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Byju's Answer
Standard XII
Mathematics
Properties of Determinants
Prove that ...
Question
Prove that
∣
∣ ∣ ∣
∣
a
2
+
1
a
b
a
c
a
b
b
2
+
1
b
c
c
a
c
b
c
2
+
1
∣
∣ ∣ ∣
∣
=
1
+
a
2
+
b
2
+
c
2
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Solution
Let , L.H.S.
Taking out common
a
,
b
,
c
=
a
b
c
∣
∣ ∣ ∣ ∣
∣
a
+
1
a
b
c
a
b
+
1
b
c
a
b
c
+
1
c
∣
∣ ∣ ∣ ∣
∣
multiplying
a
,
b
,
c
in
c
1
,
c
2
,
c
3
respectively
=
∣
∣ ∣ ∣
∣
a
2
+
1
b
2
c
2
a
2
b
2
+
1
c
2
a
2
b
2
c
2
+
1
∣
∣ ∣ ∣
∣
Applying
c
1
→
c
1
+
c
2
+
c
3
=
∣
∣ ∣ ∣
∣
1
+
a
2
+
b
2
+
c
2
b
2
c
2
1
+
a
2
+
b
2
+
c
2
b
2
+
1
c
2
1
+
a
2
+
b
2
+
c
2
b
2
c
2
+
1
∣
∣ ∣ ∣
∣
Taking out common
1
+
a
2
+
b
2
+
c
2
from
c
1
=
(
1
+
a
2
+
b
2
+
c
2
)
∣
∣ ∣ ∣
∣
1
b
2
c
2
1
b
2
+
1
c
2
1
b
2
c
2
+
1
∣
∣ ∣ ∣
∣
Applying
R
2
→
R
2
−
R
1
R
3
→
R
3
−
R
1
=
(
1
+
a
2
+
b
2
+
c
2
)
∣
∣ ∣
∣
1
b
2
c
2
0
1
0
0
0
1
∣
∣ ∣
∣
Expanding along
c
1
=
1
+
a
2
+
b
2
+
c
2
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Similar questions
Q.
Using properties of determination, prove that
∣
∣ ∣ ∣
∣
a
2
+
1
a
b
a
c
a
b
b
2
+
1
b
c
c
a
c
b
c
2
+
1
∣
∣ ∣ ∣
∣
=
1
+
a
2
+
b
2
+
c
2
Q.
Using properties of determinants, prove that
∣
∣ ∣ ∣
∣
a
2
+
1
a
b
a
c
a
b
b
2
+
1
b
c
c
a
c
b
c
2
+
1
∣
∣ ∣ ∣
∣
=
1
+
a
2
+
b
2
+
c
2
Q.
∣
∣ ∣ ∣
∣
a
2
+
1
a
b
a
c
a
b
b
2
+
1
b
c
c
a
c
b
c
2
+
1
∣
∣ ∣ ∣
∣
=
1
+
a
2
+
b
2
+
c
2
Q.
Show that: det
⎛
⎜
⎝
a
2
+
1
a
b
a
c
a
b
b
2
+
1
b
c
c
a
c
b
c
2
+
1
⎞
⎟
⎠
=
1
+
a
2
+
b
2
+
c
2
Q.
Solve:
∣
∣ ∣ ∣
∣
a
2
+
1
a
b
a
c
a
b
b
2
+
1
b
c
c
a
c
b
c
2
+
1
∣
∣ ∣ ∣
∣
=
1
+
a
2
+
b
2
+
c
2
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