Challenging +1.3 This FP3 question tests understanding of vector products and matrix algebra through conceptual proofs rather than computation. Part (a)(i) requires explaining a geometric property (perpendicularity), (a)(ii) involves manipulating the cross product to show collinearity, (b)(i) is a standard matrix cancellation proof, and (b)(ii) requires finding a specific matrix satisfying a constraint. While these are Further Maths topics (inherently harder), the individual parts are relatively straightforward applications of definitions and properties, requiring modest insight rather than extended problem-solving. The 9 marks suggest moderate scope but not exceptional difficulty.
Explain why, for any two vectors \(\mathbf{a}\) and \(\mathbf{b}\), \(\mathbf{a} \cdot \mathbf{b} \times \mathbf{a} = 0\). [2]
Given vectors \(\mathbf{a}\), \(\mathbf{b}\) and \(\mathbf{c}\) such that \(\mathbf{a} \times \mathbf{b} = \mathbf{a} \times \mathbf{c}\), where \(\mathbf{a} \neq \mathbf{0}\) and \(\mathbf{b} \neq \mathbf{c}\), show that
$$\mathbf{b} - \mathbf{c} = \lambda\mathbf{a}, \quad \text{where } \lambda \text{ is a scalar.}$$ [2]
\(\mathbf{A}\), \(\mathbf{B}\) and \(\mathbf{C}\) are \(2 \times 2\) matrices.
Given that \(\mathbf{A}\mathbf{B} = \mathbf{A}\mathbf{C}\), and that \(\mathbf{A}\) is not singular, prove that \(\mathbf{B} = \mathbf{C}\). [2]
Given that \(\mathbf{A}\mathbf{B} = \mathbf{A}\mathbf{C}\), where \(\mathbf{A} = \begin{pmatrix} 3 & 6 \\ 1 & 2 \end{pmatrix}\) and \(\mathbf{B} = \begin{pmatrix} 1 & 5 \\ 0 & 1 \end{pmatrix}\), find a matrix \(\mathbf{C}\) whose elements are all non-zero. [3]
\begin{enumerate}[label=(\alph*)]
\item
\begin{enumerate}[label=(\roman*)]
\item Explain why, for any two vectors $\mathbf{a}$ and $\mathbf{b}$, $\mathbf{a} \cdot \mathbf{b} \times \mathbf{a} = 0$. [2]
\item Given vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ such that $\mathbf{a} \times \mathbf{b} = \mathbf{a} \times \mathbf{c}$, where $\mathbf{a} \neq \mathbf{0}$ and $\mathbf{b} \neq \mathbf{c}$, show that
$$\mathbf{b} - \mathbf{c} = \lambda\mathbf{a}, \quad \text{where } \lambda \text{ is a scalar.}$$ [2]
\end{enumerate}
\item $\mathbf{A}$, $\mathbf{B}$ and $\mathbf{C}$ are $2 \times 2$ matrices.
\begin{enumerate}[label=(\roman*)]
\item Given that $\mathbf{A}\mathbf{B} = \mathbf{A}\mathbf{C}$, and that $\mathbf{A}$ is not singular, prove that $\mathbf{B} = \mathbf{C}$. [2]
\item Given that $\mathbf{A}\mathbf{B} = \mathbf{A}\mathbf{C}$, where $\mathbf{A} = \begin{pmatrix} 3 & 6 \\ 1 & 2 \end{pmatrix}$ and $\mathbf{B} = \begin{pmatrix} 1 & 5 \\ 0 & 1 \end{pmatrix}$, find a matrix $\mathbf{C}$ whose elements are all non-zero. [3]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 Q35 [9]}}