1
JEE Main 2024 (Online) 29th January Evening Shift
+4
-1

Let $$A=\left[\begin{array}{ccc}2 & 1 & 2 \\ 6 & 2 & 11 \\ 3 & 3 & 2\end{array}\right]$$ and $$P=\left[\begin{array}{lll}1 & 2 & 0 \\ 5 & 0 & 2 \\ 7 & 1 & 5\end{array}\right]$$. The sum of the prime factors of $$\left|P^{-1} A P-2 I\right|$$ is equal to

A
66
B
27
C
23
D
26
2
JEE Main 2024 (Online) 29th January Morning Shift
+4
-1

$$\text { Let } A=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & \alpha & \beta \\ 0 & \beta & \alpha \end{array}\right] \text { and }|2 \mathrm{~A}|^3=2^{21} \text { where } \alpha, \beta \in Z \text {, Then a value of } \alpha \text { is }$$

A
9
B
17
C
3
D
5
3
JEE Main 2024 (Online) 29th January Morning Shift
+4
-1

Let $$\mathrm{A}$$ be a square matrix such that $$\mathrm{AA}^{\mathrm{T}}=\mathrm{I}$$. Then $$\frac{1}{2} A\left[\left(A+A^T\right)^2+\left(A-A^T\right)^2\right]$$ is equal to

A
$$\mathrm{A}^2+\mathrm{A}^{\mathrm{T}}$$
B
$$\mathrm{A}^3+\mathrm{I}$$
C
$$\mathrm{A}^3+\mathrm{A}^{\mathrm{T}}$$
D
$$\mathrm{A}^2+\mathrm{I}$$
4
JEE Main 2024 (Online) 27th January Evening Shift
+4
-1

The values of $$\alpha$$, for which $$\left|\begin{array}{ccc}1 & \frac{3}{2} & \alpha+\frac{3}{2} \\ 1 & \frac{1}{3} & \alpha+\frac{1}{3} \\ 2 \alpha+3 & 3 \alpha+1 & 0\end{array}\right|=0$$, lie in the interval

A
$$(-2,1)$$
B
$$\left(-\frac{3}{2}, \frac{3}{2}\right)$$
C
$$(-3,0)$$
D
$$(0,3)$$
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