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CIE 9231 2025 June Paper 21 Q8

A Level / CIE / FP2

CIE 9231 2025 June Paper 21 Paper · Question 8

题目

Problem

(a) It is given that λ\lambda is an eigenvalue of the non-singular square matrix A\mathbf{A}, with corresponding eigenvector e\mathbf{e}.

Show that e\mathbf{e} is an eigenvector of A3\mathbf{A}^3 with corresponding eigenvalue λ3\lambda^3.

[2]

The matrix A\mathbf{A} is given by

A=(134010025).\mathbf{A} = \begin{pmatrix} -1 & 3 & 4\\ 0 & 1 & 0\\ 0 & -2 & 5 \end{pmatrix}.

(b) Show that the eigenvalues of A\mathbf{A} are 1-1, 11 and 55.

[2]

(c) Find a matrix P\mathbf{P} and a diagonal matrix D\mathbf{D} such that A2I=PDP1\mathbf{A} - 2\mathbf{I} = \mathbf{P}\mathbf{D}\mathbf{P}^{-1}.

[6]

(d) Use the characteristic equation of A\mathbf{A} to show that (A2I)3=aA2+bA+cI(\mathbf{A} - 2\mathbf{I})^3 = a\mathbf{A}^2 + b\mathbf{A} + c\mathbf{I} where aa, bb and cc are constants to be determined.

[3]
题目中文翻译

(a) 已知 λ\lambda 是非奇异方阵 A\mathbf{A} 的一个特征值,对应特征向量为 e\mathbf{e}

证明 e\mathbf{e}A3\mathbf{A}^3 的特征向量,对应特征值为 λ3\lambda^3

矩阵 A\mathbf{A}

A=(134010025).\mathbf{A} = \begin{pmatrix} -1 & 3 & 4\\ 0 & 1 & 0\\ 0 & -2 & 5 \end{pmatrix}.

(b) 证明 A\mathbf{A} 的特征值为 1-11155

(c) 求矩阵 P\mathbf{P} 和对角矩阵 D\mathbf{D},使得 A2I=PDP1\mathbf{A} - 2\mathbf{I} = \mathbf{P}\mathbf{D}\mathbf{P}^{-1}

(d) 使用 A\mathbf{A} 的特征方程,证明 (A2I)3=aA2+bA+cI(\mathbf{A} - 2\mathbf{I})^3 = a\mathbf{A}^2 + b\mathbf{A} + c\mathbf{I},其中 aabbcc 是需要确定的常数。

解答