MCQ
If $A$=$\left[ {\begin{array}{*{20}{c}}2&0&0\\0&2&0\\0&0&2\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}1&2&3\\0&1&3\\0&0&2\end{array}} \right],$ then $|AB|$ is equal to
  • A
    $4$
  • B
    $8$
  • $16$
  • D
    $32$

Answer

Correct option: C.
$16$
c
(c) $A = \left[ {\begin{array}{*{20}{c}}2&0&0\\0&2&0\\0&0&2\end{array}} \right] = 2I$

$\therefore $ $AB = 2IB = 2B = \left[ {\begin{array}{*{20}{c}}2&4&6\\0&2&6\\0&0&4\end{array}} \right]$

Therefore $|AB| = \left| {\,\begin{array}{*{20}{c}}2&4&6\\0&2&6\\0&0&4\end{array}\,} \right| = 2(8) = 16$

Aliter : $|A| = 2 \times 2 \times 2 = 8$, $|B| = 1 \times 1 \times 2 = 2$

$\therefore $ $|AB| = \,|A|\,|B| = 2 \times 8 = 16$.

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