Science:Math Exam Resources/Courses/MATH221/April 2013/Question 11/Solution 1

The least squares solution vector ${\hat {\mathbf {x} }}$ is a 2x1 vector. We can quickly find it using the normal equations.

${\begin{aligned}A^{T}A{\hat {\mathbf {x} }}&=A^{T}\mathbf {b} \\\left[{\begin{array}{cccc}1&1&1&1\\-1&0&1&2\end{array}}\right]\left[{\begin{array}{cc}1&-1\\1&0\\1&1\\1&2\end{array}}\right]{\hat {\mathbf {x} }}&=\left[{\begin{array}{cccc}1&1&1&1\\-1&0&1&2\end{array}}\right]\mathbf {b} \\\left[{\begin{array}{cc}4&2\\2&6\end{array}}\right]{\hat {\mathbf {x} }}&=\left[{\begin{array}{c}5\\6\end{array}}\right]\end{aligned}}$

To find ${\hat {\mathbf {x} }}$ from here we can either invert the matrix on the left hand side, or use gaussian elimination. We choose the latter.

${\begin{aligned}\left[{\begin{array}{cc|c}4&2&5\\2&6&6\end{array}}\right]\rightarrow \left[{\begin{array}{cc|c}1&3&3\\4&2&5\end{array}}\right]\rightarrow \left[{\begin{array}{cc|c}1&3&3\\0&-10&-7\end{array}}\right]\rightarrow \left[{\begin{array}{cc|c}1&0&9/10\\0&1&7/10\end{array}}\right]\end{aligned}}$

Hence

${\hat {\mathbf {x} }}=\left[{\begin{array}{c}9/10\\7/10\end{array}}\right]$

Finally, how does this translate to the linear function whose graph best fits the four given points? In other words, find m and t such that

$\displaystyle y=mx+t$

best fits the four given points (x₁,y₁), (x₂,y₂), (x₃,y₃), (x₄,y₄). To answer this we need to understand how the matrix A and the vector b were chosen. We notice that b holds the y values. Further, the x values of the points, are in the second column of A. The first column of A is ones and this is because the same constant t appears for each (x,y) pair. In matrix notation

$\left[{\begin{array}{cc}1&x_{1}\\1&x_{2}\\1&x_{3}\\1&x_{4}\end{array}}\right]\left[{\begin{array}{cc}{\hat {\mathbf {x} }}_{1}\\{\hat {\mathbf {x} }}_{2}\end{array}}\right]=\left[{\begin{array}{cc}y_{1}\\y_{2}\\y_{3}\\y_{4}\end{array}}\right]$

Reading $\mathbf {b} =A{\hat {\mathbf {x} }}$ line by line reveals

$y_{j}=1{\hat {\mathbf {x} }}_{1}+x_{j}{\hat {\mathbf {x} }}_{2}={\hat {\mathbf {x} }}_{2}x_{j}+{\hat {\mathbf {x} }}_{1}=7/10x_{j}+9/10$

for j = 1,2,3,4. Comparing to y = mx + t we obtain

$m=7/10,\quad t=9/10$

The graph below is just for illustration purposes. It is not necessary to include a graph in your answer.

Therefore, the linear function is given by

$y={\frac {7}{10}}x+{\frac {9}{10}}$