题目
Problem
The matrix A \mathbf{A} A
A = ( k 1 0 6 5 2 − 1 3 − k ) , \mathbf{A}=\begin{pmatrix}k&1&0\\6&5&2\\-1&3&-k\end{pmatrix}, A = k 6 − 1 1 5 3 0 2 − k , where k k k
(a) Show that A \mathbf{A} A
(3)
(b) Given that A − 1 = ( 3 0 − 1 1 0 0 − 23 2 1 2 3 ) \mathbf{A}^{-1}=\begin{pmatrix}3&0&-1\\1&0&0\\-\frac{23}{2}&\frac12&3\end{pmatrix} A − 1 = 3 1 − 2 23 0 0 2 1 − 1 0 3 k k k
(2)
题目中文翻译
矩阵 A \mathbf{A} A
A = ( k 1 0 6 5 2 − 1 3 − k ) , \mathbf{A}=\begin{pmatrix}k&1&0\\6&5&2\\-1&3&-k\end{pmatrix}, A = k 6 − 1 1 5 3 0 2 − k , 其中 k k k
(a) 证明 A \mathbf{A} A
(3)
(b) 已知 A − 1 = ( 3 0 − 1 1 0 0 − 23 2 1 2 3 ) \mathbf{A}^{-1}=\begin{pmatrix}3&0&-1\\1&0&0\\-\frac{23}{2}&\frac12&3\end{pmatrix} A − 1 = 3 1 − 2 23 0 0 2 1 − 1 0 3 k k k
(2)
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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沿第一行展开行列式，含k的一次项相消，结果对所有实数k恒负。
答题过程
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Expanding along the first row,
det A = k ∣ 5 2 3 − k ∣ − ∣ 6 2 − 1 − k ∣ = k ( − 5 k − 6 ) − ( − 6 k + 2 ) = − 5 k 2 − 2. \begin{align*}
\det\mathbf A=&\,k\begin{vmatrix}5&2\\3&-k\end{vmatrix}\\[4mm]
&\,-\begin{vmatrix}6&2\\-1&-k\end{vmatrix}\\[4mm]
=&\,k(-5k-6)-(-6k+2)\\[4mm]
=&\,-5k^2-2.
\end{align*} det A = = = k 5 3 2 − k − 6 − 1 2 − k k ( − 5 k − 6 ) − ( − 6 k + 2 ) − 5 k 2 − 2. Since k k k − 5 k 2 − 2 < 0 -5k^2-2<0 − 5 k 2 − 2 < 0 A \mathbf A A non-singular for every real k k k .
(b)
解法一
思路
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利用A乘其逆等于单位矩阵，只需选一个含k的元素，不必计算整个乘积。
答题过程
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The ( 1 , 1 ) (1,1) ( 1 , 1 ) A A − 1 = I \mathbf A\mathbf A^{-1}=\mathbf I A A − 1 = I
( k , 1 , 0 ) ⋅ ( 3 , 1 , − 23 / 2 ) = 1 , (k,1,0)\cdot(3,1,-23/2)=1, ( k , 1 , 0 ) ⋅ ( 3 , 1 , − 23/2 ) = 1 , so 3 k + 1 = 1 3k+1=1 3 k + 1 = 1 k = 0 \boxed{k=0} k = 0
At k = 0 k=0 k = 0
( 0 1 0 6 5 2 − 1 3 0 ) ( 3 0 − 1 1 0 0 − 23 / 2 1 / 2 3 ) = I . \begin{pmatrix}0&1&0\\6&5&2\\-1&3&0\end{pmatrix}
\begin{pmatrix}3&0&-1\\1&0&0\\-23/2&1/2&3\end{pmatrix}
=\mathbf I. 0 6 − 1 1 5 3 0 2 0 3 1 − 23/2 0 0 1/2 − 1 0 3 = I .
解法二
思路
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逆矩阵的行列式是原行列式的倒数，承接(a)即可得到关于k平方的方程。
答题过程
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Expanding the determinant of the given inverse along its second row gives
det ( A − 1 ) = − ∣ 0 − 1 1 / 2 3 ∣ = − 1 2 . \det(\mathbf A^{-1})=-\begin{vmatrix}0&-1\\1/2&3\end{vmatrix}=-\frac12. det ( A − 1 ) = − 0 1/2 − 1 3 = − 2 1 . By (a) and det ( A − 1 ) = 1 / det A \det(\mathbf A^{-1})=1/\det\mathbf A det ( A − 1 ) = 1/ det A
− 1 2 = 1 − 5 k 2 − 2 . -\frac12=\frac1{-5k^2-2}. − 2 1 = − 5 k 2 − 2 1 . Thus 5 k 2 + 2 = 2 5k^2+2=2 5 k 2 + 2 = 2 k = 0 \boxed{k=0} k = 0
