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AP EAPCET 20-Aug-2021 Shift 2 Paper
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© examsnet.com
Question : 5
Total: 160
If
k
∈
R
and
det
A
=
|
a
1
b
1
c
1
a
2
b
2
c
2
a
3
b
3
c
3
|
=
k
then
det
B
=
|
a
1
b
1
c
1
a
2
+
2
a
1
b
2
+
2
b
1
c
2
+
2
c
1
a
3
b
3
c
3
|
is equal to
0
2k
k
k
2
Validate
Solution:
det
A
=
|
a
1
b
1
c
1
a
2
b
2
c
2
a
3
b
3
c
3
|
=
k
(given)
det
B
=
|
a
1
b
1
c
1
a
2
+
2
a
1
b
2
+
2
b
1
c
2
+
2
c
1
a
3
b
3
c
3
|
=
|
a
1
b
1
c
1
a
2
b
2
c
2
a
3
b
3
c
3
|
+
|
a
1
b
1
c
1
2
a
2
2
b
2
2
c
2
a
3
b
3
c
3
|
=
k
+
0
⇒
det
B
=
k
© examsnet.com
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