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Byju's Answer
Standard XII
Mathematics
Finding Inverse Using Elementary Transformations
Solve: 1+a2...
Question
Solve:
∣
∣ ∣ ∣
∣
1
+
a
2
−
b
2
2
a
b
−
2
b
2
a
b
1
−
a
2
+
b
2
2
a
2
b
−
2
a
1
−
a
2
−
b
2
∣
∣ ∣ ∣
∣
=
(
1
+
a
2
+
b
2
)
3
.
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Solution
Now,
∣
∣ ∣ ∣
∣
1
+
a
2
−
b
2
2
a
b
−
2
b
2
a
b
1
−
a
2
+
b
2
2
a
2
b
−
2
a
1
−
a
2
−
b
2
∣
∣ ∣ ∣
∣
{
R
′
1
=
R
1
+
b
R
3
and
R
′
2
=
R
2
−
a
R
1
]
=
∣
∣ ∣ ∣
∣
1
+
a
2
+
b
2
0
−
b
(
1
+
a
2
+
b
2
)
0
1
+
a
2
+
b
2
a
(
1
+
a
2
+
b
2
)
2
b
−
2
a
1
−
a
2
−
b
2
∣
∣ ∣ ∣
∣
=
(
1
+
a
2
+
b
2
)
2
∣
∣ ∣
∣
1
0
−
b
0
1
a
2
b
−
2
a
1
−
a
2
−
b
2
∣
∣ ∣
∣
=
(
1
+
a
2
+
b
2
)
2
{
1
−
a
2
−
b
2
+
2
a
2
−
b
(
−
2
b
)
}
=
(
1
+
a
2
+
b
2
)
3
.
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0
Similar questions
Q.
∣
∣ ∣ ∣
∣
1
+
a
2
−
b
2
2
a
b
−
2
b
2
a
b
1
−
a
2
+
b
2
2
a
2
b
−
2
a
1
−
a
2
−
b
2
∣
∣ ∣ ∣
∣
=
(
1
+
a
2
+
b
2
)
3
Q.
Using properties of determinants, prove that:
∣
∣ ∣ ∣
∣
1
+
a
2
−
b
2
2
a
b
−
2
b
2
a
b
1
−
a
2
+
b
2
2
a
2
b
−
2
a
1
−
a
2
−
b
2
∣
∣ ∣ ∣
∣
=
(
1
+
a
2
+
b
2
)
3
Q.
Using properties of determinant, prove the following:
∣
∣ ∣ ∣
∣
1
+
a
2
−
b
2
2
a
b
−
2
b
2
a
b
1
−
a
2
+
b
2
2
a
2
b
−
2
a
1
−
a
2
−
b
2
∣
∣ ∣ ∣
∣
=
(
1
+
a
2
+
b
2
)
3
Q.
Find
n
if:
∣
∣ ∣ ∣
∣
1
+
a
2
−
b
2
2
a
b
−
2
b
2
a
b
1
−
a
2
+
b
2
2
a
2
b
−
2
a
1
−
a
2
−
b
2
∣
∣ ∣ ∣
∣
=
(
1
+
a
2
+
b
2
)
n
Q.
The value of determinant
∣
∣ ∣ ∣
∣
1
+
a
2
−
b
2
2
a
b
−
2
b
2
a
b
1
−
a
2
+
b
2
2
a
2
b
−
2
a
1
−
a
2
−
b
2
∣
∣ ∣ ∣
∣
is
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