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Byju's Answer
Standard XII
Mathematics
Properties of Determinants
Using propert...
Question
Using properties of determinants, prove that:
∣
∣ ∣
∣
1
+
a
1
1
1
1
+
b
1
1
1
1
+
c
∣
∣ ∣
∣
=
a
b
c
+
b
c
+
c
a
+
a
b
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Solution
△
=
∣
∣ ∣
∣
1
+
a
1
1
1
1
+
b
1
1
1
1
+
c
∣
∣ ∣
∣
R
1
→
R
1
−
R
2
R
3
→
R
3
−
R
2
=
∣
∣ ∣
∣
a
−
b
0
1
1
+
b
1
0
−
b
c
∣
∣ ∣
∣
=
a
(
c
+
b
c
+
b
)
+
b
c
=
a
b
+
b
c
+
a
c
+
a
b
c
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Similar questions
Q.
Using properties of determinants, prove the following:
∣
∣ ∣
∣
1
+
a
1
1
1
1
+
b
1
1
1
1
+
c
∣
∣ ∣
∣
=
a
b
+
b
c
+
c
a
+
a
b
c
Q.
Show that
∣
∣ ∣
∣
1
+
a
1
1
1
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b
1
1
1
1
+
c
∣
∣ ∣
∣
=
a
b
c
(
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+
1
a
+
1
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=
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Q.
Using properties of determinants, prove that:
∣
∣ ∣
∣
1
+
a
1
1
1
1
+
b
1
1
1
1
+
c
∣
∣ ∣
∣
=
a
b
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b
c
+
c
a
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a
b
Q.
Using properties of determinants, prove the following :
⎡
⎢
⎣
a
a
2
b
c
b
b
2
c
a
c
c
2
a
b
⎤
⎥
⎦
=
(
a
−
b
)
(
b
−
c
)
(
c
−
a
)
(
b
c
+
c
a
+
a
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)
Q.
Using properties of determinants, prove that:
∣
∣ ∣ ∣
∣
a
2
+
1
a
b
a
c
b
a
b
2
+
1
b
c
c
a
c
b
c
2
+
1
∣
∣ ∣ ∣
∣
=
a
2
+
b
2
+
c
2
+
1
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