ข้อสอบ A-Level คณิตประยุกต์ 1 ปี 2555 (วิชาสามัญเก่า)

$$\begin{gathered} \mathrm{จากโจทย์}\hspace{0.17em} {\mathrm{B}}_1=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right] , {\mathrm{B}}_2=\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right] , {\mathrm{B}}_3=\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right] \\ \mathrm{จะได้} {\mathrm{B}}_{\mathrm{i}}=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right] \\ {\mathrm{B}}_{\mathrm{i}}=\mathrm{I} \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} {\mathrm{X}}_1=\left[\begin{array}{c}1\\ 0\\ 5\end{array}\right] , {\mathrm{X}}_2=\left[\begin{array}{c}1\\ 2\\ 5\end{array}\right] , {\mathrm{X}}_3=\left[\begin{array}{c}1\\ 3\\ 1\end{array}\right] \\ \mathrm{จะได้} {\mathrm{X}}_{\mathrm{i}}=\left[\begin{array}{ccc}1& 1& 1\\ 0& 2& 3\\ 5& 5& 1\end{array}\right] \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} {\mathrm{AX}}_{\mathrm{i}}={\mathrm{B}}_{\mathrm{i}} \\ {\mathrm{AX}}_{\mathrm{i}}=\mathrm{I} \\ {\mathrm{AX}}_{\mathrm{i}} {\mathrm{X}}_{\mathrm{i}}^{-1}=\mathrm{I} {\mathrm{X}}_{\mathrm{i}}^{-1} \\ A\cancel{I}=\cancel{\mathrm{I} }{\mathrm{X}}_{\mathrm{i}}^{-1} \\ \mathrm{A}= {\mathrm{X}}_{\mathrm{i}}^{-1} \\ \left|\mathrm{A}\right|= \left|{\mathrm{X}}_{\mathrm{i}}^{-1}\right| \\ \left|\mathrm{A}\right|= \frac{1}{\left|{\mathrm{X}}_{\mathrm{i}}\right|}\to \boxed{1} \end{gathered}$$

$$\begin{gathered} \mathrm{โดย} \left|{\mathrm{X}}_{\mathrm{i}}\right|=\left[\begin{array}{ccc}1& 1& 1\\ 0& 2& 3\\ 5& 5& 1\end{array}\right] \\ =\left(2+15+0\right)-\left(10+15+0\right) \\ =-8 ; \mathrm{แทนใน} \boxed{1} \\ \mathrm{จาก} \boxed{1} \mathrm{จะได้} \left|\mathrm{A}\right|=\frac{1}{-8} \\ \mathrm{ดังนั้น} \det \mathrm{A}=-\frac{1}{8} \end{gathered}$$