ข้อสอบ A-Level คณิตประยุกต์ 1 ปี 2555 (วิชาสามัญเก่า)

$$\begin{gathered} \mathrm{จากสูตร} \det \left(\mathrm{A}·\mathrm{B}\right)=\det \mathrm{A}·\mathrm{detB} \\ =\left|\mathrm{A}\right|·\left|\mathrm{B}\right| \\ \mathrm{จากโจทย์} \det \left(\mathrm{A}-\sqrt{2}\mathrm{I}\right)\left(\mathrm{A}-\sqrt{3}\mathrm{I}\right)\left(\mathrm{A}-\sqrt{5}\mathrm{I}\right)\left(\mathrm{A}-\sqrt{7}\mathrm{I}\right) \\ =\left|\mathrm{A}-\sqrt{2}\mathrm{I}\right|\left|\mathrm{A}-\sqrt{3}\mathrm{I}\right|\left|\mathrm{A}-\sqrt{5}\mathrm{I}\right|\left|\mathrm{A}-\sqrt{7}\mathrm{I}\right|\to \boxed{1} \\ \underline{\mathrm{หา} \mathrm{A}-\sqrt{2}\mathrm{I}} \\ \mathrm{A}-\sqrt{2}\mathrm{I}=\left[\begin{array}{cc}\mathrm{a}& 1-\mathrm{a}\\ 1+\mathrm{a}& -\mathrm{a}\end{array}\right]-\left[\begin{array}{cc}\sqrt{2}& 0\\ 0& \sqrt{2}\end{array}\right] \\ =\left[\begin{array}{cc}\mathrm{a}-\sqrt{2}& 1-\mathrm{a}\\ 1+\mathrm{a}& -\mathrm{a}-\sqrt{2}\end{array}\right] \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} \left|\mathrm{A}-\sqrt{2}\mathrm{I}\right|=\left(\mathrm{a}-\sqrt{2}\right)\left(-\mathrm{a}-\sqrt{2}\right)-\left(1+\mathrm{a}\right)\left(1-\mathrm{a}\right) \\ =\left(-\sqrt{2}+\mathrm{a}\right)\left(-\sqrt{2}-\mathrm{a}\right)-\left(1+\mathrm{a}\right)\left(1-\mathrm{a}\right) \\ =\left[{\left(-\sqrt{2}\right)}^2-{\mathrm{a}}^2\right]-\left(1^2-{\mathrm{a}}^2\right) \\ =1 \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}; \mathrm{แทนใน} \boxed{1} \\ \underline{\mathrm{หา} \mathrm{A}-\sqrt{3}\mathrm{I}} \\ \mathrm{A}-\sqrt{3}\mathrm{I}=\left[\begin{array}{cc}\mathrm{a}& 1-\mathrm{a}\\ 1+\mathrm{a}& -\mathrm{a}\end{array}\right]-\left[\begin{array}{cc}\sqrt{3}& 0\\ 0& \sqrt{3}\end{array}\right] \\ =\left[\begin{array}{cc}\mathrm{a}-\sqrt{3}& 1-\mathrm{a}\\ 1+\mathrm{a}& -\mathrm{a}-\sqrt{3}\end{array}\right] \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} \left|\mathrm{A}-\sqrt{3}\mathrm{I}\right|=\left(\mathrm{a}-\sqrt{3}\right)\left(-\mathrm{a}-\sqrt{3}\right)-\left(1+\mathrm{a}\right)\left(1-\mathrm{a}\right) \\ =\left(-\sqrt{3}+\mathrm{a}\right)\left(-\sqrt{3}-\mathrm{a}\right)-\left(1+\mathrm{a}\right)\left(1-\mathrm{a}\right) \\ =\left[{\left(-\sqrt{3}\right)}^2-{\mathrm{a}}^2\right]-\left(1^2-{\mathrm{a}}^2\right) \\ =\left(3-{\mathrm{a}}^2\right)-\left(1-{\mathrm{a}}^2\right) \\ =2 \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}; \mathrm{แทนใน} \boxed{1} \end{gathered}$$

$$\begin{gathered} \mathrm{แสดงว่า} \left|\mathrm{A}-\sqrt{5}\mathrm{I}\right|=\left(5-{\mathrm{a}}^2\right)-\left(1-{\mathrm{a}}^2\right) \\ =4 \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}; \mathrm{แทนใน} \boxed{1} \\ \left|\mathrm{A}-\sqrt{7}\mathrm{I}\right|=\left(7-{\mathrm{a}}^2\right)-\left(1-{\mathrm{a}}^2\right) \\ =6 \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}; \mathrm{แทนใน} \boxed{1} \\ \mathrm{จาก} \boxed{1} \mathrm{จะได้} \det \left(\mathrm{A}-\sqrt{2}\mathrm{I}\right)\left(\mathrm{A}-\sqrt{3}\mathrm{I}\right)\left(\mathrm{A}-\sqrt{5}\mathrm{I}\right)\left(\mathrm{A}-\sqrt{7}\mathrm{I}\right) \\ =1\left(2\right)\left(4\right)\left(6\right) \\ =48 \end{gathered}$$