The multiplicative inverse of A = $\left[\begin{array}{cc}\cos \left(\mathrm{\theta }\right)& - \sin \left(\mathrm{\theta }\right)\\ \sin \left(\mathrm{\theta }\right)& \cos \left(\mathrm{\theta }\right)\end{array}\right]$ is

• $\left[\begin{array}{cc}- \cos \left(\mathrm{\theta }\right)& \sin \left(\mathrm{\theta }\right)\\ - \sin \left(\mathrm{\theta }\right)& - \cos \left(\mathrm{\theta }\right)\end{array}\right]$

• $\left[\begin{array}{cc}\cos \left(\mathrm{\theta }\right)& \sin \left(\mathrm{\theta }\right)\\ - \sin \left(\mathrm{\theta }\right)& \cos \left(\mathrm{\theta }\right)\end{array}\right]$

• $\left[\begin{array}{cc}- \cos \left(\mathrm{\theta }\right)& - \sin \left(\mathrm{\theta }\right)\\ \sin \left(\mathrm{\theta }\right)& - \cos \left(\mathrm{\theta }\right)\end{array}\right]$

• $\left[\begin{array}{cc}\cos \left(\mathrm{\theta }\right)& \sin \left(\mathrm{\theta }\right)\\ \sin \left(\mathrm{\theta }\right)& - \cos \left(\mathrm{\theta }\right)\end{array}\right]$

If A = $\left[\begin{array}{ccc}1& 1& 0\\ 2& 1& 5\\ 1& 2& 1\end{array}\right]$, then a₁₁A₂₁ + a₁₂A₂₂ + a₁₃A₂₃ is equal to

• 1

• 0

• - 1

• 2

If A = $\left[\begin{array}{cc}2& 2\\ - 3& 2\end{array}\right], \mathrm{B} = \left[\begin{array}{cc}0& - 1\\ 1& 0\end{array}\right]$, then (B^{- 1}A^{- 1})^{- 1} is equal to

• $\left[\begin{array}{cc}2& - 2\\ 2& 3\end{array}\right]$

• $\left[\begin{array}{cc}2& 2\\ - 2& 3\end{array}\right]$

• $\left[\begin{array}{cc}2& - 3\\ 2& 2\end{array}\right]$

• $\left[\begin{array}{cc}1& - 1\\ - 2& 3\end{array}\right]$

If matrix A = $\left[\begin{array}{cc}1& 2\\ 4& 3\end{array}\right]$ such that AX = $\mathrm{\varphi }$,then X is equal to

• $\frac{1}{5}\left[\begin{array}{cc}1& 3\\ 2& - 1\end{array}\right]$

• $\frac{1}{5}\left[\begin{array}{cc}4& 2\\ 4& - 1\end{array}\right]$

• $\frac{1}{5}\left[\begin{array}{cc}- 3& 2\\ 4& - 1\end{array}\right]$

• $\frac{1}{5}\left[\begin{array}{cc}- 1& 2\\ - 1& 4\end{array}\right]$

The inverse of the matrix $\left[\begin{array}{ccc}1& 0& 0\\ 3& 3& 0\\ 5& 2& - 1\end{array}\right]$ is

• $- \frac{1}{3}\left[\begin{array}{ccc}- 3& 0& 0\\ 3& 1& 0\\ 9& 2& - 3\end{array}\right]$

• $- \frac{1}{3}\left[\begin{array}{ccc}- 3& 0& 0\\ 3& - 1& 0\\ - 9& - 2& 3\end{array}\right]$

• $- \frac{1}{3}\left[\begin{array}{ccc}3& 0& 0\\ 3& - 1& 0\\ - 9& - 2& 3\end{array}\right]$

• $- \frac{1}{3}\left[\begin{array}{ccc}- 3& 0& 0\\ - 3& - 1& 0\\ - 9& - 2& 3\end{array}\right]$

For a invertible matrix A if A(adjA) = $\left[\begin{array}{cc}10& 0\\ 0& 10\end{array}\right]$ then $\left|\mathrm{A}\right|$ =

• 100

• - 100

• 10

• - 10

If the inverse of the matrix $\left[\begin{array}{ccc}\mathrm{\alpha }& 14& - 1\\ 2& 3& 1\\ 6& 2& 3\end{array}\right]$ does not exist, then the value of $\mathrm{\alpha }$ is

• 1

• - 1

• 0

• - 2

The system $\left(\begin{array}{ccc}1& - 1& 2\\ 3& 5& - 3\\ 2& 6& \mathrm{a}\end{array}\right)\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right) = \left(\begin{array}{c}3\\ \mathrm{b}\\ 2\end{array}\right)$ has no solutions, if

• a = - 5, $\mathrm{b} \ne 5$

• a = - 5, b = 5

• $\mathrm{a} \ne - 5, \mathrm{b} = 5$

• $\mathrm{a} \ne - 5, \mathrm{b} \ne 5$

If $\left[\begin{array}{cc}3& - 1\\ 0& 6\end{array}\right]\left[\begin{array}{c}3\mathrm{x}\\ 1\end{array}\right] + \left[\begin{array}{c}- 2\mathrm{x}\\ 3\end{array}\right] = \left[\begin{array}{c}8\\ 9\end{array}\right]$, then the value of x is

• $- \frac{3}{8}$

• 7

• $- \frac{2}{9}$

• None of these

Consider A and B two square matrices of same order. Select the correct alternative.

• $\left|\mathrm{AB}\right| \mathrm{must} \mathrm{be} \mathrm{greater} \mathrm{than} \left|\mathrm{A}\right|$

• $\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right] \mathrm{is} \mathrm{not} \mathrm{unit} \mathrm{matrix}$

• $\left|\mathrm{A} + \mathrm{B}\right| \mathrm{must} \mathrm{be} \mathrm{greater} \mathrm{than} \left|\mathrm{A}\right|$

• If AB = 0, either A or B must be zero matrix