1
BITSAT 2022
+3
-1

Let $$A = \left[ {\matrix{ 1 & { - 1} & 1 \cr 2 & 1 & { - 3} \cr 1 & 1 & 1 \cr } } \right]$$ and $$10B = \left[ {\matrix{ 4 & 2 & 2 \cr { - 5} & 0 & \alpha \cr 1 & { - 2} & 3 \cr } } \right]$$

If B is the inverse of A, then the value of $$\alpha$$ is

A
4
B
$$-$$4
C
3
D
5
2
BITSAT 2022
+3
-1

If $$\left[ {\matrix{ 1 & { - \tan \theta } \cr {\tan \theta } & 1 \cr } } \right]{\left[ {\matrix{ 1 & {\tan \theta } \cr { - \tan \theta } & 1 \cr } } \right]^{ - 1}} = \left[ {\matrix{ a & { - b} \cr b & a \cr } } \right]$$, then

A
a = 1, b = 1
B
$$a = \sin 2\theta ,b = \cos 2\theta$$
C
$$a = \cos 2\theta ,b = \sin 2\theta$$
D
None of these
3
BITSAT 2022
+3
-1

If p $$\ne$$ a, q $$\ne$$ b, r $$\ne$$ c and the system of equations

px + ay + az = 0

bx + qy + bz = 0

cx + cy + rz = 0

has a non-trivial solution, then the value of $$\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}$$ is

A
1
B
2
C
$$\frac{1}{2}$$
D
0
4
BITSAT 2021
+3
-1
If p$$\ne$$ q $$\ne$$ r and $$\left| {\matrix{ 0 & {x - p} & {x - q} \cr {x + p} & 0 & {x - r} \cr {x + q} & {x - r} & 0 \cr } } \right| = 0$$, then the value of x which satisfy the equation is
A
x = p
B
x = q
C
x = r
D
x = 0
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