ข้อสอบ PAT 1 - กุมภาพันธ์ 2562

$$\begin{gathered} \mathrm{จากโจทย์} AB = C \\ \mathrm{จะได้} A^{-1}·\left(AB\right) = A^{-1}·C \\ IB = A^{-1}·C \\ \left(IB\right)·C^{-1} = \left(A^{-1}C\right)·C^{-1} \\ \cancel{I}B{C}^{-1} = A^{-1}\cancel{I} \\ A^{-1} = B{C}^{-1} \to \boxed{1} \end{gathered}$$

$$\begin{gathered} \mathrm{หา} C^{-1} \\ \mathrm{จากสูตร} C^{-1} = \frac{1}{detC}adjC \\ = \frac{1}{detC}{\left[\begin{array}{ccc}{C}_{11}& C_{12}& C_{13}\\ C_{21}& C_{22}& C_{23}\\ C_{31}& C_{32}& C_{33}\end{array}\right]}^t \end{gathered}$$

$= \frac{1}{{\displaystyle \frac{1}{3}\left(\frac{-1}{2}\right)\left(1\right)}}{\left[\begin{array}{ccc}\frac{-1}{2}& 0& 0\\ 0& \frac{1}{3}& 0\\ 0& 0& \frac{-1}{6}\end{array}\right]}^t$

$$\begin{gathered} = -6\left[\begin{array}{ccc}\frac{-1}{2}& 0& 0\\ 0& \frac{1}{3}& 0\\ 0& 0& \frac{-1}{6}\end{array}\right] \\ = \left[\begin{array}{ccc}3& 0& 0\\ 0& -2& 0\\ 0& 0& 1\end{array}\right] \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} \boxed{1} A^{-1} = B·C^{-1} \\ -\mathrm{แทนค่า} B, C^{-1} \\ \mathrm{จะได้} A^{-1} = \left[\begin{array}{ccc}a& 2& -1\\ 3& b& 2\\ -1& 3& c\end{array}\right]·\left[\begin{array}{ccc}3& 0& 0\\ 0& -2& 0\\ 0& 0& 1\end{array}\right] \\ A^{-1} = \left[\begin{array}{ccc}3a& -4& -1\\ 9& -2b& 2\\ -3& -6& c\end{array}\right] \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} A\left[\begin{array}{c}4a+1\\ 5b+2\\ 4c+3\end{array}\right] = \left[\begin{array}{c}1\\ -2\\ 3\end{array}\right] \\ \mathrm{จะได้} A^{-1}·A\left[\begin{array}{c}4a+1\\ 5b+2\\ 4c+3\end{array}\right] = A^{-1} \left[\begin{array}{c}1\\ -2\\ 3\end{array}\right] \end{gathered}$$

$I \left[\begin{array}{c}4a+1\\ 5b+2\\ 4c+3\end{array}\right]= A^{-1} \left[\begin{array}{c}1\\ -2\\ 3\end{array}\right]$

$$\begin{gathered} - \mathrm{แทนค่า} A^{-1} \left[\begin{array}{c}4a+1\\ 5b+2\\ 4c+3\end{array}\right] = \left[\begin{array}{ccc}3a& -4& -1\\ 9& -2b& 2\\ -3& -6& c\end{array}\right]·\left[\begin{array}{c}1\\ -2\\ 3\end{array}\right] \\ \left[\begin{array}{c}4a+1\\ 5b+2\\ 4c+3\end{array}\right] = \left[\begin{array}{c}3a+5\\ 4b+15\\ 3c+9\end{array}\right] \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} \\ 4a+1 = 3a+5 5b+2 = 4b+15 4c+3 = 3c+9 \\ a = 4 b = 13 c = 6 \\ \mathrm{ดังนั้น} a+b+c = 4+13+6 \\ = 23 \end{gathered}$$