题目
Problem
(a) Find the values of a a a
3 2 x + 3 y + 8 z = 1 , a x + 3 y + 4 z = 2 , a y − z = 3 , \begin{align*}
\frac{3}{2}x+3y+8z=&\,1,\\
ax+3y+4z=&\,2,\\
ay-z=&\,3,
\end{align*} 2 3 x + 3 y + 8 z = a x + 3 y + 4 z = a y − z = 1 , 2 , 3 , does not have a unique solution.
(3)
The matrix A \mathbf{A} A
A = ( 3 2 3 8 0 3 4 0 0 − 1 ) . \mathbf{A}=\begin{pmatrix}\frac{3}{2}&3&8\\0&3&4\\0&0&-1\end{pmatrix}. A = 2 3 0 0 3 3 0 8 4 − 1 . (b) Given that B = A − 1 \mathbf{B}=\mathbf{A}^{-1} B = A − 1 A \mathbf{A} A B 2 = p I + q A \mathbf{B}^2=p\mathbf{I}+q\mathbf{A} B 2 = p I + q A p p p q q q
(4)
(c) Find a matrix P \mathbf{P} P D \mathbf{D} D A − 1 = P D P − 1 \mathbf{A}^{-1}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1} A − 1 = PD P − 1
(7)
题目中文翻译
(a) 求使方程组
3 2 x + 3 y + 8 z = 1 , a x + 3 y + 4 z = 2 , a y − z = 3 \begin{align*}
\frac{3}{2}x+3y+8z=&\,1,\\
ax+3y+4z=&\,2,\\
ay-z=&\,3
\end{align*} 2 3 x + 3 y + 8 z = a x + 3 y + 4 z = a y − z = 1 , 2 , 3 没有唯一解的 a a a
(3)
矩阵 A \mathbf{A} A
A = ( 3 2 3 8 0 3 4 0 0 − 1 ) . \mathbf{A}=\begin{pmatrix}\frac{3}{2}&3&8\\0&3&4\\0&0&-1\end{pmatrix}. A = 2 3 0 0 3 3 0 8 4 − 1 . (b) 已知 B = A − 1 \mathbf{B}=\mathbf{A}^{-1} B = A − 1 A \mathbf{A} A B 2 = p I + q A \mathbf{B}^2=p\mathbf{I}+q\mathbf{A} B 2 = p I + q A p p p q q q
(4)
(c) 求矩阵 P \mathbf{P} P D \mathbf{D} D A − 1 = P D P − 1 \mathbf{A}^{-1}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1} A − 1 = PD P − 1
(7)
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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没有唯一解时，系数行列式为零。沿第一行展开后解关于 a a a
答题过程
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The coefficient determinant is
det ( 3 2 3 8 a 3 4 0 a − 1 ) = 3 2 ( − 3 − 4 a ) + 3 a + 8 a 2 = 8 a 2 − 3 a − 9 2 . \begin{align*}
&\,\det\begin{pmatrix}
\frac32&3&8\\
a&3&4\\
0&a&-1
\end{pmatrix}\\[4mm]
=&\,\frac32(-3-4a)+3a+8a^2\\[4mm]
=&\,8a^2-3a-\frac92.
\end{align*} = = det 2 3 a 0 3 3 a 8 4 − 1 2 3 ( − 3 − 4 a ) + 3 a + 8 a 2 8 a 2 − 3 a − 2 9 . The system does not have a unique solution exactly when this determinant is zero. Solving gives
a = 3 ( 1 ± 17 ) 16 . \boxed{a=\frac{3(1\pm\sqrt{17})}{16}}. a = 16 3 ( 1 ± 17 ) .
(b)
解法一
思路
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用对角线上的特征值写出特征多项式。由 Cayley–Hamilton 定理得到矩阵恒等式，再乘 A − 2 \mathbf A^{-2} A − 2
答题过程
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The characteristic polynomial of A \mathbf A A
( λ − 3 2 ) ( λ − 3 ) ( λ + 1 ) = λ 3 − 7 2 λ 2 + 9 2 . \begin{align*}
&\,(\lambda-\tfrac32)(\lambda-3)(\lambda+1)\\[4mm]
=&\,\lambda^3-\frac72\lambda^2+\frac92.
\end{align*} = ( λ − 2 3 ) ( λ − 3 ) ( λ + 1 ) λ 3 − 2 7 λ 2 + 2 9 . By the Cayley–Hamilton theorem,
A 3 − 7 2 A 2 + 9 2 I = 0. \mathbf A^3-\frac72\mathbf A^2+\frac92\mathbf I=\mathbf0. A 3 − 2 7 A 2 + 2 9 I = 0 . Multiplying by B 2 = A − 2 \mathbf B^2=\mathbf A^{-2} B 2 = A − 2 A − 7 2 I + 9 2 B 2 = 0 \mathbf A-\frac72\mathbf I+\frac92\mathbf B^2=\mathbf0 A − 2 7 I + 2 9 B 2 = 0
B 2 = 7 9 I − 2 9 A , ( p , q ) = ( 7 9 , − 2 9 ) . \mathbf B^2=\frac79\mathbf I-\frac29\mathbf A,
\qquad (p,q)=\bigg(\frac79,-\frac29\bigg). B 2 = 9 7 I − 9 2 A , ( p , q ) = ( 9 7 , − 9 2 ) .
(c)
解法一
思路
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先用 A \mathbf A A
答题过程
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Solving ( A − λ I ) v = 0 (\mathbf A-\lambda\mathbf I)\mathbf v=\mathbf0 ( A − λ I ) v = 0
For λ = 3 2 \lambda=\frac32 λ = 2 3 ( 1 , 0 , 0 ) T (1,0,0)^{\mathsf T} ( 1 , 0 , 0 ) T
For λ = 3 \lambda=3 λ = 3 ( 2 , 1 , 0 ) T (2,1,0)^{\mathsf T} ( 2 , 1 , 0 ) T
For λ = − 1 \lambda=-1 λ = − 1 v 2 = − v 3 v_2=-v_3 v 2 = − v 3 5 2 v 1 + 3 v 2 + 8 v 3 = 0 \frac52v_1+3v_2+8v_3=0 2 5 v 1 + 3 v 2 + 8 v 3 = 0 ( 2 , 1 , − 1 ) T (2,1,-1)^{\mathsf T} ( 2 , 1 , − 1 ) T
The corresponding eigenvalues of A − 1 \mathbf A^{-1} A − 1 2 3 , 1 3 , − 1 \frac23,\frac13,-1 3 2 , 3 1 , − 1
P = ( 1 2 2 0 1 1 0 0 − 1 ) , D = ( 2 3 0 0 0 1 3 0 0 0 − 1 ) \mathbf P=\begin{pmatrix}
1&2&2\\0&1&1\\0&0&-1
\end{pmatrix},
\qquad
\mathbf D=\begin{pmatrix}
\frac23&0&0\\0&\frac13&0\\0&0&-1
\end{pmatrix} P = 1 0 0 2 1 0 2 1 − 1 , D = 3 2 0 0 0 3 1 0 0 0 − 1 give A − 1 = P D P − 1 \mathbf A^{-1}=\mathbf P\mathbf D\mathbf P^{-1} A − 1 = PD P − 1
