ข้อสอบ PAT 1 - มีนาคม 2558

$$\begin{gathered} \mathrm{จากสูตร} {\mathrm{A}}^{-1}=\frac{1}{\det \mathrm{A}}·\mathrm{adj} \mathrm{A} \\ \underline{\mathrm{หาเมทริกซ์} \mathrm{A}} \\ \mathrm{จากโจทย์}\hspace{0.17em}\mathrm{AB}=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right] \\ \mathrm{โดย} \det \left(\mathrm{AB}\right)=1\left(4\right)-2\left(3\right) \\ \det \left(\mathrm{AB}\right)=-2 \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} {\mathrm{AB}}^{-1}=\frac{1}{\det \left(\mathrm{AB}\right)}·\left[\begin{array}{cc}4& -2\\ -3& 1\end{array}\right] \\ =\frac{1}{\left(-2\right)}·\left[\begin{array}{cc}4& -2\\ -3& 1\end{array}\right] \\ =-\frac{1}{2}·\left[\begin{array}{cc}4& -2\\ -3& 1\end{array}\right] \\ \mathrm{จากโจทย์}\hspace{0.17em}\mathrm{ABA}=\left[\begin{array}{cc}-1& 2\\ -1& 4\end{array}\right] \\ \left(\mathrm{AB}\right)\mathrm{A}=\left[\begin{array}{cc}-1& 2\\ -1& 4\end{array}\right] \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} {\mathrm{AB}}^{-1}\left(\mathrm{AB}\right)\mathrm{A}={\mathrm{AB}}^{-1}\left[\begin{array}{cc}-1& 2\\ -1& 4\end{array}\right] \\ \mathrm{IA}=-\frac{1}{2}·\left[\begin{array}{cc}4& -2\\ -3& 1\end{array}\right]\left[\begin{array}{cc}-1& 2\\ -1& 4\end{array}\right] \\ \mathrm{A}=-\frac{1}{2}·\left[\begin{array}{cc}-2& 0\\ 2& -2\end{array}\right] \\ \mathrm{A}=\left[\begin{array}{cc}1& 0\\ -1& 1\end{array}\right] \\ \underline{\mathrm{หาเมทริกซ์} \mathrm{B}} \\ \mathrm{จาก} \mathrm{A}=\left[\begin{array}{cc}1& 0\\ -1& 1\end{array}\right] \\ \mathrm{detA}=1\left(1\right)-0\left(-1\right)=1 \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} {\mathrm{A}}^{-1}=\frac{1}{1}·\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right] \\ {\mathrm{A}}^{-1}=\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right] \\ \mathrm{จากโจทย์}\hspace{0.17em}\mathrm{AB}=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right] \\ \mathrm{จะได้} {\mathrm{A}}^{-1}\left(\mathrm{AB}\right)={\mathrm{A}}^{-1}\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right] \\ \mathrm{IB}=\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right] \\ \mathrm{B}=\left[\begin{array}{cc}1& 2\\ 4& 6\end{array}\right] \end{gathered}$$

$$\begin{gathered} \underline{\mathrm{พิจารณาข้อ} \left(\mathrm{ก}\right)} \\ \mathrm{จากโจทย์}\hspace{0.17em}\mathrm{BAB}=\left[\begin{array}{cc}7& 10\\ 22& 32\end{array}\right] \\ \mathbf{หาเมทริกซ์}\mathbf{ }\mathit{B}\mathit{A}\mathit{B} \\ \mathrm{จากโจทย์}\hspace{0.17em}\mathrm{AB}=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right] \\ \mathrm{จะได้} \mathrm{B}\left(\mathrm{AB}\right)=\mathrm{B}\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right] \end{gathered}$$

$$\begin{gathered} \mathrm{BAB}=\left[\begin{array}{cc}1& 2\\ 4& 6\end{array}\right]\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right] \\ =\left[\begin{array}{cc}1+6& 2+8\\ 4+18& 8+24\end{array}\right] \\ =\left[\begin{array}{cc}7& 10\\ 22& 32\end{array}\right] \\ \mathrm{ดังนั้น} \mathrm{ข้อ} \left(\mathrm{ก}\right) \mathrm{ถูก} \end{gathered}$$

$$\begin{gathered} \underline{\mathrm{พิจารณาข้อ} \left(\mathrm{ข}\right)} \\ \mathrm{จากโจทย์}\hspace{0.17em} \left(\mathrm{A}-\mathrm{B}\right)\left(\mathrm{A}+\mathrm{B}\right)\ne {\mathrm{A}}^2-{\mathrm{B}}^2 \\ \mathrm{โดย} \left(\mathrm{A}-\mathrm{B}\right)\left(\mathrm{A}+\mathrm{B}\right)={\mathrm{A}}^2-\mathrm{AB}+\mathrm{BA}-{\mathrm{B}}^2 \\ \mathrm{จะได้} {\mathrm{A}}^2-\mathrm{AB}+\mathrm{BA}-{\mathrm{B}}^2\ne {\mathrm{A}}^2-{\mathrm{B}}^2 \\ \mathrm{ดังนั้น} \mathrm{ข้อ} \left(\mathrm{ข}\right) \mathrm{ถูก} \end{gathered}$$​