题目
Problem
The matrix P \mathbf{P} P
P = ( a 1 1 0 2 a − 1 0 0 − 3 a ) . \mathbf{P} =
\begin{pmatrix}
a & 1 & 1\\
0 & 2a & -1\\
0 & 0 & -3a
\end{pmatrix}. P = a 0 0 1 2 a 0 1 − 1 − 3 a . (a) State, in terms of a a a P \mathbf{P} P
[1]
(b) (i) Find P 2 \mathbf{P}^2 P 2 a a a
[1]
(ii) Use the characteristic equation of P \mathbf{P} P P − 1 \mathbf{P}^{-1} P − 1 a a a
[3]
The 3 × 3 3 \times 3 3 × 3 A \mathbf{A} A 1 1 1 2 2 2 3 3 3
( a 0 0 ) , ( 1 2 a 0 ) , ( 1 − 1 − 3 a ) , \begin{pmatrix}
a\\
0\\
0
\end{pmatrix},
\quad
\begin{pmatrix}
1\\
2a\\
0
\end{pmatrix},
\quad
\begin{pmatrix}
1\\
-1\\
-3a
\end{pmatrix}, a 0 0 , 1 2 a 0 , 1 − 1 − 3 a , respectively.
(c) Find A \mathbf{A} A a a a
[5]
题目中文翻译
矩阵 P \mathbf{P} P
P = ( a 1 1 0 2 a − 1 0 0 − 3 a ) . \mathbf{P} =
\begin{pmatrix}
a & 1 & 1\\
0 & 2a & -1\\
0 & 0 & -3a
\end{pmatrix}. P = a 0 0 1 2 a 0 1 − 1 − 3 a . (a) 用 a a a P \mathbf{P} P
(1)
(b) (i) 求用 a a a P 2 \mathbf{P}^2 P 2
(1)
(ii) 使用 P \mathbf{P} P a a a P − 1 \mathbf{P}^{-1} P − 1
(3)
3 × 3 3 \times 3 3 × 3 A \mathbf{A} A 1 1 1 2 2 2 3 3 3
( a 0 0 ) , ( 1 2 a 0 ) , ( 1 − 1 − 3 a ) . \begin{pmatrix}
a\\
0\\
0
\end{pmatrix},
\quad
\begin{pmatrix}
1\\
2a\\
0
\end{pmatrix},
\quad
\begin{pmatrix}
1\\
-1\\
-3a
\end{pmatrix}. a 0 0 , 1 2 a 0 , 1 − 1 − 3 a . (c) 求用 a a a A \mathbf{A} A
(5)
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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上三角矩阵的特征值就是主对角线上的元素。
答题过程
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Since P \mathbf P P a , 2 a , − 3 a \boxed{a,\ 2a,\ -3a} a , 2 a , − 3 a
(b)(i)
解法一
思路
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按行乘列计算。乘积仍为上三角矩阵，只需计算对角线和上方的元素。
答题过程
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The off-diagonal entries in the product are
( P 2 ) 12 = a ( 1 ) + 1 ( 2 a ) + 1 ( 0 ) = 3 a , ( P 2 ) 13 = a ( 1 ) + 1 ( − 1 ) + 1 ( − 3 a ) = − 2 a − 1 , ( P 2 ) 23 = 0 ( 1 ) + 2 a ( − 1 ) + ( − 1 ) ( − 3 a ) = a . \begin{align*}
(P^2)_{12}=&\,a(1)+1(2a)+1(0)=3a,\\
(P^2)_{13}=&\,a(1)+1(-1)+1(-3a)=-2a-1,\\
(P^2)_{23}=&\,0(1)+2a(-1)+(-1)(-3a)=a.
