ข้อสอบ PAT 1 - มีนาคม 2559

$$\begin{gathered} \mathbf{จากสูตร}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }{\mathit{A}}^{\mathbf{-}\mathbf{1}}\mathbf{ }\mathbf{=}\mathbf{ }\frac{\mathbf{1}}{\mathbf{d}\mathbf{e}\mathbf{t}}\mathbf{·}\mathit{a}\mathit{d}\mathit{j}\mathit{A} \\ \mathrm{จากโจทย์} A^{-1} = \left[\begin{array}{cc}a& 0\\ -2& 1\end{array}\right] \\ \mathrm{จะได้} det\left(A^{-1} \right) = a\left(1\right)-0\left(-2\right) = a \end{gathered}$$

$$\begin{gathered} \mathrm{จากสูตร} A = {\left(A^{-1} \right)}^{-1} \\ = \frac{1}{det\left(A^{-1} \right)}·adj\left(A^{-1} \right) \\ \mathrm{แทนค่า} \det \left({\mathrm{A}}^{-1} \right) = \mathrm{a} \mathrm{จะได้} \\ = \frac{1}{a}·adj\left[\begin{array}{cc}a& 0\\ -2& 1\end{array}\right] = \frac{1}{a}·\left[\begin{array}{cc}1& 0\\ 2& a\end{array}\right] \\ \mathrm{ดังนั้น} A = \left[\begin{array}{cc}\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$a$}\right.& 0\\ \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$a$}\right.& 1\end{array}\right] \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} A^t = \left[\begin{array}{cc}\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$a$}\right.& \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$a$}\right.\\ 0& 1\end{array}\right] \\ det\left(A^t\right) = \left(\frac{1}{a}\right)\left(1\right)-\left(0\right)\left(2\right)=\frac{1}{a} \\ \mathbf{จากสูตร}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }{\left({\mathbf{A}}^{\mathbf{t}}\right)}^{\mathbf{-}\mathbf{1}}\mathbf{ }\mathbf{=}\mathbf{ }\frac{\mathbf{1}}{\mathbf{d}\mathbf{e}\mathbf{t}\left({\mathbf{A}}^{\mathbf{t}}\right)}\mathbf{·}\mathit{a}\mathit{d}\mathit{j}\left({\mathbf{A}}^{\mathbf{t}}\right) \\ = \frac{1}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$a$}\right.}}\left[\begin{array}{cc}1& \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$a$}\right.\\ 0& \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$a$}\right.\end{array}\right] = a\left[\begin{array}{cc}1& \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$a$}\right.\\ 0& \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$a$}\right.\end{array}\right] \\ \mathrm{ดังนั้น} {\left({\mathrm{A}}^{\mathrm{t}}\right)}^{-1} = \left[\begin{array}{cc}a& -2\\ 0& -1\end{array}\right] \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} B^{-1} = \left[\begin{array}{cc}1& 0\\ b& 1\end{array}\right] \\ \mathrm{จะได้} det\left(B^{-1}\right) = 1\left(1\right)-0\left(b\right) = 1 \\ \mathrm{จากสูตร} \mathbf{ }\mathit{B}\mathbf{ }\mathbf{=}\mathbf{ }{\left({\mathbf{B}}^{\mathbf{-}\mathbf{1}}\right)}^{\mathbf{-}\mathbf{1}}\mathbf{ }\mathbf{=}\mathbf{ }\frac{\mathbf{1}}{\mathbf{d}\mathbf{e}\mathbf{t}\left({\mathbf{B}}^{\mathbf{-}\mathbf{1}}\right)}\mathbf{·}\mathit{a}\mathit{d}\mathit{j}\mathbf{ }\left({\mathbf{B}}^{\mathbf{-}\mathbf{1}}\right) \\ \mathrm{ดังนั้น} B \mathbf{ }\mathbf{ }= \frac{1}{1}\left[\begin{array}{cc}1& 0\\ -b& 1\end{array}\right] = \left[\begin{array}{cc}1& 0\\ -b& 1\end{array}\right] \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} {\left(A^t\right)}^{-1}B = \left[\begin{array}{cc}8& -2\\ -3& 1\end{array}\right] \\ \left[\begin{array}{cc}a& -2\\ 0& -1\end{array}\right]\left[\begin{array}{cc}1& 0\\ -b& 1\end{array}\right] = \left[\begin{array}{cc}8& -2\\ -3& 1\end{array}\right] \\ \left[\begin{array}{cc}a+2b& -2\\ -b& 1\end{array}\right] = \left[\begin{array}{cc}8& -2\\ -3& 1\end{array}\right] \\ \mathrm{จากคุณสมบัติความเท่ากันของเมทริกซ์} \\ \mathrm{จะได้} -b = -3 \to b=3 \\ \mathrm{และ} a+2b = 8 \end{gathered}$$

$$\begin{gathered} \mathrm{แทนค่า} \mathrm{b}=3 ; a+2\left(3\right) = 8 \\ a = 2 \\ \mathrm{ดังนั้น} A = \left[\begin{array}{cc}\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.& 0\\ \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$2$}\right.& 1\end{array}\right] = \left[\begin{array}{cc}\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.& 0\\ 1& 1\end{array}\right] \\ B = \left[\begin{array}{cc}1& 0\\ -3& 1\end{array}\right] \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} \mathrm{ค่าของ} det\left(2A+B\right) \mathrm{เท่ากับข้อใด} \\ \mathbf{หาค่า} det\left(2A+B\right) \\ \mathrm{จะได้} 2A+B = 2\left[\begin{array}{cc}\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.& 0\\ 1& 1\end{array}\right]+\left[\begin{array}{cc}1& 0\\ -3& 1\end{array}\right] \\ = \left[\begin{array}{cc}2& 0\\ -1& 3\end{array}\right] \\ \mathrm{ดังนั้น} det\left(2A+B\right) = 2\left(3\right)-0\left(-1\right) \\ = 6 \\ \mathbf{ดังนั้น}\mathbf{ }\mathbf{ตอบ}\mathbf{ }\mathbf{2}\mathbf{)}\mathbf{ }\mathbf{ }\mathbf{6} \end{gathered}$$