 3 & 5 & 7 \\
2 & 4 & 6 & 8
\end{array}
}\right]$$\left[\vphantom{
\begin{array}{cc}
1 & 2 \\
3 & 4 \\
5 & 6 \\
7 & 8
\end{array}
}\right.$$\begin{array}{cc}
1 & 2 \\
3 & 4 \\
5 & 6 \\
7 & 8
\end{array}$$\left.\vphantom{
\begin{array}{cc}
1 & 2 \\
3 & 4 \\
5 & 6 \\
7 & 8
\end{array}
}\right]$$\left.\vphantom{ \left[
\begin{array}{cccc}
1 & 3 & 5 & 7 \\
2 & 4 & 6 & ...
...}{cc}
1 & 2 \\
3 & 4 \\
5 & 6 \\
7 & 8
\end{array}
\right] }\right)$$\left[\vphantom{
\begin{array}{c}
x \\
y
\end{array}
}\right.$$\begin{array}{c}
x \\
y
\end{array}$$\left.\vphantom{
\begin{array}{c}
x \\
y
\end{array}
}\right]$


                 = $\left(\vphantom{ \left[
\begin{array}{cc}
84 & 100 \\
100 & 120
\end{array}
\right] }\right.$$\left[\vphantom{
\begin{array}{cc}
84 & 100 \\
100 & 120
\end{array}
}\right.$$\begin{array}{cc}
84 & 100 \\
100 & 120
\end{array}$$\left.\vphantom{
\begin{array}{cc}
84 & 100 \\
100 & 120
\end{array}
}\right]$$\left.\vphantom{ \left[
\begin{array}{cc}
84 & 100 \\
100 & 120
\end{array}
\right] }\right)$$\left[\vphantom{
\begin{array}{c}
x \\
y
\end{array}
}\right.$$\begin{array}{c}
x \\
y
\end{array}$$\left.\vphantom{
\begin{array}{c}
x \\
y
\end{array}
}\right]$


ATB = $\left[\vphantom{
\begin{array}{cccc}
1 & 3 & 5 & 7 \\
2 & 4 & 6 & 8
\end{array}
}\right.$$\begin{array}{cccc}
1 & 3 & 5 & 7 \\
2 & 4 & 6 & 8
\end{array}$$\left.\vphantom{
\begin{array}{cccc}
1 & 3 & 5 & 7 \\
2 & 4 & 6 & 8
\end{array}
}\right]$$\left[\vphantom{
\begin{array}{c}
3 \\
7 \\
11 \\
15
\end{array}
}\right.$$\begin{array}{c}
3 \\
7 \\
11 \\
15
\end{array}$$\left.\vphantom{
\begin{array}{c}
3 \\
7 \\
11 \\
15
\end{array}
}\right]$ = $\left[\vphantom{
\begin{array}{c}
184 \\
220
\end{array}
}\right.$$\begin{array}{c}
184 \\
220
\end{array}$$\left.\vphantom{
\begin{array}{c}
184 \\
220
\end{array}
}\right]$

This calculation gives the following equation, which has an exact solution.

$\blacktriangleright$ Evaluate

$\left[\vphantom{
\begin{array}{cc}
84 & 100 \\
100 & 120
\end{array}
}\right.$$\begin{array}{cc}
84 & 100 \\
100 & 120
\end{array}$$\left.\vphantom{
\begin{array}{cc}
84 & 100 \\
100 & 120
\end{array}
}\right]$$\left[\vphantom{
\begin{array}{c}
1 \\
1
\end{array}
}\right.$$\begin{array}{c}
1 \\
1
\end{array}$$\left.\vphantom{
\begin{array}{c}
1 \\
1
\end{array}
}\right]$ = $\left[\vphantom{
\begin{array}{c}
184 \\
220
\end{array}
}\right.$$\begin{array}{c}
184 \\
220
\end{array}$$\left.\vphantom{
\begin{array}{c}
184 \\
220
\end{array}
}\right]$