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Byju's Answer
Standard XII
Mathematics
Properties of Determinants
Prove that: ...
Question
Prove that:
∣
∣ ∣
∣
1
x
y
+
z
1
y
z
+
x
1
z
x
+
y
∣
∣ ∣
∣
=
0
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Solution
To prove:
∣
∣ ∣
∣
1
x
y
+
z
1
y
z
+
x
1
z
x
+
y
∣
∣ ∣
∣
=
0
Lets solve
∣
∣ ∣
∣
1
x
y
+
z
1
y
z
+
x
1
z
x
+
y
∣
∣ ∣
∣
as
=
1
[
y
(
x
+
y
)
−
z
(
z
+
x
)
]
−
x
[
x
+
y
−
(
z
+
x
)
]
+
(
y
+
z
)
[
z
−
y
]
=
[
x
y
+
y
2
−
z
2
−
z
x
]
−
[
x
2
+
x
y
−
x
z
−
x
2
]
+
[
y
z
−
y
2
+
z
2
−
z
y
]
=
x
y
+
y
2
−
z
2
−
z
x
−
x
y
+
x
z
+
y
z
−
y
2
+
z
2
−
z
y
=
0
Therefore,
∣
∣ ∣
∣
1
x
y
+
z
1
y
z
+
x
1
z
x
+
y
∣
∣ ∣
∣
=
0
Hence, proved.
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0
Similar questions
Q.
The equation,
[
1
x
y
]
⎡
⎢
⎣
1
3
1
0
2
−
1
0
0
1
⎤
⎥
⎦
⎡
⎢
⎣
1
x
y
⎤
⎥
⎦
=
[
0
]
has for
(i)
y
=
0
, (p) rational roots
(ii)
y
=
−
1
(q) irrational roots
(r) integral roots
Q.
If
x
>
y
>
z
>
0
, then find the value of
cot
−
1
x
y
+
1
x
−
y
+
cot
−
1
y
z
+
1
y
−
z
+
cot
−
1
z
x
+
1
z
−
x
Q.
Solve:
d
y
d
x
+
√
(
x
2
−
1
)
(
y
2
−
1
)
x
y
=
0
Q.
cot
−
1
x
y
+
1
x
−
y
+
cot
1
y
z
+
1
y
−
z
+
cot
−
1
x
z
+
1
z
−
x
Q.
The equation of the bisectors of angle between the lines represented by equation
(
y
−
m
x
)
2
=
(
x
+
m
y
)
2
is