MCQ
Let $A^{-1}=\left[\begin{array}{lll}1 & 2017 & 2 \\ 1 & 2017 & 4 \\ 1 & 2018 & 8\end{array}\right]$ Then, $|2 A|-\left|2 A^{-1}\right|$ is equal to
- A$3$
- B$-3$
- ✓$12$
- D$-12$
Given,
$\begin{aligned} A^{-1} &=\left[\begin{array}{lll}1 & 2017 & 2 \\ 1 & 2017 & 4 \\ 1 & 2018 & 8\end{array}\right] \\\left|A^{-1}\right| &=\left[\begin{array}{lll}1 & 2017 & 2 \\ 1 & 2017 & 4 \\ 1 & 2018 & 8\end{array}\right] \end{aligned}$
Apply $R_1 \rightarrow R_1-R_2$
$\left|A^{-1}\right|=\left[\begin{array}{ccc} 0 & 0 & -2 \\ 1 & 2017 & 4 \\ 1 & 2018 & 8 \end{array}\right]$
Expand along $R_1$
$\begin{aligned} \left|A^{-1}\right| &=-2(2018-2017)=-2 \\ A A^{-1} &=I \\ |A|\left|A^{-1}\right| &=1 \\ |A| &=\frac{1}{\left|A^{-1}\right|^2}=-\frac{1}{2} \\ \text { Now, }|2 A| &-\left|2 A^{-1}\right|=8|A|-8\left|A^{-1}\right| \\ &=8\left[\frac{-1}{2}+2\right]=12\end{aligned}$
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$\overrightarrow{\mathrm{a}} \times\{(\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{a}}\}+\overrightarrow{\mathrm{b}} \times\{(\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{c}}) \times \overrightarrow{\mathrm{b}}\}+\overrightarrow{\mathrm{c}} \times\{(\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{a}}) \times \overrightarrow{\mathrm{c}}\}=\overrightarrow{0}$
then $\overrightarrow{\mathrm{r}}$ is equal to:
Then for the objective function $z=-x+2 y$
$(i)$ Maximum value of $z$ has at $\ldots \ldots \ldots . . .$
$(ii)$ Minimum value of $z$ has at $\ldots \ldots \ldots . . .$
$(iii)$ The maximum value of $z$ is $\ldots \ldots \ldots . . .$
$(iv)$ The minimum value of $z$ is $\ldots \ldots \ldots . . .$