Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
If |x&x+y&x+y+z 2x&3x+2y&4x+3y+2z 3x&6x+3y&10x+6y+3z|= 64 then the real value of x is :
Q. If
∣
∣
x
2
x
3
x
x
+
y
3
x
+
2
y
6
x
+
3
y
x
+
y
+
z
4
x
+
3
y
+
2
z
10
x
+
6
y
+
3
z
∣
∣
=
64
then the real value of x is :
3363
186
Determinants
Report Error
A
2
21%
B
3
37%
C
4
32%
D
6
11%
Solution:
Let
Δ
=
∣
∣
x
2
x
3
x
x
+
y
3
x
+
2
y
6
x
+
3
y
x
+
y
+
z
4
x
+
3
y
+
2
z
10
x
+
6
y
+
3
z
∣
∣
=
∣
∣
x
0
0
x
+
y
x
3
x
x
+
y
+
z
2
x
+
y
7
x
+
3
y
∣
∣
using
R
2
→
R
2
−
2
R
1
R
3
→
R
3
−
3
R
1
=
∣
∣
x
0
0
x
+
y
x
0
x
+
y
+
z
2
x
+
y
x
∣
∣
using
R
3
→
R
3
−
3
R
2
=
x
3
Giver
Δ
=
64
∴
x
3
=
64
⇒
x
=
4