\end{align*} ( P 2 ) 12 = ( P 2 ) 13 = ( P 2 ) 23 = a ( 1 ) + 1 ( 2 a ) + 1 ( 0 ) = 3 a , a ( 1 ) + 1 ( − 1 ) + 1 ( − 3 a ) = − 2 a − 1 , 0 ( 1 ) + 2 a ( − 1 ) + ( − 1 ) ( − 3 a ) = a . Thus
P 2 = ( a 2 3 a − 2 a − 1 0 4 a 2 a 0 0 9 a 2 ) . \mathbf P^2=\begin{pmatrix}
a^2&3a&-2a-1\\
0&4a^2&a\\
0&0&9a^2
\end{pmatrix}. P 2 = a 2 0 0 3 a 4 a 2 0 − 2 a − 1 a 9 a 2 .
(b)(ii)
解法一
思路
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由三个特征值写出特征多项式。代入 P \mathbf P P P − 1 \mathbf P^{-1} P − 1 P 2 \mathbf P^2 P 2
答题过程
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The characteristic polynomial is
( λ − a ) ( λ − 2 a ) ( λ + 3 a ) = λ 3 − 7 a 2 λ + 6 a 3 . (\lambda-a)(\lambda-2a)(\lambda+3a)=\lambda^3-7a^2\lambda+6a^3. ( λ − a ) ( λ − 2 a ) ( λ + 3 a ) = λ 3 − 7 a 2 λ + 6 a 3 . By the Cayley–Hamilton theorem, P 3 − 7 a 2 P + 6 a 3 I = 0 \mathbf P^3-7a^2\mathbf P+6a^3\mathbf I=\mathbf 0 P 3 − 7 a 2 P + 6 a 3 I = 0 P \mathbf P P a ≠ 0 a\ne0 a = 0 P − 1 \mathbf P^{-1} P − 1
P − 1 = 7 a 2 I − P 2 6 a 3 = 1 6 a 3 ( 6 a 2 − 3 a 2 a + 1 0 3 a 2 − a 0 0 − 2 a 2 ) . \begin{align*}
\mathbf P^{-1}=&\,\frac{7a^2\mathbf I-\mathbf P^2}{6a^3}\\
=&\,\frac1{6a^3}\begin{pmatrix}
6a^2&-3a&2a+1\\
0&3a^2&-a\\
0&0&-2a^2
\end{pmatrix}.
\end{align*} P − 1 = = 6 a 3 7 a 2 I − P 2 6 a 3 1 6 a 2 0 0 − 3 a 3 a 2 0 2 a + 1 − a − 2 a 2 .
(c)
解法一
思路
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三个特征向量正好依序组成 P \mathbf P P A = P D P − 1 \mathbf A=\mathbf P\mathbf D\mathbf P^{-1} A = PD P − 1
答题过程
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Let D = diag ( 1 , 2 , 3 ) \mathbf D=\operatorname{diag}(1,2,3) D = diag ( 1 , 2 , 3 ) A P = P D \mathbf A\mathbf P=\mathbf P\mathbf D AP = PD
A = P D P − 1 = 1 6 a 3 ( a 2 3 0 4 a − 3 0 0 − 9 a ) ( 6 a 2 − 3 a 2 a + 1 0 3 a 2 − a 0 0 − 2 a 2 ) = ( 1 1 2 a − 4 a + 1 6 a 2 0 2 1 3 a 0 0 3 ) . \begin{align*}
\mathbf A
=&\,\mathbf P\mathbf D\mathbf P^{-1}\\
=&\,\frac1{6a^3}\begin{pmatrix}a&2&3\\0&4a&-3\\0&0&-9a\end{pmatrix}
\begin{pmatrix}6a^2&-3a&2a+1\\0&3a^2&-a\\0&0&-2a^2\end{pmatrix}\\
=&\,\begin{pmatrix}
1&\frac1{2a}&-\frac{4a+1}{6a^2}\\
0&2&\frac1{3a}\\
0&0&3
\end{pmatrix}.
\end{align*} A = = = PD P − 1 6 a 3 1 a 0 0 2 4 a 0 3 − 3 − 9 a 6 a 2 0 0 − 3 a 3 a 2 0 2 a + 1 − a − 2 a 2 1 0 0 2 a 1 2 0 − 6 a 2 4 a + 1 3 a 1 3 . 