ข้อสอบ PAT 1 - มีนาคม 2557

$$\begin{gathered} \mathrm{จากโจทย์} det\left(A^2+xI\right) = 0\to \boxed{1} \\ \mathbf{ต้องหาค่าของ}\mathbf{ }{\mathit{A}}^{\mathbf{2}}\mathbf{+}\mathit{x}\mathit{I} \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{A}^2+xI=\left[\begin{array}{cc}1& -2\\ 0& -1\end{array}\right]\left[\begin{array}{cc}1& -2\\ 0& -1\end{array}\right]+x\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ =\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}x& 0\\ 0& x\end{array}\right] \\ =\left[\begin{array}{cc}1+x& 0\\ 0& 1+x\end{array}\right] \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} \boxed{1} det\left(A^2 + xI\right) = 0 ; \mathrm{แทนค่า} A^2 + xI \\ \mathrm{จะได้} \hspace{0.17em}\hspace{0.17em} \left|\begin{array}{cc}1+\mathrm{x}& 0\\ 0& 1+\mathrm{x}\end{array}\right|=0 \\ {\left(x+1\right)}^2=0 \\ \mathrm{ดังนั้น} x=-1 \end{gathered}$$​

$$\begin{gathered} \left(ก\right) \mathrm{จากโจทย์} det\left(A+xI\right)=0\to \boxed{2} \\ \mathrm{ต้องหาค่าของ} A+xI \\ A+xI=\left[\begin{array}{cc}1& -2\\ 0& -1\end{array}\right]+\left(-1\right)\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ =\left[\begin{array}{cc}0& -2\\ 0& -2\end{array}\right] \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} \boxed{2} det\left(A+xI\right)=0 ; \mathrm{แทนค่า} A+xI \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \left|\begin{array}{cc}0& -2\\ 0& -2\end{array}\right|=0 \\ \left(0\right)\left(-2\right)-\left(0\right)\left(-2\right)=0 \\ 0=0 \mathrm{ข้อ} \left(ก\right) \mathrm{ถูก} \end{gathered}$$

$$\begin{gathered} \left(ข\right) \mathrm{จาก} A^2+xI=\left[\begin{array}{cc}1+x& 0\\ 0& 1+x\end{array}\right] ; \mathrm{แทน} x=-1 \\ A^2+xI=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \\ =0 \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} det\left(A^2+xI\mathit{-}B\right) = det\left(B^t\right) \\ -\mathrm{แทนค่า} A^2+xI=0 \\ \mathrm{จะได้} det \left(0-B\right)=det B \\ det \left(-B\right)=det B \\ \hspace{0.17em}{\left(-1\right)}^2\det B=det B \\ det B=det B \mathrm{ข้อ} \left(ข\right) \mathrm{ถูก} \end{gathered}$$