(x + y - z) + x + (z+ x- y) y- (x + y - 2)z
Answers
∣
(x+y)
2
zx
zy
zx
(z+y)
2
xy
zy
xy
(z+x)
2
∣
∣
∣
∣
∣
=2xyz(x+y+z)
3
.
Easy
Solution
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LHS :
∣
∣
∣
∣
∣
∣
∣
∣
(x+y)
2
zx
zy
zx
(z+y)
2
xy
zy
xy
(z+x)
2
∣
∣
∣
∣
∣
∣
∣
∣
Applying R
1
→zR
1
,R
2
→xR
2
,R
3
→yR
3
, we get,
=
xyz
1
∣
∣
∣
∣
∣
∣
∣
∣
z(x+y)
2
zx
2
zy
2
z
2
x
x(z+y)
2
xy
2
z
2
y
x
2
y
y(z+x)
2
∣
∣
∣
∣
∣
∣
∣
∣
Applying C
1
→
z
1
C
1
,C
2
→
x
1
C
2
,C
3
→
y
1
C
3
, we get,
=
∣
∣
∣
∣
∣
∣
∣
∣
(x+y)
2
x
2
y
2
z
2
(z+y)
2
y
2
z
2
x
2
(z+x)
2
∣
∣
∣
∣
∣
∣
∣
∣
Applying C
1
→C
1
−C
3
,C
2
→C
2
−C
3
, we get,
=(x+y+z)
2
∣
∣
∣
∣
∣
∣
∣
∣
x+y−z
0
y−z−x
0
z+y−x
y−z−x
z
2
x
2
(z+x)
2
∣
∣
∣
∣
∣
∣
∣
∣
Applying R
3
→R
3
−(R
1
+R
2
), we get,
=(x+y+z)
2
∣
∣
∣
∣
∣
∣
∣
∣
x+y−z
0
−2x
0
z+y−x
−2z
z
2
x
2
2xz
∣
∣
∣
∣
∣
∣
∣
∣
Applying C
1
→C
1
+
z
1
C
3
,C
2
→C
2
+
x
1
C
3
, we get,
=(x+y+z)
2
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
x+y
z
x
2
0
x
z
2
z+y
0
z
2
x
2
2xz
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
=(x+y+z)
2
[2xz(xz+xy+yz+y
2
−xz)]
=(x+y+z)
2
[2xyz(x+y+z)]
=2xyz(x+y+z)
3
= RHS
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SIMILAR QUESTIONS
star-struck
Using properties of determinants prove that :
∣
∣
∣
∣
∣
∣
∣
∣
a+b+c
−c
−b
−c
a+b+c
−a
−b
−a
a+b+c
∣
∣
∣
∣
∣
∣
∣
∣
=2(a+b)(b+c)(c+a)
Medium
View solution
>
Prove that:
∣
∣
∣
∣
∣
∣
∣
∣
1
x
yz
1
y
zx
1
z
xy
∣
∣
∣
∣
∣
∣
∣
∣
=(x−y)(y−z)(z−x).