题目
Problem
(a) Find the value of a a a
3 x + a y = 0 , 5 x − y = 0 , x + 3 y + 2 z = 0 , \begin{align*}
3x+ay=&\,0,\\
5x-y=&\,0,\\
x+3y+2z=&\,0,
\end{align*} 3 x + a y = 5 x − y = x + 3 y + 2 z = 0 , 0 , 0 , does not have a unique solution.
(2)
The matrix A \mathbf{A} A
A = ( 3 0 0 5 − 1 0 1 3 2 ) . \mathbf{A}=\begin{pmatrix}3&0&0\\5&-1&0\\1&3&2\end{pmatrix}. A = 3 5 1 0 − 1 3 0 0 2 . (b) Find a matrix P \mathbf{P} P D \mathbf{D} D A 2 = P D P − 1 \mathbf{A}^2=\mathbf{P}\mathbf{D}\mathbf{P}^{-1} A 2 = PD P − 1
(7)
(c) Use the characteristic equation of A \mathbf{A} A
( A + 6 I ) 2 = A 4 ( A + b I ) 2 , (\mathbf{A}+6\mathbf{I})^2=\mathbf{A}^4(\mathbf{A}+b\mathbf{I})^2, ( A + 6 I ) 2 = A 4 ( A + b I ) 2 , where b b b
(4)
题目中文翻译
(a) 求使方程组
3 x + a y = 0 , 5 x − y = 0 , x + 3 y + 2 z = 0 \begin{align*}
3x+ay=&\,0,\\
5x-y=&\,0,\\
x+3y+2z=&\,0
\end{align*} 3 x + a y = 5 x − y = x + 3 y + 2 z = 0 , 0 , 0 没有唯一解的 a a a
(2)
矩阵 A \mathbf{A} A
A = ( 3 0 0 5 − 1 0 1 3 2 ) . \mathbf{A}=\begin{pmatrix}3&0&0\\5&-1&0\\1&3&2\end{pmatrix}. A = 3 5 1 0 − 1 3 0 0 2 . (b) 求矩阵 P \mathbf{P} P D \mathbf{D} D A 2 = P D P − 1 \mathbf{A}^2=\mathbf{P}\mathbf{D}\mathbf{P}^{-1} A 2 = PD P − 1
(7)
(c) 利用 A \mathbf{A} A
( A + 6 I ) 2 = A 4 ( A + b I ) 2 , (\mathbf{A}+6\mathbf{I})^2=\mathbf{A}^4(\mathbf{A}+b\mathbf{I})^2, ( A + 6 I ) 2 = A 4 ( A + b I ) 2 , 其中 b b b
(4)
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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系数矩阵的第三列只有一个非零项。沿该列展开，令行列式为零。
答题过程
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The coefficient determinant is 2 det ( 3 a 5 − 1 ) = − 6 − 10 a 2\det\begin{pmatrix}3&a\\5&-1\end{pmatrix}=-6-10a 2 det ( 3 5 a − 1 ) = − 6 − 10 a a = − 3 / 5 \boxed{a=-3/5} a = − 3/5
(b)
解法一
思路
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由下三角矩阵的对角线读出特征值，再求各自的特征向量。平方矩阵保留这些向量，特征值平方。
答题过程
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The matrix is lower triangular, so its characteristic equation is ( λ − 3 ) ( λ + 1 ) ( λ − 2 ) = 0 (\lambda-3)(\lambda+1)(\lambda-2)=0 ( λ − 3 ) ( λ + 1 ) ( λ − 2 ) = 0 A \mathbf A A 3 , − 1 , 2 3,-1,2 3 , − 1 , 2
For λ = 3 \lambda=3 λ = 3 5 v 1 = 4 v 2 5v_1=4v_2 5 v 1 = 4 v 2 v 3 = v 1 + 3 v 2 v_3=v_1+3v_2 v 3 = v 1 + 3 v 2 ( 4 , 5 , 19 ) T (4,5,19)^{\mathsf T} ( 4 , 5 , 19 ) T
For λ = − 1 \lambda=-1 λ = − 1 v 1 = 0 v_1=0 v 1 = 0 v 3 = − v 2 v_3=-v_2 v 3 = − v 2 ( 0 , − 1 , 1 ) T (0,-1,1)^{\mathsf T} ( 0 , − 1 , 1 ) T
For λ = 2 \lambda=2 λ = 2 v 1 = v 2 = 0 v_1=v_2=0 v 1 = v 2 = 0 ( 0 , 0 , 1 ) T (0,0,1)^{\mathsf T} ( 0 , 0 , 1 ) T
These vectors form a basis, and A 2 \mathbf A^2 A 2 9 , 1 , 4 9,1,4 9 , 1 , 4
P = ( 4 0 0 5 − 1 0 19 1 1 ) , D = ( 9 0 0 0 1 0 0 0 4 ) . \begin{align*}
\mathbf P=&\,\begin{pmatrix}4&0&0\\5&-1&0\\19&1&1\end{pmatrix},\\
\mathbf D=&\,\begin{pmatrix}9&0&0\\0&1&0\\0&0&4\end{pmatrix}.
\end{align*} P = D = 4 5 19 0 − 1 1 0 0 1 , 9 0 0 0 1 0 0 0 4 .
(c)
解法一
思路
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用 Cayley–Hamilton 定理把 A + 6 I \mathbf A+6\mathbf I A + 6 I A 2 \mathbf A^2 A 2
答题过程
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The characteristic polynomial is
( λ − 3 ) ( λ + 1 ) ( λ − 2 ) = λ 3 − 4 λ 2 + λ + 6. (\lambda-3)(\lambda+1)(\lambda-2)=\lambda^3-4\lambda^2+\lambda+6. ( λ − 3 ) ( λ + 1 ) ( λ − 2 ) = λ 3 − 4 λ 2 + λ + 6. By the Cayley–Hamilton theorem,
A 3 − 4 A 2 + A + 6 I = 0 , A + 6 I = − A 3 + 4 A 2 = − A 2 ( A − 4 I ) . \begin{align*}
\mathbf A^3-4\mathbf A^2+\mathbf A+6\mathbf I=&\,\mathbf0,\\
\mathbf A+6\mathbf I=&\,-\mathbf A^3+4\mathbf A^2\\
=&\,-\mathbf A^2(\mathbf A-4\mathbf I).
\end{align*} A 3 − 4 A 2 + A + 6 I = A + 6 I = = 0 , − A 3 + 4 A 2 − A 2 ( A − 4 I ) . The factors commute, so
( A + 6 I ) 2 = A 4 ( A − 4 I ) 2 . (\mathbf A+6\mathbf I)^2=\mathbf A^4(\mathbf A-4\mathbf I)^2. ( A + 6 I ) 2 = A 4 ( A − 4 I ) 2 . Hence b = − 4 \boxed{b=-4} b = − 4
