## Linear Algebra and Its Applications, exercise 1.2.9

Exercise 1.2.9. We can re-express the system of equations from 1.2.8 as follows: $u \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix} + v \begin{bmatrix}1 \\ 2 \\ 1\end{bmatrix} + w \begin{bmatrix}1 \\ 3 \\ 2\end{bmatrix} = b$

Demonstrate that the third column can be expressed as a linear combination of the first two columns, so that the three columns lie in the same plane. Then assume that b = (0, 0, 0) and categorize the solutions (u, v, w) for that special case.

Answer: We are looking for $c_1$ and $c_2$ such that $c_1 \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix} + c_2 \begin{bmatrix}1 \\ 2 \\ 1\end{bmatrix} = \begin{bmatrix}1 \\ 3 \\ 2\end{bmatrix}$

Note that the last entry in the first column is 0, so that $c_1 \cdot 0 + c_2 \cdot 1 = 2 \Rightarrow c_2 = 2$

From the first entry in each column we have $c_1 \cdot 1 + c_2 \cdot 1 = 1 \Rightarrow c_1 + 2 = 1 \Rightarrow c_1 = -1$

Finally, from the second entry in each column we have $c_1 \cdot 1 + c_2 \cdot 2 = -1 \cdot 1 + 2 \cdot 2 = -1 + 4 = 3$

So we have the third column as a linear combination of the first two columns: $-1 \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix} + 2 \begin{bmatrix}1 \\ 2 \\ 1\end{bmatrix} = \begin{bmatrix}1 \\ 3 \\ 2\end{bmatrix}$

Now assume b = (0, 0, 0) in the original system: $u \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix} + v \begin{bmatrix}1 \\ 2 \\ 1\end{bmatrix} + w \begin{bmatrix}1 \\ 3 \\ 2\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}$

From the equation above we have $-1 \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix} + 2 \begin{bmatrix}1 \\ 2 \\ 1\end{bmatrix} = \begin{bmatrix}1 \\ 3 \\ 2\end{bmatrix} \Rightarrow -1 \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix} + 2 \begin{bmatrix}1 \\ 2 \\ 1\end{bmatrix} - 1 \begin{bmatrix}1 \\ 3 \\ 2\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}$

So one solution for (u, v, w) is (-1, 2, -1). Another solution is (0, 0, 0). We can multiply the equation above by an arbitrary constant c and the right-hand side will remain zero, so the complete set of solutions is all vectors of the form c(-1, 2, -1), i.e., a line through the points (-1, 2, -1) and the origin.

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, Fourth Edition and the accompanying free online course, and Dr Strang’s other books .

This entry was posted in linear algebra. Bookmark the permalink.