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]