ข้อสอบ PAT 1 - มีนาคม 2558

$$\begin{gathered} \mathrm{จากสูตร} {\mathrm{A}}^{-1}=\frac{1}{\mathrm{detA}}·\mathrm{adjA} \mathrm{จะได้ว่า} \mathrm{adjA}=\mathrm{detA}·{\mathrm{A}}^{-1} \\ \mathrm{จากโจทย์}\hspace{0.17em} \det \left(\mathrm{A}·\mathrm{adjA}\right)-2{\left(\mathrm{detA}\right)}^2-3\mathrm{detA}=0\to \boxed{1} \\ \underline{\mathrm{พิจารณา} \det \left(\mathrm{A} \mathrm{adjA}\right)} \end{gathered}$$
$$\begin{gathered} \mathrm{จะได้} \det \left(\mathrm{A}·\mathrm{adjA}\right)=\det \left(\mathrm{A}·\mathrm{detA}·A^{-1}\right) ; \mathrm{โดย} \mathrm{A}·{\mathrm{A}}^{-1}=\mathrm{I} \\ =\det \left(\mathrm{detA}·\mathrm{A}·{\mathrm{A}}^{-1}\right) ; \mathrm{มอง} \det \mathrm{A}=\mathrm{ตัวเลข} \\ =\det \left(\mathrm{detA}·\mathrm{I}\right) ; \mathrm{สูตร} \det \left({\mathrm{kM}}_{\mathrm{ij}}\right)={\mathrm{k}}^{\mathrm{n}}{\mathrm{detM}}_{\mathrm{ij}} \\ ={\left(\mathrm{detA}\right)}^3\det \mathrm{I} \\ ={\left(\mathrm{detA}\right)}^3\left(1\right) \\ ={\left(\mathrm{detA}\right)}^3 ; \mathrm{แทนใน} \boxed{1} \end{gathered}$$
$$\begin{gathered} \mathrm{จะได้} (\mathrm{detA}{)}^2-2{\left(\mathrm{detA}\right)}^2-3 \det \mathrm{A}=0 \\ \mathrm{กำหนดให้} \mathrm{x}=\mathrm{detA} \\ \mathrm{จะได้} {\mathrm{x}}^3-2{\mathrm{x}}^2-3\mathrm{x}=0 \\ \mathrm{x}\left({\mathrm{x}}^2-2\mathrm{x}-3\right)=0 \\ \mathrm{x}\left(\mathrm{x}-3\right)\left(\mathrm{x}+1\right)=0 \\ \mathrm{x}=0,3,-1 \\ \mathrm{detA}=0,3,-1 \\ \mathrm{จากโจทย์}\hspace{0.17em} \mathrm{detA}>0 \\ \mathrm{จะได้} \mathrm{detA}=3 \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์}\hspace{0.17em} \mathrm{AB}=\mathrm{I} \mathrm{take} \det \mathrm{ทั้งสองข้าง} \\ \mathrm{จะได้} \det \left(\mathrm{AB}\right)=\det \left(\mathrm{I}\right) \\ \mathrm{detA}·\mathrm{detB}=1 \\ \left(3\right)\mathrm{detB}=1 \\ \mathrm{detB}=\frac{1}{3} \\ \underline{\mathrm{พิจารณา} \left(\mathrm{ก}\right)} \\ \mathrm{จากโจทย์}\hspace{0.17em} 7\mathrm{detB}-{\mathrm{detA}}^{\mathrm{t}}<0 \end{gathered}$$

$$\begin{gathered} \mathbf{หาค่า}\mathbf{ }\mathbf{7}\mathit{d}\mathit{e}\mathit{t}\mathit{B}\mathbf{-}\mathit{d}\mathit{e}\mathit{t}{\mathit{A}}^{\mathbf{t}} \\ \mathrm{จะได้} 7\mathrm{detB}-\det \left({\mathrm{A}}^{\mathrm{t}}\right)=7\mathrm{detB}-\mathrm{detA} ; \mathrm{โดย} \det \left({\mathrm{A}}^{\mathrm{t}}\right)=\mathrm{detA} \\ =7\left(\frac{1}{3}\right)-3 \\ =-\frac{2}{3} \\ \mathrm{ดังนั้น} \mathrm{ข้อ} \left(\mathrm{ก}\right) \mathrm{ถูก} \\ \underline{\mathrm{พิจารณา} \left(\mathrm{ข}\right)} \\ \mathrm{จากโจทย์}\hspace{0.17em} \det \left(2\mathrm{A}-3\mathrm{adj} \mathrm{B}\right)=2 \\ \mathbf{หาค่า}\mathbf{ }\mathit{a}\mathit{d}\mathit{j}\mathit{B} \mathrm{ก่อน} \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์}\hspace{0.17em} \mathrm{AB}=\mathrm{I} \\ \mathrm{A}={\mathrm{B}}^{-1}\mathrm{I} ; \mathrm{จากสมบัติ} \mathrm{BI}=\mathrm{B} \\ \mathrm{ทำให้} {\mathrm{B}}^{-1}\mathrm{I}={\mathrm{B}}^{-1} \\ \mathrm{ดังนั้น} \mathrm{A}={\mathrm{B}}^{-1} \\ \mathrm{จากสูตร} {\mathrm{B}}^{-1}=\frac{1}{\mathrm{detB}}·\mathrm{adjB} \\ \mathrm{adjB}=\mathrm{detB}·{\mathrm{B}}^{-1} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em} ; \mathrm{โดย} \det \mathrm{B}=\frac{1}{3}, {\mathrm{B}}^{-1}=\mathrm{A} \\ \mathrm{adjB}=\left(\frac{1}{3}\right)\mathrm{A} \end{gathered}$$

$$\begin{gathered} \mathrm{แสดงว่า} \det \left(2\mathrm{A}-3\mathrm{adj} \mathrm{B}\right)=\det \left[2\mathrm{A}-3\left(\frac{1}{3}\right)\mathrm{A}\right] \\ =\det \left(2\mathrm{A}-\mathrm{A}\right) \\ =\mathrm{detA} \\ =3 \\ \mathrm{ดังนั้น} \mathrm{ข้อ} \left(\mathrm{ข}\right) \mathrm{ผิด} \end{gathered}$$