ข้อสอบ A-Level คณิตประยุกต์ 1 ปี 2562 (วิชาสามัญเก่า)

$$\begin{gathered} \mathrm{จากโจทย์} A{B}^t = 2I \\ \frac{1}{2} A{B}^t = I \\ A\frac{1}{2}{B}^t = I ; take A^{-1} \mathrm{ทั้ง} 2 \mathrm{ข้าง} \end{gathered}$$

$$\begin{gathered} A^{-1}· A\frac{1}{2}{B}^t = A^{-1}·I \\ I\frac{1}{2}{B}^t = A^{-1}·I \\ \frac{1}{2}{B}^t = A^{-1} \to \boxed{1} \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} A\left[\begin{array}{c}\begin{array}{c}x\\ y\end{array}\\ z\end{array}\right] = \left[\begin{array}{c}\begin{array}{c}-1\\ 5\end{array}\\ 3\end{array}\right] \\ take A^{-1} \mathrm{ทั้ง} 2 \mathrm{ข้าง} \\ \mathrm{จะได้} A^{-1}·A\left[\begin{array}{c}\begin{array}{c}x\\ y\end{array}\\ z\end{array}\right] = A^{-1} \left[\begin{array}{c}\begin{array}{c}-1\\ 5\end{array}\\ 3\end{array}\right] \\ I\left[\begin{array}{c}\begin{array}{c}x\\ y\end{array}\\ z\end{array}\right] = A^{-1} \left[\begin{array}{c}\begin{array}{c}-1\\ 5\end{array}\\ 3\end{array}\right] \end{gathered}$$

$$\begin{gathered} - \mathrm{นำ} A^{-1} \mathrm{จาก} \boxed{1} \mathrm{มาแทนค่า} \\ \mathrm{จะได้} \left[\begin{array}{c}\begin{array}{c}x\\ y\end{array}\\ z\end{array}\right] = \frac{1}{2}{B}^t\left[\begin{array}{c}\begin{array}{c}-1\\ 5\end{array}\\ 3\end{array}\right] \\ = \frac{1}{2}{\left[\begin{array}{ccc}2& 1& 2\\ 1& 1& -1\\ 3& 2& -2\end{array}\right]}^t\left[\begin{array}{c}\begin{array}{c}-1\\ 5\end{array}\\ 3\end{array}\right] \\ = \frac{1}{2}\left[\begin{array}{ccc}2& 1& 3\\ 1& 1& 2\\ 2& -1& -2\end{array}\right]\left[\begin{array}{c}\begin{array}{c}-1\\ 5\end{array}\\ 3\end{array}\right] \end{gathered}$$

$$\begin{gathered} \mathrm{ดังนั้น} x = \frac{1}{2}\left[2\left(-1\right)+1\left(5\right)+3\left(3\right)\right] \\ x = \frac{1}{2}\left(12\right) = 6 \end{gathered}$$