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Byju's Answer
Standard XII
Mathematics
Properties of Determinants
By using prop...
Question
By using properties of determination, show that
∣
∣ ∣ ∣
∣
1
+
a
2
−
b
2
2
a
b
−
2
b
2
a
b
1
−
a
2
+
b
2
−
2
b
2
b
−
2
a
1
−
a
2
−
b
2
∣
∣ ∣ ∣
∣
=
(
1
+
a
2
+
b
2
)
3
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Solution
∣
∣ ∣ ∣
∣
1
+
a
2
−
b
2
2
a
b
−
2
b
2
a
b
1
−
a
2
+
b
2
−
2
b
2
b
−
2
a
1
−
a
2
−
b
2
∣
∣ ∣ ∣
∣
c
1
⇒
c
1
−
b
c
3
;
c
2
⇒
c
2
+
a
c
3
∣
∣ ∣ ∣
∣
1
+
a
2
+
b
2
0
−
2
b
0
1
+
a
2
+
b
2
2
a
2
b
−
b
(
1
−
a
2
−
b
2
)
−
2
a
+
a
(
1
−
a
2
−
b
2
)
1
−
a
2
−
b
2
∣
∣ ∣ ∣
∣
=
∣
∣ ∣ ∣
∣
1
+
a
2
+
b
2
0
−
2
b
0
1
+
a
2
+
b
2
2
a
2
b
−
b
+
b
(
a
2
+
b
2
)
−
2
a
+
a
−
a
(
a
2
+
b
2
)
1
−
a
2
−
b
2
∣
∣ ∣ ∣
∣
=
∣
∣ ∣ ∣
∣
1
+
a
2
+
b
2
0
−
2
b
0
1
+
a
2
+
b
2
2
a
b
(
1
+
a
2
+
b
2
)
−
a
(
1
+
a
2
+
b
2
)
1
−
a
2
−
b
2
∣
∣ ∣ ∣
∣
=
(
1
+
a
2
+
b
2
)
2
∣
∣ ∣ ∣
∣
1
0
−
2
b
0
1
2
a
b
−
a
1
−
a
2
−
b
2
∣
∣ ∣ ∣
∣
=
(
1
+
a
2
+
b
2
)
2
(
1
+
a
2
+
b
2
)
=
(
1
+
a
2
+
b
2
)
3
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Similar questions
Q.
Using properties of determinants, prove that:
∣
∣ ∣ ∣
∣
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+
a
2
−
b
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∣
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+
a
2
−
b
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1
−
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∣
∣ ∣ ∣
∣
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Q.
By using properties of determinants, show that: