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Byju's Answer
Standard VIII
Mathematics
Laws of Exponents
Evaluate the ...
Question
Evaluate the following determinant:
(i)
1
+
a
1
1
1
1
+
a
1
1
1
1
+
a
=
a
3
+
3
a
2
(ii)
a
2
+
2
a
2
a
+
1
1
2
a
+
1
a
+
2
1
3
3
1
=
a
-
1
3
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Solution
(ii) To Prove:
a
2
+
2
a
2
a
+
1
1
2
a
+
1
a
+
2
1
3
3
1
=
a
-
1
3
LHS
=
a
2
+
2
a
2
a
+
1
1
2
a
+
1
a
+
2
1
3
3
1
Applying
R
1
→
R
1
-
R
2
=
a
2
+
2
a
-
2
a
-
1
2
a
+
1
-
a
-
2
1
-
1
2
a
+
1
a
+
2
1
3
3
1
=
a
2
-
1
a
-
1
0
2
a
+
1
a
+
2
1
3
3
1
Taking
a
-
1
common
from
R
1
=
a
-
1
a
+
1
1
0
2
a
+
1
a
+
2
1
3
3
1
Applying
R
2
→
R
2
-
R
3
=
a
-
1
a
+
1
1
0
2
a
+
1
-
3
a
+
2
-
3
1
-
1
3
3
1
=
a
-
1
a
+
1
1
0
2
a
-
2
a
-
1
0
3
3
1
Taking
a
-
1
common
from
R
2
=
a
-
1
2
a
+
1
1
0
2
1
0
3
3
1
Expanding
through
C
3
=
a
-
1
2
1
1
a
+
1
-
2
=
a
-
1
2
1
a
+
1
-
2
=
a
-
1
2
a
-
1
=
a
-
1
3
=
RHS
Hence,
a
2
+
2
a
2
a
+
1
1
2
a
+
1
a
+
2
1
3
3
1
=
a
-
1
3
.
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0
Similar questions
Q.
Using properties of determinants, prove that
∣
∣ ∣
∣
a
2
+
2
a
2
a
+
1
1
2
a
+
1
a
+
2
1
3
3
1
∣
∣ ∣
∣
=
(
a
−
1
)
3
Q.
Prove that:
∣
∣ ∣
∣
a
2
+
2
a
2
a
+
1
1
2
a
+
1
a
+
2
1
3
3
1
∣
∣ ∣
∣
=
(
a
−
1
)
2
Q.
Prove that:
∣
∣ ∣
∣
a
2
+
2
a
2
a
+
1
1
2
a
+
1
a
+
2
1
3
3
1
∣
∣ ∣
∣
=
(
a
−
1
)
C
then
C
=
?
Q.
1
+
a
1
1
1
1
+
a
a
1
1
1
+
a
=
a
3
+
3
a
2
Q.
The value of the determinant
∣
∣ ∣
∣
1
+
a
1
1
1
1
+
a
1
1
1
1
+
a
∣
∣ ∣
∣
is
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