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Shubhrajyoti Ghosh
Subject: Maths
, asked on 13/3/18
Question) Solve it.
P
r
o
v
e
t
h
a
t
(
a
+
b
)
2
c
a
b
c
c
a
(
b
+
c
)
2
a
b
b
c
a
b
(
c
+
a
)
2
=
2
b
c
(
a
+
b
+
c
)
3
Answer
1
Arnapurna Paikaray
Subject: Maths
, asked on 28/2/18
Can we solve this by using some properties of determinant?
Answer
1
Arnapurna Paikaray
Subject: Maths
, asked on 28/2/18
How can we take common xyz from the determinant ? All xyz are in different columns and rows.
Answer
1
Arnapurna Paikaray
Subject: Maths
, asked on 28/2/18
Q 25 main .Please don't send links.
25
.
If
x
+
y
+
z
=
0
,
prove
that
:
xa
yb
zc
yc
za
xb
zb
xc
ya
=
xyz
a
b
c
c
a
b
b
c
a
Answer
1
Arnapurna Paikaray
Subject: Maths
, asked on 28/2/18
25
)
I
f
x
+
y
+
z
=
0
,
p
r
o
v
e
t
h
a
t
:
x
a
y
b
z
c
y
c
z
a
x
b
z
b
x
c
y
a
=
x
y
z
a
b
c
c
a
b
b
c
a
O
R
E
v
a
l
u
a
t
e
3
x
-
x
+
y
-
x
+
z
x
-
y
3
y
z
-
y
x
-
z
y
-
z
3
z
Answer
1
Dhinesh Vijayan
Subject: Maths
, asked on 23/2/18
Q.25.
x
a
y
b
z
c
y
c
z
a
x
b
z
b
x
c
y
a
=
x
y
c
a
b
c
c
a
b
b
c
a
Answer
1
Bhavya Jain
Subject: Maths
, asked on 23/2/18
Please help with this question ....
Q Using properties of determinants, prove that
a
+
b
2
c
c
c
a
b
+
c
2
a
a
b
b
c
+
a
2
b
=
2
a
+
b
+
c
3
Answer
1
Vibhav
Subject: Maths
, asked on 14/2/18
Is |A||A^-1| = 1
We know that A.A^-1 = I
Taking dterminant both sides we get |A||A^-1| = 1
However A^-1 = 1/|A| * (ADJ. A)
If I take determinant on both sides I get
|A^-1| = |Adj.A|/|A|
|A||A^-1| = |Adj.A|
Clearly |Adj.A| will NOT always be equal to 1
So what is |A||A^-1| and What is the ERROR in one of the approaches???
Answer
1
Tithi
Subject: Maths
, asked on 5/2/18
Solve this:
p
r
o
v
e
t
h
a
t
a
2
2
a
b
b
2
b
2
a
2
2
a
b
2
a
b
b
2
a
2
=
a
3
+
b
3
2
u
sin
g
p
r
o
p
e
r
t
i
e
s
o
f
d
e
t
e
r
m
i
n
a
t
e
s
.
Answer
1
Raghav Soni
Subject: Maths
, asked on 25/1/18
If [.] denotes the greatest integer less than or equal to real number under consideration and -1<=x<0, 0<=y<1, 1<=z<2 then find the value of the determinant
x
+
1
y
z
x
y
+
1
z
x
y
z
+
1
Answer
1
Vibhav
Subject: Maths
, asked on 24/1/18
Cramer's rule of solving linear equations by determinants please explain what is it??
Answer
1
Tithi
Subject: Maths
, asked on 22/1/18
Using properties of determinants, prove that
a
2
+
1
a
b
a
c
a
b
b
2
+
1
b
c
c
a
c
b
c
2
+
1
=
1
+
a
2
+
b
2
Answer
1
Shubhrajyoti Ghosh
Subject: Maths
, asked on 15/1/18
Q. For any matrix A , if
A
-
1
exists then which of the following is not true?
a
.
A
-
1
-
1
=
A
b
.
A
T
-
1
=
A
-
1
T
c
.
A
2
-
1
=
A
-
1
2
d
.
A
-
1
=
A
-
1
Answer
1
Shubhrajyoti Ghosh
Subject: Maths
, asked on 13/1/18
If
a
+
i
b
c
+
i
d
-
c
+
i
d
a
-
i
b
×
α
-
i
β
c
-
i
d
-
γ
-
i
δ
α
+
i
β
=
A
-
i
B
C
-
i
D
-
C
-
i
D
A
+
i
B
, write down the values of
A, B, C, D
. Hence, show that the product of two sums, each of four squares, can be expressed as the sum of four squares
Answer
1
Shubhrajyoti Ghosh
Subject: Maths
, asked on 13/1/18
No links please
Q.36.
1
1
x
p
+
3
p
+
1
p
+
x
3
x
+
1
x
+
1
=
0
Ans : 1 and 2
Answer
1
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What are you looking for?
Q Using properties of determinants, prove that
We know that A.A^-1 = I
Taking dterminant both sides we get |A||A^-1| = 1
However A^-1 = 1/|A| * (ADJ. A)
If I take determinant on both sides I get
|A^-1| = |Adj.A|/|A|
|A||A^-1| = |Adj.A|
Clearly |Adj.A| will NOT always be equal to 1
So what is |A||A^-1| and What is the ERROR in one of the approaches???
Q. For any matrix A , if exists then which of the following is not true?
Q.36.
Ans : 1 and 2