Question

Vectors $\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right],\mathit{}\left[\begin{array}{c}2\\ 1\\ 3\end{array}\right],\mathit{}\left[\begin{array}{c}3\\ 2\\ 1\end{array}\right]$ form a basis of$R^3$ .

Short answer

The above statement is true.

Vectors$\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right],\mathit{}\left[\begin{array}{c}2\\ 1\\ 3\end{array}\right],\mathit{}\left[\begin{array}{c}3\\ 2\\ 1\end{array}\right]$ form a basis of$R^3$.

Step-by-step solution

Definition of a basis

Any subset S of a vector space V(K) is called a basis of V(K), if -

1. S is linearly independent.
2. S generates V

i.e. L(S) = V

To check whether the given vectors form a basis of R3

Let$\mathit{S}=\left\{\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right],\mathit{}\left[\begin{array}{c}2\\ 1\\ 3\end{array}\right],\mathit{}\left[\begin{array}{c}3\\ 2\\ 1\end{array}\right]\right\}$

Again let, $a_1,a_2,a_3\in R^3$be such that

${\mathit{a}}_1\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]+{\mathit{a}}_2\left[\begin{array}{c}2\\ 1\\ 3\end{array}\right]+{\mathit{a}}_3\left[\begin{array}{c}3\\ 2\\ 1\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

$$\begin{gathered} \Rightarrow a_1+2a_2+3a_3=0 \\ {a}_2+2a_3=0 \\ {a}_3=0 \end{gathered}$$

$\Rightarrow a_1=0,a_2=0,a_3=0$

Hence, the vectors are L.I. and the only solution to these equations is

$a_1=0,a_2=0,a_3=0$

Final Answer

For a set $S=\left\{\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}2\\ 1\\ 3\end{array}\right],\left[\begin{array}{c}3\\ 2\\ 1\end{array}\right]\right\},$we can clearly see that the vectors of S are L.I. and

$L\left(S\right)=R^3.$

Therefore,

The vectors $\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}2\\ 1\\ 3\end{array}\right],\left[\begin{array}{c}3\\ 2\\ 1\end{array}\right]$form a basis of $R^3$.