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3 and 4 . determinant and metrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths3 and 4 . determinant and metrices1 Mark
MCQ
$\left| {\,\begin{array}{*{20}{c}}{{{\log }_3}512}&{{{\log }_4}3}\\{{{\log }_3}8}&{{{\log }_4}9}\end{array}\,} \right| \times \left| {\,\begin{array}{*{20}{c}}{{{\log }_2}3}&{{{\log }_8}3}\\{{{\log }_3}4}&{{{\log }_3}4}\end{array}\,} \right|=$
  • A
    $7$
  • ✓
    $10$
  • C
    $13$
  • D
    $17$
✓

Answer

Correct option: B.
$10$
b
(b) $\left| {\,\begin{array}{*{20}{c}}{{{\log }_2}512}&{{{\log }_8}3}\\{{{\log }_3}8}&{{{\log }_3}9}\end{array}\,} \right|$  $×$  $\left| {\,\begin{array}{*{20}{c}}{{{\log }_2}3}&{{{\log }_8}3}\\{{{\log }_3}4}&{{{\log }_3}4}\end{array}\,} \right|$

$ = \left( {\frac{{\log 512}}{{\log 3}} \times \frac{{\log 9}}{{\log 4}} - \frac{{\log 3}}{{\log 4}} \times \frac{{\log 8}}{{\log 3}}} \right)$

$×$ $\left( {\frac{{\log 3}}{{\log 2}} \times \frac{{\log 4}}{{\log 3}} - \frac{{\log 3}}{{\log 8}} \times \frac{{\log 4}}{{\log 3}}} \right)$

$= \left( {\frac{{\log {2^9}}}{{\log 3}} \times \frac{{\log {3^2}}}{{\log {2^2}}} - \frac{{\log {2^3}}}{{\log {2^2}}}} \right)$ $×$ $\left( {\frac{{\log {2^2}}}{{\log 2}} - \frac{{\log {2^2}}}{{\log {2^3}}}} \right)$

$= \left( {\frac{{9 \times 2}}{2} - \frac{3}{2}} \right)\,\left( {2 - \frac{2}{3}} \right) = \frac{{15}}{2} \times \frac{4}{3} = 10$.

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