ข้อสอบ A-Level คณิตประยุกต์ 1 ปี 2555 (วิชาสามัญเก่า)

$$\begin{gathered} \mathrm{จากโจทย์} \left|\begin{array}{ccc}2& -1& -2\\ 2& 2& 4\\ \mathrm{a}& \mathrm{b}& \mathrm{c}\end{array}\right|=24 \\ \mathrm{จะได้} \left(4\mathrm{c}-4\mathrm{a}-4\mathrm{b}\right)-\left(-4\mathrm{a}+8\mathrm{b}-2\mathrm{c}\right)=24 \\ 4\mathrm{c}\cancel{-4\mathrm{a}}-4\mathrm{b}\cancel{+4\mathrm{a}}-8\mathrm{b}+2\mathrm{c}=24 \\ -12\mathrm{b}+6\mathrm{b}=24 \\ 6\left(-2\mathrm{b}+\mathrm{c}\right)=24 \\ -2\mathrm{b}+\mathrm{c}=4\to \boxed{1} \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} \mathrm{x},\mathrm{y},\mathrm{z} \mathrm{สอดคล้องกับระบบสมการ} \\ \mathrm{AX}=\mathrm{B} \\ \left[\begin{array}{ccc}2& 1& 2\\ 1& 1& -1\\ 3& 2& -2\end{array}\right]\left[\begin{array}{c}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]=\left[\begin{array}{c}\mathrm{a}\\ \mathrm{b}\\ \mathrm{c}\end{array}\right] \\ \mathrm{จะได้} \det \mathrm{A}=\left|\begin{array}{c}\begin{array}{ccc}2& 1& 2\\ 1& 1& -1\\ 3& 2& -2\end{array}\end{array}\right| \\ =\left( \cancel{-4}-3+\cancel{4}\right)-\left(6-4-2\right) \\ =-3 \end{gathered}$$

$$\begin{gathered} \mathbf{โดย} \mathrm{x}=\frac{\left|\begin{array}{c}\begin{array}{ccc}\mathrm{a}& 1& 2\\ \mathrm{b}& 1& -1\\ \mathrm{c}& 2& -2\end{array}\end{array}\right|}{\det \mathrm{A}} \\ =\frac{\left(-2\mathrm{a}-\mathrm{c}+4\mathrm{b}\right)-\left(2\mathrm{c}-2\mathrm{a}-2\mathrm{b}\right)}{-3} \\ =\frac{6\mathrm{b}-3\mathrm{c}}{-3}=\frac{\cancel{3}\left(2\mathrm{b}-\mathrm{c}\right)}{-\cancel{3}} \\ =-2\mathrm{b}+\mathrm{c} ; \mathrm{จาก} \boxed{1} -2\mathrm{b}+\mathrm{c}=4 \\ =4 \end{gathered}$$