Linear Algebra and Its Applications, Exercise 3.1.10

Exercise 3.1.10. Given the two vectors $(1, 1, 2)$ and $(1, 2, 3)$ find a homogeneous system in three unknowns whose solutions are the linear combinations of the vectors.

Answer: In the previous exercise 3.1.9 we showed that the plane spanned by the vectors $(1, 1, 2)$ and $(1, 2, 3)$ was orthogonal to the line passing through the origin and $(-1, -1, 1)$ . This means that for any vector $x$ in that plane the inner product of $x$ and $(-1, -1, 1)$ is zero, or

$(-1) \cdot x_1 + (-1) \cdot x_2 + 1 \cdot x_3 = 0$

So we have $-x_1 -x_2 + x_3 = 0$ for all vectors $x$ in the plane spanned by $(1, 1, 2)$ and $(1, 2, 3)$ . But the plane spanned by $(1, 1, 2)$ and $(1, 2, 3)$ is simply the set of all vectors that are linear combinations of $(1, 1, 2)$ and $(1, 2, 3)$ . So we have found a homogeneous system in three unknowns whose solutions are the linear combinations of $(1, 1, 2)$ and $(1, 2, 3)$ .

(Note that we can multiply the system by -1 on both sides to obtain an equivalent system $x_1 + x_2 - x_3 = 0$ .)

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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