1
MHT CET 2025 25th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $\mathrm{A}=\left[\begin{array}{cc}5 \mathrm{a} & -\mathrm{b} \\ 3 & 2\end{array}\right]$ and A .adj $\mathrm{A}=\mathrm{AA}^{\mathrm{T}}$, then $5 \mathrm{a}+\mathrm{b}=$

A
7
B
9
C
13
D
5
2
MHT CET 2025 25th April Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $A=\left[\begin{array}{lll}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{array}\right]_{3 \times 3}$, then $A^{-1}=$

A
A
B
$\mathrm{A}^2$
C
$\mathrm{A}^3$
D
$\mathrm{A}^4$
3
MHT CET 2025 23rd April Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $A=\left[\begin{array}{rrr}1 & -1 & 1 \\ 0 & 2 & -3 \\ 2 & 1 & 0\end{array}\right], B=\operatorname{adj} A$ and $C=5 A$, then $\frac{|\operatorname{adjB}|}{|\mathrm{C}|}=$

A
2
B
4
C
1
D
5
4
MHT CET 2025 23rd April Morning Shift
MCQ (Single Correct Answer)
+2
-0

If A and B are non-singular matrices of order 2 such that $\quad(A B)^{-1}=\frac{1}{6}\left[\begin{array}{cc}-1 & -3 \\ 2 & 3\end{array}\right] \quad$ and $A^{-1}=\frac{1}{3}\left[\begin{array}{cc}4 & 3 \\ -1 & 0\end{array}\right]$ then $B^{-1}=$

A
$\frac{1}{2}\left[\begin{array}{cc}2 & 3 \\ 1 & -1\end{array}\right]$
B
$\frac{1}{2}\left[\begin{array}{ll}3 & 1 \\ 2 & 4\end{array}\right]$
C
$\frac{1}{2}\left[\begin{array}{cc}-1 & 3 \\ 1 & 2\end{array}\right]$
D
$\frac{1}{6}\left[\begin{array}{ll}1 & 1 \\ 2 & 3\end{array}\right]$
MHT CET Subjects
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