题目
Problem
The cubic equation x 3 + 5 x 2 + 10 x − 2 = 0 x^3+5x^2+10x-2=0 x 3 + 5 x 2 + 10 x − 2 = 0 α \alpha α β \beta β γ \gamma γ
(a) Find the value of α 2 + β 2 + γ 2 \alpha^2+\beta^2+\gamma^2 α 2 + β 2 + γ 2
(3)
(b) Show that the matrix ( 1 α β α 1 γ β γ 1 ) \begin{pmatrix}1&\alpha&\beta\\\alpha&1&\gamma\\\beta&\gamma&1\end{pmatrix} 1 α β α 1 γ β γ 1
(4)
题目中文翻译
三次方程 x 3 + 5 x 2 + 10 x − 2 = 0 x^3+5x^2+10x-2=0 x 3 + 5 x 2 + 10 x − 2 = 0 α \alpha α β \beta β γ \gamma γ
(a) 求 α 2 + β 2 + γ 2 \alpha^2+\beta^2+\gamma^2 α 2 + β 2 + γ 2
(3)
(b) 证明矩阵 ( 1 α β α 1 γ β γ 1 ) \begin{pmatrix}1&\alpha&\beta\\\alpha&1&\gamma\\\beta&\gamma&1\end{pmatrix} 1 α β α 1 γ β γ 1
(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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由根与系数关系求根的和与两两乘积和，再使用平方和恒等式。
答题过程
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For x 3 + 5 x 2 + 10 x − 2 = 0 x^3+5x^2+10x-2=0 x 3 + 5 x 2 + 10 x − 2 = 0
α + β + γ = − 5 , α β + β γ + γ α = 10. \begin{align*}
\alpha+\beta+\gamma=&\,-5,\\[4mm]
\alpha\beta+\beta\gamma+\gamma\alpha=&\,10.
\end{align*} α + β + γ = α β + β γ + γ α = − 5 , 10. Hence
α 2 + β 2 + γ 2 = ( α + β + γ ) 2 − 2 ( α β + β γ + γ α ) = 25 − 20 = 5 . \begin{align*}
\alpha^2+\beta^2+\gamma^2
=&\,(\alpha+\beta+\gamma)^2\\[4mm]
&\,-2(\alpha\beta+\beta\gamma+\gamma\alpha)\\[4mm]
=&\,25-20=\boxed5.
\end{align*} α 2 + β 2 + γ 2 = = ( α + β + γ ) 2 − 2 ( α β + β γ + γ α ) 25 − 20 = 5 .
解法二
思路
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也可建立以三个原根的平方为根的新三次方程，再读出它们的和。
答题过程
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For any original root x x x
x ( x 2 + 10 ) = 2 − 5 x 2 . x(x^2+10)=2-5x^2. x ( x 2 + 10 ) = 2 − 5 x 2 . Put y = x 2 y=x^2 y = x 2
y ( y + 10 ) 2 = ( 2 − 5 y ) 2 , y 3 + 20 y 2 + 100 y = 4 − 20 y + 25 y 2 , y 3 − 5 y 2 + 120 y − 4 = 0. \begin{align*}
y(y+10)^2=&\,(2-5y)^2,\\[4mm]
y^3+20y^2+100y=&\,4-20y+25y^2,\\[4mm]
y^3-5y^2+120y-4=&\,0.
\end{align*} y ( y + 10 ) 2 = y 3 + 20 y 2 + 100 y = y 3 − 5 y 2 + 120 y − 4 = ( 2 − 5 y ) 2 , 4 − 20 y + 25 y 2 , 0. Its roots are α 2 , β 2 , γ 2 \alpha^2,\beta^2,\gamma^2 α 2 , β 2 , γ 2 y 2 y^2 y 2
α 2 + β 2 + γ 2 = 5 . \boxed{\alpha^2+\beta^2+\gamma^2=5}. α 2 + β 2 + γ 2 = 5 .
(b)
解法一
思路
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展开行列式后，正好出现(a)的平方和与原三次方程的三根乘积。
答题过程
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Let the given matrix be M \mathbf M M
det M = ( 1 − γ 2 ) − α ( α − β γ ) + β ( α γ − β ) = 1 − ( α 2 + β 2 + γ 2 ) + 2 α β γ . \begin{align*}
\det\mathbf M
=&\,(1-\gamma^2)-\alpha(\alpha-\beta\gamma)\\[4mm]
&\,+\beta(\alpha\gamma-\beta)\\[4mm]
=&\,1-(\alpha^2+\beta^2+\gamma^2)+2\alpha\beta\gamma.
\end{align*} det M = = ( 1 − γ 2 ) − α ( α − β γ ) + β ( α γ − β ) 1 − ( α 2 + β 2 + γ 2 ) + 2 α β γ . From the original cubic, α β γ = 2 \alpha\beta\gamma=2 α β γ = 2 5 5 5
det M = 1 − 5 + 2 ( 2 ) = 0. \det\mathbf M=1-5+2(2)=0. det M = 1 − 5 + 2 ( 2 ) = 0. Therefore M is singular \boxed{\mathbf M\text{ is singular}} M is singular
