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Byju's Answer
Standard XII
Mathematics
Properties of Determinants
begin vmatrix...
Question
∣
∣ ∣
∣
a
+
b
a
+
2
b
a
+
3
b
a
+
2
b
a
+
3
b
a
+
4
b
a
+
4
b
a
+
5
b
a
+
6
b
∣
∣ ∣
∣
=
A
a
2
+
b
2
+
c
2
−
3
a
b
c
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B
3ab
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C
3a+5b
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Solution
∣
∣ ∣
∣
a
+
b
a
+
2
b
a
+
3
b
a
+
2
b
a
+
3
b
a
+
4
b
a
+
4
b
a
+
5
b
a
+
6
b
∣
∣ ∣
∣
=
∣
∣ ∣
∣
a
+
b
a
+
2
b
a
+
3
b
b
b
b
2
b
2
b
2
b
∣
∣ ∣
∣
=
0
b
y
{
R
2
→
R
2
−
R
1
R
3
→
R
3
−
R
2
}
Trick : Putting a=1=b.
The determinant will be
∣
∣ ∣
∣
2
3
4
3
4
5
5
6
7
∣
∣ ∣
∣
=
0
.
Suggest Corrections
1
Similar questions
Q.
∣
∣ ∣
∣
a
+
b
a
+
2
b
a
+
3
b
a
+
2
b
a
+
3
b
a
+
4
b
a
+
4
b
a
+
5
b
a
+
6
b
∣
∣ ∣
∣
=
Q.
The value of determinant is
∣
∣ ∣
∣
a
+
b
a
+
2
b
a
+
3
b
a
+
2
b
a
+
3
b
a
+
4
b
a
+
4
b
a
+
5
b
a
+
6
b
∣
∣ ∣
∣
is ________.
Q.
∣
∣ ∣
∣
a
a
+
b
a
+
b
+
c
2
a
3
a
+
2
b
4
a
+
3
b
+
2
c
3
a
6
a
+
3
b
10
a
+
6
b
+
3
a
∣
∣ ∣
∣
=
a
3
Q.
Prove that
∣
∣ ∣
∣
a
a
+
b
a
+
b
+
c
2
a
3
a
+
2
b
4
a
+
3
b
+
2
c
3
a
6
a
+
3
b
10
a
+
6
b
+
3
c
∣
∣ ∣
∣
=
a
3
.
Q.
Show that
∣
∣ ∣
∣
a
a
+
b
a
+
2
b
a
+
2
b
a
a
+
b
a
+
b
a
+
2
b
a
∣
∣ ∣
∣
=
9
b
2
(
a
+
b
)
.
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