4.16.E: Exercises for Section 4.11

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Page ID: 197446

Doli Bambhania, Fatemeh Yarahmadi, and Bill Wilson
De Anza College

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Exercise $\PageIndex{1}$

Find the least squares solution to the following system. \[\begin{aligned}x+2y&=1 \\ 2x+3y&=2 \\ 3x+5y&=4\end{aligned}\]

Answer: \[\begin{aligned}\left[\begin{array}{cc}1&2\\2&3\\3&5\end{array}\right]^T\left[\begin{array}{cc}1&2\\2&3\\3&5\end{array}\right]&=\left[\begin{array}{cc}14&23\\23&38\end{array}\right]\left[\begin{array}{cc}14&23\\23&38\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right] \\ &=\left[\begin{array}{cc}1&2\\2&3\\3&5\end{array}\right]^T\left[\begin{array}{c}1\\2\\4\end{array}\right]=\left[\begin{array}{c}17\\28\end{array}\right]\end{aligned}\] \[\begin{aligned}\left[\begin{array}{cc}14&23\\23&38\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right]&=\left[\begin{array}{c}17\\28\end{array}\right] \\ \left[\begin{array}{cc}14&23\\23&38\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right]&=\left[\begin{array}{c}17\\28\end{array}\right],\end{aligned}\] Solution is: $\left[\begin{array}{c}\frac{2}{3}\\ \frac{1}{3}\end{array}\right]$

Exercise $\PageIndex{2}$

You are doing experiments and have obtained the ordered pairs, \[(0, 1),(1, 2),(2, 3.5),(3, 4)\nonumber\] Find $m$ and $b$ such that $y = mx+b$ approximates these four points as well as possible.

Exercise $\PageIndex{3}$

Given the ordered pairs $(2, 1)$, $(1, 2)$, $(0, 1)$ $,$ $(-1, 2)$ $,$ and $(-2, 1)$, find $m$ and $b$ such that $y = mx+b$ approximates these five points as well as possible.

Answer: $y=1.4$

Exercise $\PageIndex{4}$

Suppose you have several ordered triples, $(x_i , y_i ,z_i)$. Describe how to find a polynomial such as \[z = a+bx+cy+dxy+ex^2 + fy^2\nonumber\] giving the best fit to the given ordered triples.

Exercise $\PageIndex{5}$

Find the least squares solution for the system \[\begin{aligned}x+y+z&=2 \\ 2x+3y+4z&=5 \\ 3x+5y+7z&=8\\ 4x+7y+10z&=11\end{aligned}\]

Exercise $\PageIndex{6}$

In general, you have a point $(x_0, y_0,z_0)$ and a scalar equation for a plane $ax+by+cz = d$ where $a^2 +b^2 +c^2 > 0$. Determine a formula for the closest point on the plane to the given point. Then use this point to get a formula for the distance from the given point to the plane.

Answer: Find the line perpendicular to the plane which goes through the given point: $(x,y,z) = (x_0, y_0,z_0) + t(a,b, c)$. Now require that this point satisfy the equation for the plane to determine $t$.

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