题目

Problem

$\mathbf{M} = \begin{pmatrix} 3x & 7 \\ 4x + 1 & 2 - x \end{pmatrix}$

Find the range of values of $x$ for which the determinant of the matrix $\mathbf{M}$ is positive.

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题目中文翻译

$\mathbf{M} = \begin{pmatrix} 3x & 7 \\ 4x + 1 & 2 - x \end{pmatrix}$

求使矩阵 $\mathbf{M}$ 的行列式为正值的 $x$ 的取值范围。

解答

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先按二阶矩阵行列式公式展开并化简，再把二次不等式因式分解。找出两个临界值后，根据二次项系数为负判断表达式在两根之间为正；由于题目要求严格大于零，端点不包括在内。

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The determinant of $\mathbf M$ is

\begin{align*} \det\mathbf M =&\,(3x)(2-x)-7(4x+1)\\ =&\,6x-3x^2-28x-7\\ =&\,-3x^2-22x-7. \end{align*}

For the determinant to be positive,

-3x^2-22x-7>0.

The corresponding quadratic equation factorises as

\begin{align*} -3x^2-22x-7=&\,0,\\ (-3x-1)(x+7)=&\,0. \end{align*}

Thus the critical values are

x=-7 \qquad\text{and}\qquad x=-\frac13.

Since the coefficient of $x^2$ is negative, the quadratic is positive between its roots. Therefore,

\boxed{-7<x<-\frac13}.