CAIE FP1 2011 June — Question 8 11 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2011
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
Topic3x3 Matrices
TypeFind P and D for diagonalization / matrix powers
DifficultyChallenging +1.2 This is a standard Further Maths diagonalization question requiring eigenvalue calculation via characteristic equation (3×3 determinant), finding three eigenvectors, then constructing P and D. While mechanically lengthy with multiple steps, it follows a completely routine algorithm with no conceptual surprises or problem-solving insight needed—typical FP1 examination fare but harder than core A-level due to 3×3 matrices and the diagonalization framework.
Spec4.03l Singular/non-singular matrices4.03m det(AB) = det(A)*det(B)4.03n Inverse 2x2 matrix4.03o Inverse 3x3 matrix
8 Find the eigenvalues and corresponding eigenvectors of the matrix \(\mathbf { A } = \left( \begin{array} { r r r } 4 & - 1 & 1 \\ - 1 & 0 & - 3 \\ 1 & - 3 & 0 \end{array} \right)\). Find a non-singular matrix \(\mathbf { P }\) and a diagonal matrix \(\mathbf { D }\) such that \(\mathbf { A } ^ { 5 } = \mathbf { P D P } ^ { - 1 }\).
Question 8:
Characteristic equation:
AnswerMarks
\(\text{Det}(\mathbf{A} - \lambda\mathbf{I}) = 0 \Rightarrow \lambda^3 - 4\lambda^2 - 11\lambda + 30 = 0\)M1A1
\(\Rightarrow \lambda = -3, 2, 5\) (All three)A1A1
Eigenvectors:
AnswerMarks Guidance
\(\lambda = -3\): \(\mathbf{e}_1 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 7 & -1 & 1 \\ -1 & 3 & -3 \end{vmatrix} = \begin{pmatrix}0\\20\\20\end{pmatrix} = t\begin{pmatrix}0\\1\\1\end{pmatrix}\)M1A1
\(\lambda = 2\): \(\mathbf{e}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 1 \\ -1 & -2 & -3 \end{vmatrix} = \begin{pmatrix}5\\5\\-5\end{pmatrix} = t\begin{pmatrix}1\\1\\-1\end{pmatrix}\)A1 Part mark: 8
\(\lambda = 5\): \(\mathbf{e}_3 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & -1 & 1 \\ -1 & -5 & -3 \end{vmatrix} = \begin{pmatrix}8\\-4\\4\end{pmatrix} = t\begin{pmatrix}2\\-1\\1\end{pmatrix}\)A1
Matrix P and D:
AnswerMarks Guidance
\(\mathbf{P} = \begin{pmatrix}0 & 1 & 2\\1 & 1 & -1\\1 & -1 & 1\end{pmatrix}\), \(\mathbf{D} = \begin{pmatrix}-243 & 0 & 0\\0 & 32 & 0\\0 & 0 & 3125\end{pmatrix}\)B1\(\sqrt{}\), M1, A1\(\sqrt{}\) Columns of P and D can be permuted but must match; Part mark: 3
Total: [11]
## Question 8:

**Characteristic equation:**

$\text{Det}(\mathbf{A} - \lambda\mathbf{I}) = 0 \Rightarrow \lambda^3 - 4\lambda^2 - 11\lambda + 30 = 0$ | M1A1 |

$\Rightarrow \lambda = -3, 2, 5$ (All three) | A1A1 |

**Eigenvectors:**

$\lambda = -3$: $\mathbf{e}_1 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 7 & -1 & 1 \\ -1 & 3 & -3 \end{vmatrix} = \begin{pmatrix}0\\20\\20\end{pmatrix} = t\begin{pmatrix}0\\1\\1\end{pmatrix}$ | M1A1 |

$\lambda = 2$: $\mathbf{e}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 1 \\ -1 & -2 & -3 \end{vmatrix} = \begin{pmatrix}5\\5\\-5\end{pmatrix} = t\begin{pmatrix}1\\1\\-1\end{pmatrix}$ | A1 | Part mark: 8

$\lambda = 5$: $\mathbf{e}_3 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & -1 & 1 \\ -1 & -5 & -3 \end{vmatrix} = \begin{pmatrix}8\\-4\\4\end{pmatrix} = t\begin{pmatrix}2\\-1\\1\end{pmatrix}$ | A1 |

**Matrix P and D:**

$\mathbf{P} = \begin{pmatrix}0 & 1 & 2\\1 & 1 & -1\\1 & -1 & 1\end{pmatrix}$, $\mathbf{D} = \begin{pmatrix}-243 & 0 & 0\\0 & 32 & 0\\0 & 0 & 3125\end{pmatrix}$ | B1$\sqrt{}$, M1, A1$\sqrt{}$ | Columns of P and D can be permuted but must match; Part mark: 3

**Total: [11]**

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8 Find the eigenvalues and corresponding eigenvectors of the matrix $\mathbf { A } = \left( \begin{array} { r r r } 4 & - 1 & 1 \\ - 1 & 0 & - 3 \\ 1 & - 3 & 0 \end{array} \right)$.

Find a non-singular matrix $\mathbf { P }$ and a diagonal matrix $\mathbf { D }$ such that $\mathbf { A } ^ { 5 } = \mathbf { P D P } ^ { - 1 }$.

\hfill \mbox{\textit{CAIE FP1 2011 Q8 [11]}}
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