Show that the following vectors form a basis for the vector space ${\mathbb{R}^3}$ . $$\left( \begin{array}{l} 1\\ 2\\ 3 \end{array} \right);\left( \begin{array}{l} 2\\ 3\\ 1 \end{array} \right);\left( \begin{array}{l} 5\\ 2\\ 5 \end{array} \right)$$

Express the following vector as a linear combination of the above vectors. $$\left( \begin{array}{l} 12\\ 14\\ 16 \end{array} \right)$$

let ${\boldsymbol{A}} = \left| {\begin{array}{*{20}{c}}   1&2&5 \\   2&3&2 \\   3&1&5 \end{array}} \right|$ and consider $\det (\boldsymbol{A}) = - 30$     (M1)A1

the vectors form a basis because the determinant is non-zero (or because the matrix is non-singular)     R1

[3 marks]

let $\left( \begin{array}{l} 12\\ 14\\ 16 \end{array} \right) = \lambda \left( \begin{array}{l} 1\\ 2\\ 3 \end{array} \right) + \mu \left( \begin{array}{l} 2\\ 3\\ 1 \end{array} \right) + v\left( \begin{array}{l} 5\\ 2\\ 5 \end{array} \right)$     M1A1

so that

EITHER

$\begin{array}{l} \lambda  + 2\mu  + 5v = 12\\ 2\lambda  + 3\mu  + 2v = 14\\ 3\lambda  + \mu  + 5v = 16 \end{array}$     M1

OR

$\left( {\begin{array}{*{20}{c}} 1&2&5\\ 2&3&2\\ 3&1&5 \end{array}} \right)\left( \begin{array}{l} \lambda \\ \mu \\ v \end{array} \right) = \left( \begin{array}{l} 12\\ 14\\ 16 \end{array} \right)$     M1

THEN

giving $\lambda = 3$, $\mu = 2$, $v = 1$     (A1)

hence $\left( \begin{array}{l} 12\\ 14\\ 16 \end{array} \right) = 3\left( \begin{array}{l} 1\\ 2\\ 3 \end{array} \right) + 2\left( \begin{array}{l} 2\\ 3\\ 1 \end{array} \right) + 1\left( \begin{array}{l} 5\\ 2\\ 5 \end{array} \right)$     A1

[5 marks]