## 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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