Standard +0.3 This is a straightforward Further Maths question requiring determinant calculation to show non-invertibility, then solving a consistent system with infinitely many solutions. While it involves 3×3 matrices (a Further Maths topic), the techniques are routine: compute det=0, then use row reduction or substitution to find the general solution with a parameter. Slightly above average difficulty due to the Further Maths context and multi-step nature, but no novel insight required.
2 Show that the matrix \(\left( \begin{array} { r r r } 1 & 4 & 2 \\ 3 & 0 & - 2 \\ 3 & - 3 & - 4 \end{array} \right)\) has no inverse.
Solve the system of equations
$$\begin{array} { r }
x + 4 y + 2 z = 0 \\
3 x - 2 z = 4 \\
3 x - 3 y - 4 z = 5
\end{array}$$
2 Show that the matrix $\left( \begin{array} { r r r } 1 & 4 & 2 \\ 3 & 0 & - 2 \\ 3 & - 3 & - 4 \end{array} \right)$ has no inverse.
Solve the system of equations
$$\begin{array} { r }
x + 4 y + 2 z = 0 \\
3 x - 2 z = 4 \\
3 x - 3 y - 4 z = 5
\end{array}$$
\hfill \mbox{\textit{CAIE FP1 2013 Q2 [6]}}