Question
Evaluate $\begin{bmatrix}1&0&1\\0&0&1\\1&0&1\end{bmatrix}$ is:
- 2
- 0
- 1
- -1
Evaluate $\begin{bmatrix}1&0&1\\0&0&1\\1&0&1\end{bmatrix}$ is:
Solution:
$\triangle=\begin{bmatrix}1&0&1\\0&0&1\\1&0&1\end{bmatrix}$
$\triangle=1\begin{bmatrix}0&1\\0&1\end{bmatrix}-0\begin{bmatrix}0&1\\1&1\end{bmatrix}+1\begin{bmatrix}0&0\\1&0\end{bmatrix}$
$\triangle=1(0-0)-0(0-1)+1(0-0)$
$\triangle=0-0+0=0.$
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Statement-$2$ : ${\tan ^{ - 1}}\left[ {\frac{{1 + \log {x^2}}}{{1 - \log {x^2}}}} \right]$ = ${\tan ^{ - 1}}\,1 + \,{\tan ^{ - 1}}\left( {\log {x^2}} \right)$
$\frac{15}{16}$
$\frac{3}{16}$
$-\frac{3}{16}$
$-\frac{16}{3}$