ข้อสอบ PAT 1 - กุมภาพันธ์ 2563

$$\begin{gathered} \mathrm{จากสูตร} det\left(adj\left(A\right)\right) = {\left[det\left(A\right)\right]}^{n-1} \\ \left|\begin{array}{ccc}-4& -1& x\\ -2& x& -2\\ 1& -5& 1\end{array}\right| = {\left(-7\right)}^{3-1} \end{gathered}$$

$$\begin{gathered} (-4x+2+10x)-(x^2-40+2) = 49 \\ 0 = x^2-6x+9 \\ 0 = {(x-3)}^2 \\ x = 3 \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} det\left(xadj\left(A\right)\right) = det\left(3adj\left(A\right)\right) \\ = 3^{n}det\left(adj\left(A\right)\right) \\ = 3^n{\left(det\left(A\right)\right)}^{n-1} \\ = 3^3{\left(-7\right)}^{3-1} \\ = 27\left(49\right) \\ = 1323 \end{gathered}$$