Prove that the vectors i, i+j, i+j+3k are independent in R3

Let

$A = \mathbf{i}, \qquad B = \mathbf{j}, \qquad C = \mathbf{i} + \mathbf{j} + 3 \mathbf{k}$

be vectors in $\mathbb{R}^3$ .

Prove that these three vectors are independent.
Express the unit coordinate vectors $\mathbf{j}$ and $\mathbf{k}$ as linear combinations of these three vectors.
Express the vector $2 \mathbf{i} - 3 \mathbf{j} + 5 \mathbf{k}$ as a linear combination of $A,B,C$ .
Prove that this set is a basis for $\mathbb{R}^3$ .

Proof. The equation $xA + yB + zC = O$ implies
$x+y+z = 0, \quad x +y = 0, \quad 3z = 0.$

But this implies $x = y = z = 0$ . Hence, $A,B,C$ are linearly independent $. \qquad \blacksquare$
First, we have
$\begin{align*} xA + yB + zC = \mathbf{j} && \implies && x+y+z &= 0 \\ &&&& y+z &= 1 \\ &&&& 3z &= 0. \end{align*}$

From these we have $z = 0$ , $y = 1$ and $x = -1$ . Thus, $\mathbf{j} = -A + B$ .
Next, we have

$\begin{align*} xA + yB + zC = \mathbf{k} && \implies && x+y+z &= 0 \\ &&&& y+z &= 0 \\ &&&& 3z &= 1. \end{align*}$

This implies $z = \frac{1}{3}$ , $y = -\frac{1}{3}$ and $x = 0$ . Therefore, $\mathbf{k} = -\frac{1}{3} B + \frac{1}{3} C$ .
We compute,
$\begin{align*} 2 \mathbf{i} - 3 \mathbf{j} + 5 \mathbf{k} = xA + yB + zC && \implies && x + y + z &= 2 \\ &&&& y+z &= -3 \\ &&&& 3z &= 5. \end{align*}$

Therefore, $z = \frac{5}{3}$ , $y = -\frac{14}{3}$ and $x = 5$ . Hence,

$2 \mathbf{i} - 3 \mathbf{j} + 5 \mathbf{k} = \frac{1}{3} (15A - 14B + 5C).$
Proof. By Theorem 12.10, any set of 3 linearly independent (part (a)) vectors in $\mathbb{R}^3$ is a basis for $\mathbb{R}^3. \qquad \blacksquare$