1
MHT CET 2021 20th September Morning Shift
MCQ (Single Correct Answer)
+2
-0

$$A(\propto)=\left[\begin{array}{cc}\cos \propto & \sin \propto \\ -\sin \propto & \cos \propto\end{array}\right]$$, then $$\left[A^2(\propto)\right]^{-1}=$$

A
$$\mathrm{A}(\propto)$$
B
$$\mathrm{A}^2(\propto)$$
C
$$\mathrm{A}(-2 \propto)$$
D
$$\mathrm{A}(2 \propto)$$
2
MHT CET 2021 20th September Morning Shift
MCQ (Single Correct Answer)
+2
-0

If inverse of $$\left[\begin{array}{ccc}1 & 2 & x \\ 4 & -1 & 7 \\ 2 & 4 & -6\end{array}\right]$$ does not exist, then $$x=$$

A
$$-$$3
B
2
C
3
D
0
3
MHT CET 2021 20th September Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $$A = \left[ {\matrix{ 3 & 2 & 4 \cr 1 & 2 & 1 \cr 3 & 2 & 6 \cr } } \right]$$ and A$$_{ij}$$ are cofactors of the elements a$$_{ij}$$ of A, then $${a_{11}}{A_{11}} + {a_{12}}{A_{12}} + {a_{13}}{A_{13}}$$ is equal to

A
8
B
6
C
4
D
0
4
MHT CET 2020 19th October Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $A=\left[\begin{array}{ll}4 & 5 \\ 2 & 1\end{array}\right]$ and $A^2-5 A-6 I=0$, then $A^{-1}=$

A
$\frac{1}{6}\left[\begin{array}{cc}-1 & 5 \\ -2 & -4\end{array}\right]$.
B
$\frac{1}{6}\left[\begin{array}{cc}-1 & 5 \\ 2 & 4\end{array}\right]$
C
$\frac{1}{6}\left[\begin{array}{cc}-1 & 5 \\ 2 & -4\end{array}\right]$
D
$\frac{1}{6}\left[\begin{array}{cc}1 & 5 \\ 2 & -4\end{array}\right]$
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