1
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
$\frac{1}{6}\left[\begin{array}{cc}1 & 5 \\ 2 & -4\end{array}\right]$
2
MHT CET 2020 19th October Evening Shift
MCQ (Single Correct Answer)
+2
-0

The cofactors of the elements of the first column of the matrix $A=\left[\begin{array}{ccc}2 & 0 & -1 \\ 3 & 1 & 2 \\ -1 & 1 & 2\end{array}\right]$ are

A
$0,-7,2$
B
$-1,3,-2$
C
$0,-8,4$
D
$0,-1,1$
3
MHT CET 2020 16th October Evening Shift
MCQ (Single Correct Answer)
+2
-0

The matrix $$A=\left[\begin{array}{rrr}a & -1 & 4 \\ -3 & 0 & 1 \\ -1 & 1 & 2\end{array}\right]$$ is not invertible only if $$a=$$

A
17
B
$$-$$16
C
$$-$$17
D
16
4
MHT CET 2020 16th October Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $$A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right], \quad B=\left[\begin{array}{ll}1 & 0 \\ 3 & 1\end{array}\right]$$, then $$B^{-1} A^{-1}=$$

A
$$\left[\begin{array}{rr}-2 & -3 \\ -7 & 11\end{array}\right]$$
B
$$\left[\begin{array}{rr}2 & -3 \\ -7 & 11\end{array}\right]$$
C
$$\left[\begin{array}{rr}2 & 3 \\ 7 & 11\end{array}\right]$$
D
$$\left[\begin{array}{cc}-2 & -3 \\ -7 & -11\end{array}\right]$$
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