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Byju's Answer
Standard XII
Mathematics
Properties of Determinants
Without expan...
Question
Without expanding, prove that
Δ
=
∣
∣ ∣
∣
x
+
y
y
+
z
z
+
x
z
x
y
1
1
1
∣
∣ ∣
∣
=
0
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Solution
Applying
R
1
→
R
1
+
R
2
to
Δ
, we get
Δ
=
∣
∣ ∣
∣
x
+
y
+
z
x
+
y
+
z
x
+
y
+
z
z
x
y
1
1
1
∣
∣ ∣
∣
=
0
since the elements of
R
1
and
R
3
are proportional,
Δ
=
0
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