Perhatikan perkalian matriks berikut:
\begin{aligned} P = \begin{pmatrix} {}^{\sqrt{3}} \! \log 2 & {}^{1/2} \! \log 3 \\[8pt] {}^{1/9} \! \log 4 & {}^{2\sqrt{2}} \! \log 9 \end{pmatrix} \begin{pmatrix} {}^4 \! \log 3 & {}^{1/2} \! \log 3\sqrt{3} \\[8pt] {}^{1/3} \! \log 8 & {}^9 \! \log \frac{1}{8} \end{pmatrix} \end{aligned}
Nilai \( \det P \) adalah…
- -23/4
- -35/4
- -43/2
- -83/2
- -143/4
Pembahasan:
Untuk mengerjakan soal ini Anda perlu memahami sifat determinan dan logaritma.
\begin{aligned} |P| &= \begin{vmatrix} {}^{\sqrt{3}} \! \log 2 & {}^{1/2} \! \log 3 \\[8pt] {}^{1/9} \! \log 4 & {}^{2\sqrt{2}} \! \log 9 \end{vmatrix} \begin{vmatrix} {}^4 \! \log 3 & {}^{1/2} \! \log 3\sqrt{3} \\[8pt] {}^{1/3} \! \log 8 & {}^9 \! \log \frac{1}{8} \end{vmatrix} \\[8pt] &= \left( {}^{\sqrt{3}} \! \log 2 \cdot {}^{2\sqrt{2}} \! \log 9 - {}^{1/2} \! \log 3 \cdot {}^{1/9} \! \log 4 \right) \left( {}^4 \! \log 3 \cdot {}^9 \! \log \frac{1}{8} - {}^{1/2} \! \log 3\sqrt{3} \cdot {}^{1/3} \! \log 8 \right) \\[8pt] &= \left( {}^{3^\frac{1}{2}} \! \log 2 \cdot {}^{2^\frac{3}{2}} \! \log 3^2 - {}^{2^{-1}} \! \log 3 \cdot {}^{3^{-2}} \! \log 2^2 \right) \left( {}^{2^2} \! \log 3 \cdot {}^{3^2} \! \log 2^{-3} - {}^{2^{-1}} \! \log 3^{\frac{3}{2}} \cdot {}^{3^{-1}} \! \log 2^3 \right) \\[8pt] &= \left( \frac{1}{\frac{1}{2}} \ {}^3 \! \log 2 \cdot \frac{2}{\frac{3}{2}} \ {}^2 \! \log 3 - \frac{1}{-1} \ {}^2 \! \log 3 \cdot \frac{2}{-2} \ {}^3 \! \log 2 \right)\left( \frac{1}{2} \ {}^2 \! \log 3 \cdot \frac{-3}{2} \ {}^3 \! \log 2 - \frac{\frac{3}{2}}{-1} \ {}^2 \! \log 3 \cdot \frac{3}{-1} \ {}^3 \! \log 2 \right) \\[8pt] &= \left( 2 \cdot \frac{4}{3} - (-1) \cdot (-1) \right)\left( \frac{1}{2} \cdot \frac{-3}{2} -\left(-\frac{3}{2}\right) (-3) \right) \\[8pt] &= \left( \frac{8}{3}-1 \right) \left( -\frac{3}{4}-\frac{9}{2} \right) \\[8pt] &= \frac{5}{3} \cdot \frac{21}{4} = -\frac{35}{4} \end{aligned}
Catatan: Ingat bahwa \( {}^2 \! \log 3 \cdot {}^3 \! \log 2 = 1 \)
Jawaban B.