ข้อสอบ A-Level คณิตประยุกต์ 1 ปี 2566 (หลักสูตรใหม่)

$$\begin{gathered} \mathrm{จากสูตร} A = \left[\begin{array}{cc}a& b\\ c& d\end{array}\right] \to detA=ad-bc \\ \mathrm{จากโจทย์} A = \left[\begin{array}{cc}x& -1\\ 0& 3\end{array}\right] \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} detA = 3x-0\left(-1\right) \\ detA = 3x \to \boxed{1} \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} B = \left[\begin{array}{cc}-2& 0\\ x& 1\end{array}\right] \\ \mathrm{จะได้} detB = -2\left(1\right)-0\left(x\right) \\ detB = -2 \to \boxed{2} \end{gathered}$$

$$\begin{gathered} \mathrm{จากสูตร} det\left(AB\right) = detAdetB \\ det\left(A^{-1}\right) = \frac{1}{detA} \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} det\left(B^{-1}A\right) = -6 \\ \mathrm{จะได้} det\left(B^{-1}\right)·det\left(A\right) = -6 \\ \frac{1}{det\left(B\right)}·det\left(A\right) = -6 \end{gathered}$$

$$\begin{gathered} -\mathrm{จาก} \boxed{1}, \boxed{2} \mathrm{แทนค่า} \\ \mathrm{จะได้} \frac{1}{\left(-2\right)}·3x = -6 \\ x = 4 \end{gathered}$$