题目
Problem
The matrix A \mathbf{A} A
A = ( 6 − 9 5 5 − 8 5 1 − 1 2 ) . \mathbf{A}=\begin{pmatrix}6&-9&5\\5&-8&5\\1&-1&2\end{pmatrix}. A = 6 5 1 − 9 − 8 − 1 5 5 2 . (a) Find the eigenvalues of A \mathbf{A} A
(4)
(b) Use the characteristic equation of A \mathbf{A} A A − 1 = p A 2 + q I \mathbf{A}^{-1}=p\mathbf{A}^2+q\mathbf{I} A − 1 = p A 2 + q I p p p q q q
(3)
题目中文翻译
矩阵 A \mathbf{A} A
A = ( 6 − 9 5 5 − 8 5 1 − 1 2 ) . \mathbf{A}=\begin{pmatrix}6&-9&5\\5&-8&5\\1&-1&2\end{pmatrix}. A = 6 5 1 − 9 − 8 − 1 5 5 2 . (a) 求 A \mathbf{A} A
(4)
(b) 利用 A \mathbf{A} A A − 1 = p A 2 + q I \mathbf{A}^{-1}=p\mathbf{A}^2+q\mathbf{I} A − 1 = p A 2 + q I p p p q q q
(3)
原卷通用作答要求
Answer all questions. You should use a calculator where appropriate.
You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。
解答
(a)
解法一
思路
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计算 λ I − A \lambda\mathbf I-\mathbf A λ I − A
答题过程
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The characteristic determinant is
det ( λ − 6 9 − 5 − 5 λ + 8 − 5 − 1 1 λ − 2 ) . \det\begin{pmatrix}\lambda-6&9&-5\\-5&\lambda+8&-5\\-1&1&\lambda-2\end{pmatrix}. det λ − 6 − 5 − 1 9 λ + 8 1 − 5 − 5 λ − 2 . Expanding along the first row gives
det ( λ I − A ) = ( λ − 6 ) [ ( λ + 8 ) ( λ − 2 ) + 5 ] − 9 [ − 5 ( λ − 2 ) − 5 ] − 5 [ − 5 + ( λ + 8 ) ] = ( λ − 6 ) ( λ 2 + 6 λ − 11 ) + 45 λ − 45 − 5 λ − 15 = λ 3 − 7 λ + 6 = ( λ − 1 ) ( λ − 2 ) ( λ + 3 ) . \begin{align*}
\det(\lambda\mathbf I-\mathbf A)
=&\,(\lambda-6)\big[(\lambda+8)(\lambda-2)+5\big]\\
&\,\hspace{2pt}-9[-5(\lambda-2)-5]-5[-5+(\lambda+8)]\\
=&\,(\lambda-6)(\lambda^2+6\lambda-11)\\
&\,\hspace{2pt}+45\lambda-45-5\lambda-15\\
=&\,\lambda^3-7\lambda+6\\
=&\,(\lambda-1)(\lambda-2)(\lambda+3).
\end{align*} det ( λ I − A ) = = = = ( λ − 6 ) [ ( λ + 8 ) ( λ − 2 ) + 5 ] − 9 [ − 5 ( λ − 2 ) − 5 ] − 5 [ − 5 + ( λ + 8 )] ( λ − 6 ) ( λ 2 + 6 λ − 11 ) + 45 λ − 45 − 5 λ − 15 λ 3 − 7 λ + 6 ( λ − 1 ) ( λ − 2 ) ( λ + 3 ) . The eigenvalues are 1 , 2 , − 3 \boxed{1,2,-3} 1 , 2 , − 3
(b)
解法一
思路
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由特征方程用 Cayley–Hamilton 定理得到矩阵恒等式。再乘 A − 1 \mathbf A^{-1} A − 1
答题过程
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By the Cayley–Hamilton theorem,
A 3 − 7 A + 6 I = 0. \mathbf A^3-7\mathbf A+6\mathbf I=\mathbf0. A 3 − 7 A + 6 I = 0 . The eigenvalues are non-zero, so A − 1 \mathbf A^{-1} A − 1 A 2 − 7 I + 6 A − 1 = 0 \mathbf A^2-7\mathbf I+6\mathbf A^{-1}=\mathbf0 A 2 − 7 I + 6 A − 1 = 0
6 A − 1 = − A 2 + 7 I , A − 1 = − 1 6 A 2 + 7 6 I . \begin{align*}
6\mathbf A^{-1}=&\,-\mathbf A^2+7\mathbf I,\\
\mathbf A^{-1}=&\,-\frac16\mathbf A^2+\frac76\mathbf I.
\end{align*} 6 A − 1 = A − 1 = − A 2 + 7 I , − 6 1 A 2 + 6 7 I . Thus ( p , q ) = ( − 1 / 6 , 7 / 6 ) (p,q)=(-1/6,7/6) ( p , q ) = ( − 1/6 , 7/6 )
