Standard +0.3 This is a structured multi-part eigenvalue question with clear guidance at each step. Verifying given eigenvectors requires only matrix multiplication (routine), finding remaining eigenvalues uses standard characteristic equation methods, and the final part about AB's eigenvector follows directly from the property that if e is an eigenvector of both A and B, it's also an eigenvector of AB with eigenvalue being the product. While it involves 3×3 matrices and multiple parts, each step is algorithmic with no novel insight required, making it slightly easier than average.
6 The matrix \(\mathbf { A }\) is given by
$$\mathbf { A } = \left( \begin{array} { l l l }
4 & - 5 & 3 \\
3 & - 4 & 3 \\
1 & - 1 & 2
\end{array} \right)$$
Show that \(\mathbf { e } = \left( \begin{array} { l } 1 \\ 1 \\ 1 \end{array} \right)\) is an eigenvector of \(\mathbf { A }\) and state the corresponding eigenvalue.
Find the other two eigenvalues of \(\mathbf { A }\).
The matrix \(\mathbf { B }\) is given by
$$\mathbf { B } = \left( \begin{array} { r r r }
- 1 & 4 & 0 \\
- 1 & 3 & 1 \\
1 & - 1 & 3
\end{array} \right)$$
Show that \(\mathbf { e }\) is an eigenvector of \(\mathbf { B }\) and deduce an eigenvector of the matrix \(\mathbf { A B }\), stating the corresponding eigenvalue.
\(\mathbf{Ae}=2\mathbf{e} \Rightarrow \mathbf{e}\) is an eigenvector with eigenvalue 2
M1A1
Shows e is eigenvector of A, gives eigenvalue
\(\lambda^3-2\lambda^2-\lambda+2=0\)
M1A1
Finds characteristic equation
\(\Rightarrow(\lambda-2)(\lambda^2-1)=0\)
A1
Factorises
Other eigenvalues are \(-1\) and \(1\)
A1
States other eigenvalues
\(\mathbf{Be}=3\mathbf{e} \Rightarrow \mathbf{e}\) is an eigenvector with eigenvalue 3
B1
Repeats for B
\(\mathbf{ABe}=\mathbf{A}\cdot3\mathbf{e}=3\mathbf{Ae}=3\cdot2\mathbf{e}=6\mathbf{e}\); AB has eigenvector e with eigenvalue 6
M1A1
States result for AB
Total: [9]
## Question 6:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{Ae}=2\mathbf{e} \Rightarrow \mathbf{e}$ is an eigenvector with eigenvalue 2 | M1A1 | Shows **e** is eigenvector of **A**, gives eigenvalue |
| $\lambda^3-2\lambda^2-\lambda+2=0$ | M1A1 | Finds characteristic equation |
| $\Rightarrow(\lambda-2)(\lambda^2-1)=0$ | A1 | Factorises |
| Other eigenvalues are $-1$ and $1$ | A1 | States other eigenvalues |
| $\mathbf{Be}=3\mathbf{e} \Rightarrow \mathbf{e}$ is an eigenvector with eigenvalue 3 | B1 | Repeats for **B** |
| $\mathbf{ABe}=\mathbf{A}\cdot3\mathbf{e}=3\mathbf{Ae}=3\cdot2\mathbf{e}=6\mathbf{e}$; **AB** has eigenvector **e** with eigenvalue 6 | M1A1 | States result for **AB** |
**Total: [9]**
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6 The matrix $\mathbf { A }$ is given by
$$\mathbf { A } = \left( \begin{array} { l l l }
4 & - 5 & 3 \\
3 & - 4 & 3 \\
1 & - 1 & 2
\end{array} \right)$$
Show that $\mathbf { e } = \left( \begin{array} { l } 1 \\ 1 \\ 1 \end{array} \right)$ is an eigenvector of $\mathbf { A }$ and state the corresponding eigenvalue.
Find the other two eigenvalues of $\mathbf { A }$.
The matrix $\mathbf { B }$ is given by
$$\mathbf { B } = \left( \begin{array} { r r r }
- 1 & 4 & 0 \\
- 1 & 3 & 1 \\
1 & - 1 & 3
\end{array} \right)$$
Show that $\mathbf { e }$ is an eigenvector of $\mathbf { B }$ and deduce an eigenvector of the matrix $\mathbf { A B }$, stating the corresponding eigenvalue.
\hfill \mbox{\textit{CAIE FP1 2013 Q6 [9]}}