## Linear Algebra and Its Applications, Exercise 3.3.24

Exercise 3.3.24. Given the measurements $y = 2, 0, -3, -5$ at $t= -1, 0, 1, 2$ use least squares to find the line of best fit.

Answer: This corresponds to a system of the form $Ax = b$ as follows: $\begin{bmatrix} 1&-1 \\ 1&0 \\ 1&1 \\ 1&2 \end{bmatrix} \begin{bmatrix} C \\ D \end{bmatrix} = \begin{bmatrix} 2 \\ 0 \\ -3 \\ -5 \end{bmatrix}$

To find the least squares solution we multiply both sides by $A^T$ to create a system of the form $A^TA\bar{x} = A^Tb$. We have $A^TA = \begin{bmatrix} 1&1&1&1 \\ -1&0&1&2 \end{bmatrix} \begin{bmatrix} 1&-1 \\ 1&0 \\ 1&1 \\ 1&2 \end{bmatrix} = \begin{bmatrix} 4&2 \\ 2&6 \end{bmatrix}$

and $A^Tb = \begin{bmatrix} 1&1&1&1 \\ -1&0&1&2 \end{bmatrix} \begin{bmatrix} 2 \\ 0 \\ -3 \\ -5 \end{bmatrix} = \begin{bmatrix} -6 \\ -15 \end{bmatrix}$

so that the new system is $\begin{bmatrix} 4&2 \\ 2&6 \end{bmatrix} \begin{bmatrix} \bar{C} \\ \bar{D} \end{bmatrix} = \begin{bmatrix} -6 \\ -15 \end{bmatrix}$

We can multiply the first equation by $\frac{1}{2}$ and subtract it from the second equation to form the system $\begin{bmatrix} 4&2 \\ 0&5 \end{bmatrix} \begin{bmatrix} \bar{C} \\ \bar{D} \end{bmatrix} = \begin{bmatrix} -6 \\ -12 \end{bmatrix}$

From the second equation we have $\bar{D} = -\frac{12}{5}$ and can substitute into the first equation to get $4\bar{C} - \frac{24}{5} = -6$ or $\bar{C} = \frac{1}{4} (-6 + \frac{24}{5}) = \frac{1}{4} (-\frac{6}{5}) = -\frac{3}{10}$

The line of best fit is thus $-\frac{3}{10} - \frac{12}{5}t$.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition , Dr Strang’s introductory textbook Introduction to Linear Algebra, Fifth Edition and the accompanying free online course, and Dr Strang’s other books .

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