linear algebra

(a)

$\left( \begin{matrix} 1 & 2 & 3 \\ 1 & -2 & 1 \\ 2 & 1 & -1 \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) = \left( \begin{matrix} 14 \\ 0 \\ 1 \end{matrix} \right)$

(i)

In Gaussian elimination, reduce the matrix to upper triangular form, then back-substitute.

Eliminate:

$\begin{aligned} R_2 &\to R_2 - R_1 \\ R_3 &\to R_3 - 2R_1\end{aligned}$

$\left( \begin{matrix} 1 & 2 & 3 \\ 0& -4 & -2 \\ 0 & -3 & -7 \end{matrix} \right| \left. \begin{matrix} 14 \\ -14 \\ -27 \end{matrix} \right)$

$R_3 \to R_3 - (-3)/(-4)R_2$

$\left( \begin{matrix} 1 & 2 & 3 \\ 0& -4 & -2 \\ 0 & 0 & -11/2 \end{matrix} \right| \left. \begin{matrix} 14 \\ -14 \\ -33/2 \end{matrix} \right)$

Back-substitute:

$R_3 \Rightarrow z = 3$

$R_2 \Rightarrow \begin{aligned} -4y -2(3) &= -14 \\ y &= 2 \end{aligned}$

$R_1 \Rightarrow \begin{aligned} x +2(2) +3(3) &= 14 \\ x &= 1 \end{aligned}$

The solution is

$\left( \begin{matrix} x \\ y \\ z \end{matrix} \right) = \left( \begin{matrix} 1 \\ 2 \\ 3 \end{matrix} \right)$

NST First Year Maths

Some unofficial solutions to Cambridge NST IA maths papers

2013 Paper 1 Question 11