题目

Problem

Find the values of $k$ for which the system of equations

\begin{align*} x - y + z =&\, 0,\\ x + ky + 3z =&\, 0,\\ x + 2y + kz =&\, 0, \end{align*}

does not have a unique solution.

[3]

题目中文翻译

求 $k$ 的值，使得方程组

\begin{align*} x - y + z =&\, 0,\\ x + ky + 3z =&\, 0,\\ x + 2y + kz =&\, 0 \end{align*}

没有唯一解。

(3)

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

解法一

思路

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没有唯一解时，齐次方程组的系数矩阵必须奇异。先用第一行消去其余两行的首项，再求行列式。

答题过程

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The homogeneous system has a unique solution exactly when its coefficient matrix is invertible. Subtracting the first row from the second and third gives

\begin{align*} \det\begin{pmatrix}1&-1&1\\1&k&3\\1&2&k\end{pmatrix} =&\,\det\begin{pmatrix}1&-1&1\\0&k+1&2\\0&3&k-1\end{pmatrix}\\ =&\,(k+1)(k-1)-6\\ =&\,k^2-7. \end{align*}

The determinant is zero for $\boxed{k=\pm\sqrt7}$ .

CIE 9231 2025 Nov Paper 21 Q1

题目

解答

解法一

思路

答题过程