Standard +0.8 This is a standard FP1 matrix determinant question requiring students to form a 3×3 coefficient matrix, calculate its determinant (involving parameter λ), set it equal to zero, and solve the resulting equation. While it involves multiple steps and algebraic manipulation, it's a textbook application of a core FP1 technique with no novel insight required—moderately above average difficulty due to the algebraic complexity but well within expected FP1 scope.
By using the determinant of an appropriate matrix, find the values of \(\lambda\) for which the simultaneous equations
\begin{align}
3x + 2y + 4z &= 5,
\lambda y + z &= 1,
x + \lambda y + \lambda z &= 4,
\end{align}
do not have a unique solution for \(x\), \(y\) and \(z\). [6]
Answer: \(3r^2-7r+2\) and \(\frac{1}{3}\) or \(2\)
M1, M1, A1, B1*, DM1, A1 [6]
Show correct expansion process for correct 3 x 3; Correct evaluation of any 2 x 2; Obtain correct 3 term quadratic; Equate their det to 0; Attempt to solve a quadratic equation; Obtain correct answers
Answer: $3r^2-7r+2$ and $\frac{1}{3}$ or $2$ | M1, M1, A1, B1*, DM1, A1 [6] | Show correct expansion process for correct 3 x 3; Correct evaluation of any 2 x 2; Obtain correct 3 term quadratic; Equate their det to 0; Attempt to solve a quadratic equation; Obtain correct answers
By using the determinant of an appropriate matrix, find the values of $\lambda$ for which the simultaneous equations
\begin{align}
3x + 2y + 4z &= 5, \\
\lambda y + z &= 1, \\
x + \lambda y + \lambda z &= 4,
\end{align}
do not have a unique solution for $x$, $y$ and $z$. [6]
\hfill \mbox{\textit{OCR FP1 2013 Q5 [6]}}