MCQ
If $A = \left[ {\begin{array}{*{20}{c}}1&3\\2&1\end{array}} \right]$, then determinant of ${A^2} - 2A$ is
- A$5$
- ✓$25$
- C$-5$
- D$-25$
${A^2} = \left[ {\begin{array}{*{20}{c}}1&3\\2&1\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}1&3\\2&1\end{array}} \right]\, = \,\left[ {\begin{array}{*{20}{c}}7&6\\4&7\end{array}} \right]$
and ${A^2} - 2A = \left[ {\begin{array}{*{20}{c}}5&0\\0&5\end{array}} \right]\,,{\rm{det }}({A^2} - 2A) = \left| {\,\begin{array}{*{20}{c}}5&0\\0&5\end{array}\,} \right| = 25$.
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$\left| {\begin{array}{*{20}{c}}a&{a + 1}&{a - 1}\\{ - b}&{b + 1}&{b - 1}\\c&{c - 1}&{c + 1}\end{array}} \right| + \left| {\begin{array}{*{20}{c}}{a + 1}&{b + 1}&{c - 1}\\{a - 1}&{b - 1}&{c + 1}\\{{{\left( { - 1} \right)}^{n + 2}} \cdot a}&{{{\left( { - 1} \right)}^{n + 1}} \cdot b}&{{{\left( { - 1} \right)}^n} \cdot c}\end{array}} \right| = 0$ then $n$ equals to
$A_1=\left\{(x, y): x^2+2 y^2 \leq 1\right\}$
$A_2=\left\{(x, y):\left|x^3\right|+2 \sqrt{2} \mid y^3 \leq 1\right\}$
$A_3=\{(x, y): \max (|x|, \sqrt{2}|y|) \leq 1\}$ Then,