1
COMEDK 2024 Evening Shift
+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
$$\left(\begin{array}{ccc} 5 & -5 & 10 \\ 15 & 0 & 20 \end{array}\right)$$
2
COMEDK 2024 Evening Shift
+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
3
COMEDK 2024 Evening Shift
+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
$$\left(\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right)$$
4
COMEDK 2024 Evening Shift
+1
-0

If $$A=\left[\begin{array}{ccc}0 & x & 16 \\ x & 5 & 7 \\ 0 & 9 & x\end{array}\right]$$ is a singular matrix then $$x$$ is equal to

A
$$-12$$
B
21
C
$$-144$$
D
144
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