ข้อสอบ PAT 1 - มีนาคม 2556

$$\begin{gathered} \mathrm{จากโจทย์}\hspace{0.17em} \mathrm{AB} + 3\mathrm{A} =2\mathrm{I} \\ \mathrm{A}\left(\mathrm{B}+3\mathrm{I}\right)=2\mathrm{I} \\ \mathrm{take} \det \mathrm{ทั้งสองข้าง} \\ \mathrm{จะได้} \det \left[\mathrm{A}\left(\mathrm{B}+3\mathrm{I}\right)\right]=\det \left(2\mathrm{I}\right) \\ \det \mathrm{A} \det \left(\mathrm{B}+3\mathrm{I}\right)={\left(2\right)}^3\det \left(\mathrm{I}\right) ; \mathrm{โดย} \det \mathrm{A}=2 \\ \left(2\right)\det \left(\mathrm{B}+3\mathrm{I}\right)=8\left(1\right) \\ \det \left(\mathrm{B}+3\mathrm{I}\right)=4\to \boxed{1} \end{gathered}$$

$$\begin{gathered} \underline{\mathrm{พิจารณา} \mathrm{B}+3\mathrm{I}} \\ \mathrm{B}+3\mathrm{I}= \left[\begin{array}{ccc}1& 3& 2\\ 0& -1& \mathrm{x}\\ 0& -2& \mathrm{y}\end{array}\right]+3\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]= \left[\begin{array}{ccc}4& 3& 2\\ 0& 2& \mathrm{x}\\ 0& -2& \mathrm{y}+3\end{array}\right] \end{gathered}$$  

$$\begin{gathered} \underline{\mathrm{หาค่า} \det \left(\mathrm{B}+3\mathrm{I}\right)} \\ \det \left(\mathrm{B}+3\mathrm{I}\right)=\left[\begin{array}{ccc}4& 3& 2\\ 0& 2& \mathrm{x}\\ 0& -2& \mathrm{y}+3\end{array}\right] \\ =\left(8\mathrm{y}+24\right)-\left(-8\mathrm{x}\right) \\ \det \left(\mathrm{B}+3\mathrm{I}\right)=8\mathrm{x}+8\mathrm{y}+24 \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} \boxed{1} \det \left(\mathrm{B}+3\mathrm{I}\right)=4 \\ \mathrm{จะได้} 8\mathrm{x}+8\mathrm{y}+24=4 \\ 8\mathrm{x}+8\mathrm{y}=-20 \\ 8\left(\mathrm{x}+\mathrm{y}\right)=-20 \\ \mathrm{ดังนั้น} \mathrm{x}+\mathrm{y}=-2.5 \end{gathered}$$