Edexcel CP1 2021 June — Question 4 9 marks

Exam BoardEdexcel
ModuleCP1 (Core Pure 1)
Year2021
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
Topic3x3 Matrices
TypeMatrix equation solving (AB = C)
DifficultyStandard +0.3 This is a straightforward Core Pure 1 matrices question testing basic concepts: matrix multiplication/addition compatibility (routine recall), finding constants from matrix equation (simple arithmetic), deducing inverse from AB=λI (direct application of definition), and finding when determinant equals zero (standard calculation with trigonometric simplification). All parts are textbook exercises requiring no novel insight, making it slightly easier than average.
Spec4.03a Matrix language: terminology and notation4.03b Matrix operations: addition, multiplication, scalar4.03l Singular/non-singular matrices4.03o Inverse 3x3 matrix

  1. \(\mathbf{A}\) is a 2 by 2 matrix and \(\mathbf{B}\) is a 2 by 3 matrix. Giving a reason for your answer, explain whether it is possible to evaluate
    1. \(\mathbf{AB}\)
    2. \(\mathbf{A} + \mathbf{B}\)
    [2]
  2. Given that $$\begin{pmatrix} -5 & 3 & 1 \\ a & 0 & 0 \\ b & a & b \end{pmatrix}\begin{pmatrix} 0 & 5 & 0 \\ 2 & 12 & -1 \\ -1 & -11 & 3 \end{pmatrix} = \lambda\mathbf{I}$$ where \(a\), \(b\) and \(\lambda\) are constants,
    1. determine
    2. Hence deduce the inverse of the matrix \(\begin{pmatrix} -5 & 3 & 1 \\ a & 0 & 0 \\ b & a & b \end{pmatrix}\)
    [3]
  3. Given that $$\mathbf{M} = \begin{pmatrix} 1 & 1 & 1 \\ 0 & \sin\theta & \cos\theta \\ 0 & \cos 2\theta & \sin 2\theta \end{pmatrix} \quad \text{where } 0 \leq \theta < \pi$$ determine the values of \(\theta\) for which the matrix \(\mathbf{M}\) is singular. [4]

\begin{enumerate}[label=(\roman*)]
\item $\mathbf{A}$ is a 2 by 2 matrix and $\mathbf{B}$ is a 2 by 3 matrix.

Giving a reason for your answer, explain whether it is possible to evaluate
\begin{enumerate}[label=(\alph*)]
\item $\mathbf{AB}$
\item $\mathbf{A} + \mathbf{B}$
\end{enumerate}
[2]

\item Given that

$$\begin{pmatrix} -5 & 3 & 1 \\ a & 0 & 0 \\ b & a & b \end{pmatrix}\begin{pmatrix} 0 & 5 & 0 \\ 2 & 12 & -1 \\ -1 & -11 & 3 \end{pmatrix} = \lambda\mathbf{I}$$

where $a$, $b$ and $\lambda$ are constants,
\begin{enumerate}[label=(\alph*)]
\item determine
\begin{itemize}
\item the value of $\lambda$
\item the value of $a$
\item the value of $b$
\end{itemize}

\item Hence deduce the inverse of the matrix $\begin{pmatrix} -5 & 3 & 1 \\ a & 0 & 0 \\ b & a & b \end{pmatrix}$
\end{enumerate}
[3]

\item Given that

$$\mathbf{M} = \begin{pmatrix} 1 & 1 & 1 \\ 0 & \sin\theta & \cos\theta \\ 0 & \cos 2\theta & \sin 2\theta \end{pmatrix} \quad \text{where } 0 \leq \theta < \pi$$

determine the values of $\theta$ for which the matrix $\mathbf{M}$ is singular.
[4]
\end{enumerate}

\hfill \mbox{\textit{Edexcel CP1 2021 Q4 [9]}}