ข้อสอบ A-Level คณิตประยุกต์ 1 ปี 2558 (วิชาสามัญเก่า)

$$\begin{gathered} \mathrm{จากโจทย์} \mathrm{A} \mathrm{เป็นเมทริกซ์ขนาด} 3\times 3 \mathrm{ซึ่ง} \mathrm{A}=\lfloor {\mathrm{a}}_{\mathrm{ij}}{\rfloor }_{3\times 3} \\ \mathrm{จะได้} \mathrm{A}=\left[\begin{array}{ccc}{\mathrm{a}}_{11}& {\mathrm{a}}_{12}& {\mathrm{a}}_{13}\\ {\mathrm{a}}_{21}& {\mathrm{a}}_{22}& {\mathrm{a}}_{23}\\ {\mathrm{a}}_{31}& {\mathrm{a}}_{32}& {\mathrm{a}}_{33}\end{array}\right] \\ \mathbf{ } \\ \mathbf{หาค่า}\mathbf{ }\mathbf{A}\mathbf{+}\mathbf{B} \\ \mathrm{A}+\mathrm{B}=\left[\begin{array}{ccc}{\mathrm{a}}_{11}& {\mathrm{a}}_{12}& {\mathrm{a}}_{13}\\ {\mathrm{a}}_{21}& {\mathrm{a}}_{22}& {\mathrm{a}}_{23}\\ {\mathrm{a}}_{31}& {\mathrm{a}}_{32}& {\mathrm{a}}_{33}\end{array}\right]+\left[\begin{array}{ccc}0& 0& 0\\ 2{\mathrm{a}}_{11}& 2{\mathrm{a}}_{12}& 2{\mathrm{a}}_{13}\\ {\mathrm{a}}_{31}& {\mathrm{a}}_{32}& {\mathrm{a}}_{33}\end{array}\right] \end{gathered}$$
$\mathrm{A}+\mathrm{B}=\left[\begin{array}{ccc}{\mathrm{a}}_{11}& {\mathrm{a}}_{12}& {\mathrm{a}}_{13}\\ {\mathrm{a}}_{21}+2{\mathrm{a}}_{11}& {\mathrm{a}}_{22}+2{\mathrm{a}}_{12}& {\mathrm{a}}_{23}+2{\mathrm{a}}_{13}\\ 2{\mathrm{a}}_{31}& 2{\mathrm{a}}_{32}& 2{\mathrm{a}}_{33}\end{array}\right]\stackrel{{\mathrm{R}}_2-2{\mathrm{R}}_1}{\to }\left[\begin{array}{ccc}{\mathrm{a}}_{11}& {\mathrm{a}}_{12}& {\mathrm{a}}_{13}\\ {\mathrm{a}}_{21}& {\mathrm{a}}_{22}& {\mathrm{a}}_{23}\\ 2{\mathrm{a}}_{31}& 2{\mathrm{a}}_{32}& 2{\mathrm{a}}_{33}\end{array}\right]$
$$\begin{gathered} \mathrm{ดังนั้น} \mathrm{A}+\mathrm{B}=\left[\begin{array}{ccc}{\mathrm{a}}_{11}& {\mathrm{a}}_{12}& {\mathrm{a}}_{13}\\ {\mathrm{a}}_{21}& {\mathrm{a}}_{22}& {\mathrm{a}}_{23}\\ 2{\mathrm{a}}_{31}& 2{\mathrm{a}}_{32}& 2{\mathrm{a}}_{33}\end{array}\right] \\ \mathrm{จะได้} \det \left(\mathrm{A}+\mathrm{B}\right)=\left|\begin{array}{ccc}{\mathrm{a}}_{11}& {\mathrm{a}}_{12}& {\mathrm{a}}_{13}\\ {\mathrm{a}}_{21}& {\mathrm{a}}_{22}& {\mathrm{a}}_{23}\\ 2{\mathrm{a}}_{31}& 2{\mathrm{a}}_{32}& 2{\mathrm{a}}_{33}\end{array}\right|\begin{array}{}\\ \\ \end{array} \\ \mathrm{ใช้สมบัติ} \det \mathrm{ดึง} 2 \mathrm{ออกจากแถวที่} 3 ; \\ =2\left|\begin{array}{ccc}{\mathrm{a}}_{11}& {\mathrm{a}}_{12}& {\mathrm{a}}_{13}\\ {\mathrm{a}}_{21}& {\mathrm{a}}_{22}& {\mathrm{a}}_{23}\\ {\mathrm{a}}_{31}& {\mathrm{a}}_{32}& {\mathrm{a}}_{33}\end{array}\right|\begin{array}{}\\ \\ \end{array} \end{gathered}$$

$$\begin{gathered} =2\mathrm{detA} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}; \mathrm{จากโจทย์} \mathrm{detA}=10 \\ =2\left(10\right) \\ =20 \\ \mathrm{ดังนั้น} \det \left(\mathrm{A}+\mathrm{B}\right)=20 \end{gathered}$$