SPS SPS FM Pure 2024 June — Question 1 6 marks

Exam BoardSPS
ModuleSPS FM Pure (SPS FM Pure)
Year2024
SessionJune
Marks6
Topic3x3 Matrices
TypeMatrix equation solving (AB = C)
DifficultyStandard +0.3 This is a straightforward matrix equation problem requiring students to recognize that M must be a 1×2 matrix (from dimension analysis), set up equations by comparing entries after multiplication, and solve a simple system. While it involves 3D thinking about matrix dimensions and requires careful algebraic manipulation, it's a standard textbook exercise with a clear method and no novel insight required.
Spec4.03b Matrix operations: addition, multiplication, scalar4.03c Matrix multiplication: properties (associative, not commutative)

1. The matrix \(\mathbf { M }\) is such that \(\mathbf { M } \left( \begin{array} { r r r } 1 & 0 & k \\ 2 & - 1 & 1 \end{array} \right) = \left( \begin{array} { l l l } 1 & - 2 & 0 \end{array} \right)\).
Find
  • the matrix \(\mathbf { M }\),
  • the value of the constant \(k\).
    [0pt]

1.

The matrix $\mathbf { M }$ is such that $\mathbf { M } \left( \begin{array} { r r r } 1 & 0 & k \\ 2 & - 1 & 1 \end{array} \right) = \left( \begin{array} { l l l } 1 & - 2 & 0 \end{array} \right)$.\\
Find

\begin{itemize}
  \item the matrix $\mathbf { M }$,
  \item the value of the constant $k$.\\[0pt]

\end{itemize}

\hfill \mbox{\textit{SPS SPS FM Pure 2024 Q1 [6]}}