1
COMEDK 2025 Morning Shift
MCQ (Single Correct Answer)
+1
-0
If $A=\left[\begin{array}{ccc}0 & 1 & -2 \\ -1 & 0 & 3 \\ 2 & -3 & 0\end{array}\right]$ then $A^{-1}$
A
equal to $-\frac{1}{12}(\operatorname{adj} A)$
B
equal to $-$12
C
equal to $\frac{1}{12}(\operatorname{adj} A)$
D
doesn't exit
2
COMEDK 2024 Evening Shift
MCQ (Single Correct Answer)
+1
-0

$$ \text { If } 3 A+4 B^t=\left(\begin{array}{ccc} 7 & -10 & 17 \\ 0 & 6 & 31 \end{array}\right) \text { and } 2 B-3 A^t=\left(\begin{array}{cc} -1 & 18 \\ 4 & -6 \\ -5 & -7 \end{array}\right) \text { then }(5 B)^t= $$

A
$$ \left(\begin{array}{ccc} 5 & 5 & 10 \\ 15 & 0 & 20 \end{array}\right) $$
B
$$ \left(\begin{array}{ccc} -5 & 5 & 10 \\ -15 & 0 & 20 \end{array}\right) $$
C
$$ \left(\begin{array}{ccc} 5 & -5 & -10 \\ 15 & 0 & -20 \end{array}\right) $$
D
$$ \left(\begin{array}{ccc} 5 & -5 & 10 \\ 15 & 0 & 20 \end{array}\right) $$
3
COMEDK 2024 Evening Shift
MCQ (Single Correct Answer)
+1
-0

$$ \text { If } A=\left[\begin{array}{cc} 5 a & -b \\ 3 & 2 \end{array}\right] \text { and } A \operatorname{adj} A=A A^t \text {, then } 5 a+b \text { is equal to } $$

A
5
B
$$-$$1
C
4
D
13
4
COMEDK 2024 Evening Shift
MCQ (Single Correct Answer)
+1
-0

If the matrix $A$ is such that $$A\left(\begin{array}{cc}-1 & 2 \\ 3 & 1\end{array}\right)=\left(\begin{array}{cc}-4 & 1 \\ 7 & 7\end{array}\right)$$ then $$A$$ is equal to

A
$$\left(\begin{array}{cc}1 & 1 \\ 2 & -3\end{array}\right)$$
B
$$\left(\begin{array}{cc}-1 & 1 \\ 2 & 3\end{array}\right)$$
C
$$\left(\begin{array}{cc}1 & 1 \\ -2 & 3\end{array}\right)$$
D
$$\left(\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right)$$
COMEDK Subjects
EXAM MAP