ข้อสอบ PAT 1 - มีนาคม 2555

$$\begin{gathered} \mathrm{จากโจทย์} \mathrm{A} =\left[\begin{array}{cc}1& \mathrm{x}\\ \mathrm{y}& -1\end{array}\right] \\ {\mathrm{A}}^2=\left[\begin{array}{cc}1& \mathrm{x}\\ \mathrm{y}& -1\end{array}\right]\left[\begin{array}{cc}1& \mathrm{x}\\ \mathrm{y}& -1\end{array}\right] \\ =\left[\begin{array}{cc}1+\mathrm{xy}& \mathrm{x}-\mathrm{x}\\ \mathrm{y}-\mathrm{y}& \mathrm{xy}+1\end{array}\right] \\ =\left[\begin{array}{cc}1+\mathrm{xy}& 0\\ 0& \mathrm{xy}+1\end{array}\right] \\ \mathrm{จากโจทย์} {\mathrm{A}}^2=\mathrm{I} \\ \mathrm{จะได้} \mathrm{xy}=0 \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} \mathrm{AB}=2\mathrm{C} ; \mathrm{take} \det \mathrm{ทั้ง} 2 \mathrm{ข้าง} \\ \mathrm{จะได้} \det \mathrm{AB}=\det 2\mathrm{C} \\ \det \mathrm{A} \det \mathrm{B}=2^2 \det \mathrm{C} ; \mathrm{แทนค่า} \det \mathrm{A} , \det \mathrm{C} \\ \det \left[\begin{array}{cc}1& \mathrm{x}\\ \mathrm{y}& -1\end{array}\right]\det \mathrm{B}=4\det \left[\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right] \\ \left[1\left(-1\right)+\mathrm{xy}\right]\det \mathrm{B}=4\left[1\left(-1\right)\right] \\ -\det \mathrm{B}=-4 \\ \det \mathrm{B}=4 \end{gathered}$$

$$\begin{gathered} \mathrm{ดังนั้น} \det {\mathrm{B}}^{-1}=\frac{1}{\det \mathrm{B}} \\ =\frac{1}{4} \\ =0.25 \end{gathered}$$