## 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.

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