ข้อสอบ PAT 1 - กุมภาพันธ์ 2562

$$\begin{gathered} \mathrm{สูตรเมทริกซ์} A = \left[\begin{array}{cc}a& b\\ c& d\end{array}\right] \\ A^{-1} = \frac{1}{detA}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right] \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} CA = AB \\ \left(CA\right)A^{-1} = \left(AB\right)A^{-1} \\ CI = \left(AB\right)A^{-1} \\ C = \left[\begin{array}{cc}1& 5\\ 1& 1\end{array}\right]\left[\begin{array}{cc}-2& 0\\ 0& 2\end{array}\right]\frac{1}{1-5}\left[\begin{array}{cc}1& -5\\ -1& 1\end{array}\right] \end{gathered}$$

$$\begin{gathered} = \frac{1}{-4}\left[\begin{array}{cc}-2& 10\\ -2& 2\end{array}\right]\left[\begin{array}{cc}1& -5\\ -1& 1\end{array}\right] \\ = \frac{-1}{2}\left[\begin{array}{cc}-1& 5\\ -1& 1\end{array}\right]\left[\begin{array}{cc}1& -5\\ -1& 1\end{array}\right] \\ = \frac{-1}{2}\left[\begin{array}{cc}-6& 10\\ -2& 6\end{array}\right] = \left[\begin{array}{cc}3& -5\\ 1& -3\end{array}\right] \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} det\left(C^2+xB\right) = -20 \to \boxed{1} \\ \mathrm{หาค่า} C^2+xB = \left[\begin{array}{cc}3& -5\\ 1& -3\end{array}\right]\left[\begin{array}{cc}3& -5\\ 1& -3\end{array}\right]+x\left[\begin{array}{cc}-2& 0\\ 0& 2\end{array}\right] \\ = \left[\begin{array}{cc}4& 0\\ 0& 4\end{array}\right]+\left[\begin{array}{cc}-2x& 0\\ 0& 2x\end{array}\right] \\ = \left[\begin{array}{cc}4-2x& 0\\ 0& 4+2x\end{array}\right] \end{gathered}$$

$$\begin{gathered} det\left(C^2+xB\right) = \left(4-2x\right)\left(4+2x\right)-0\left(0\right) \\ = 16-4x^2 \\ \mathrm{จาก} \boxed{1} det\left(C^2+xB\right) = -20 \\ 16-4x^2 = -20 \\ 4x^2 = 36 \\ x = \pm 3 \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} x \mathrm{เป็นจำนวนจริงบวก} \\ \mathrm{แสดงว่า} x=3 \\ \mathrm{ดังนั้น} x^2+x+1 = 3^2+3+1 \\ = 13 \end{gathered}$$