For a^x = (x + y + z)^y, a^y = (x + y + z)^z, a^z = (x + y + z)^x, then find the values of x,y and z. Where a > 0 and a ¹ 1.
Answers
Answer:
Answer
Given,
A=
⎣
⎢
⎢
⎡
0
x
x
2y
y
−y
z
−z
z
⎦
⎥
⎥
⎤
A
′
=
⎣
⎢
⎢
⎡
0
2y
z
x
y
−z
x
−y
z
⎦
⎥
⎥
⎤
I=
⎣
⎢
⎢
⎡
1
0
0
0
1
0
0
0
1
⎦
⎥
⎥
⎤
Now, A
′
A=I
Putting values
⎣
⎢
⎢
⎡
0
2y
z
x
y
−z
x
−y
z
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
0
x
x
2y
y
−y
z
−z
z
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
1
0
0
0
1
0
0
0
1
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
0(0)+x(x)+x(x)
2y(0)+y(x)−y(x)
z(0)−z(x)+z(x)
0(2y)+x(y)+x(−y)
2y(2y)+y(y)−y(−y)
z(2y)−z(y)+z(−y)
0(z)+x(−z)+x(z)
2y(z)+y(−z)−y(z)
z(z)−z(−z)+z(z)
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
1
0
0
0
1
0
0
0
1
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
0+x
2
+x
2
0+xy−xy
0−xz+xz
0+xy−xy
4y
2
+y
2
+y
2
2zy−zy−zy
0−xz+xz
2zy−zy−zy
z
2
+z
2
+z
2
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
1
0
0
0
1
0
0
0
1
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
2x
2
0
0
0
6y
2
0
0
0
3z
2
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
1
0
0
0
1
0
0
0
1
⎦
⎥
⎥
⎤
Since matrices are equal,
corresponding elements are equal,
2x
2
=1
x
2
=
2
1
x=±
2
1
x=±
2
1
6y
2
=1
y
2
=
6
1
y=±
6
1
y=±
6
1
3z
2
=1
z
2
=
3
1
z=±
3
1
z=±
3
1
Thus, x=±
2
1
,y=±
6
1
,z=±
3
1