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Byju's Answer
Standard XII
Mathematics
Properties of Determinants
Evaluate: 1...
Question
Evaluate:
∣
∣ ∣ ∣
∣
1
b
c
a
(
b
+
c
)
1
c
a
b
(
c
+
a
)
1
a
b
c
(
a
+
b
)
∣
∣ ∣ ∣
∣
A
0
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B
1
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C
a
b
c
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D
a
2
+
b
2
+
c
2
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Solution
The correct option is
A
0
Performing
C
2
→
C
2
a
b
c
C
3
→
C
3
a
b
c
and then
C
3
→
C
3
+
C
2
∣
∣ ∣ ∣ ∣
∣
1
1
a
1
a
+
1
b
+
1
c
1
1
b
1
a
+
1
b
+
1
c
1
1
c
1
a
+
1
b
+
1
c
∣
∣ ∣ ∣ ∣
∣
=
(
1
a
+
1
b
+
1
c
)
∣
∣ ∣ ∣ ∣
∣
1
1
a
1
1
1
b
1
1
1
c
1
∣
∣ ∣ ∣ ∣
∣
=
0
(Two columns are identical on taking
1
a
+
1
b
+
1
c
common)
Suggest Corrections
0
Similar questions
Q.
Show that
∣
∣ ∣ ∣
∣
1
b
c
a
(
b
+
c
)
1
c
a
b
(
c
+
a
)
1
a
b
c
(
a
+
b
)
∣
∣ ∣ ∣
∣
=
0
Q.
∣
∣ ∣ ∣
∣
1
b
c
a
(
b
+
c
)
1
c
a
b
(
c
+
a
)
1
a
b
c
(
a
+
b
)
∣
∣ ∣ ∣
∣
=
Q.
Solve
∣
∣ ∣ ∣
∣
1
b
c
a
(
b
+
c
)
1
c
a
b
(
c
+
a
)
1
a
b
c
(
a
+
b
)
∣
∣ ∣ ∣
∣
Q.
What is the value of the determinant
∣
∣ ∣ ∣
∣
1
b
c
a
(
b
+
c
)
1
c
a
b
(
c
+
a
)
1
a
b
c
(
a
+
b
)
∣
∣ ∣ ∣
∣
?
Q.
Using the properties of determinant and without expanding , prove that:
∣
∣ ∣ ∣
∣
1
b
c
a
(
b
+
c
)
1
c
a
b
(
c
+
a
)
1
a
b
c
(
a
+
b
)
∣
∣ ∣ ∣
∣
=
0
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