## Linear Algebra and Its Applications, Review Exercise 2.14

Review exercise 2.14. Do the three vectors $(1, 1, 3)$, $(2, 3, 6)$, and $(1, 4, 3)$ form a basis for the vector space $\mathbb{R}^3$?

Answer: In order for the three vectors to be a basis for $\mathbb{R}^3$ they must be linearly independent. We can test this by doing elimination on the following matrix with the three vectors as the columns: $\begin{bmatrix} 1&2&1 \\ 1&3&4 \\ 3&6&3 \end{bmatrix}$

Subtracting 1 times the first row from the second row we have $\begin{bmatrix} 1&2&1 \\ 1&3&4 \\ 3&6&3 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&2&1 \\ 0&1&3 \\ 3&6&3 \end{bmatrix}$

and subtracting 3 times the first row from the third row we have $\begin{bmatrix} 1&2&1 \\ 0&1&3 \\ 3&6&3 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&2&1 \\ 0&1&3 \\ 0&0&0 \end{bmatrix}$

Since there are only two pivots, the rank of the vector’s column space is $r = 2$. This means that the three columns are not linearly independent and thus the original three vectors cannot be a basis for $\mathbb{R}^3$.

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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