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$32$
$\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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Statement $-2$ : $f(x) = \frac{1}{{\sqrt {1 - {x^2}} }} + \left[ {\frac{{{x^2} + x + 1}}{4}} \right]$ , where $[.]$ is greatest integer function. Function $f(x)$ is even function