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Tardigrade
Question
Mathematics
If | beginmatrix 1 1 0 2 0 3 5 -6 x endmatrix |=29, then x is
Q. If
∣
∣
1
2
5
1
0
−
6
0
3
x
∣
∣
=
29
,
then
x
is
2561
218
J & K CET
J & K CET 2007
Determinants
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A
1
11%
B
2
59%
C
3
18%
D
4
12%
Solution:
Given, that,
∣
∣
1
2
5
1
0
−
6
0
3
x
∣
∣
=
29
⇒
1
(
0
+
18
)
−
1
(
2
x
−
15
)
=
29
⇒
18
−
2
x
+
15
=
29
⇒
2
x
=
4
⇒
x
=
2