Example.1
If f, g, h are differentiable functions of x, and D(x)
=
, prove that D'(x)
= ![](http://www.quizsolver.com/radix/dth/notif/Determinanant%20SUB%204_files/image002.gif)
Solution
D(x) =
Operating (i) R2
→ R2
– R1 (ii) R3 → R3 –4 R2
+ 2R1 and shifting x of R2 to R3
D (x) = ![](http://www.quizsolver.com/radix/dth/notif/Determinanant%20SUB%204_files/image004.gif)
&⇒ D¢ (x) = 0 +0 +
. Hence Proved
Example.2
Prove without
expansion that
.
Solution
Rewriting the given
determinant as under
![](http://www.quizsolver.com/radix/dth/notif/Determinanant%20SUB%204_files/image007.gif)
by operating in second determinant C3→ C3
+ bC2 , we get
=
D1 -
=
D1 +![](http://www.quizsolver.com/radix/dth/notif/Determinanant%20SUB%204_files/image010.gif)
=
=
.
Example.3
If A, B, C are angles of a triangles then
prove that
= – 4
Solution
Since A + B + C = p and eip =
cosp +
i sinp =
-1,
ei(B+C) = ei(p -
A)
= – e-iA
e –i(B+C) = –eiA
By taking eiA ,
e iB, eiC common from R1 , R2
and R3 respectively, we have
D = -![](http://www.quizsolver.com/radix/dth/notif/Determinanant%20SUB%204_files/image014.gif)
![](http://www.quizsolver.com/radix/dth/notif/Determinanant%20SUB%204_files/image015.gif)
By taking eiA, e
iB, eiC common from C1, C2 and C3
respectively, we have
D =
= –
4.
Example.4
Consider a determinant of order three
whose all elements are 1 or –1, prove that maximum value of the determinant is
4. .
Solution
Let D =
be
a determinant of order 3,
Where a1, a2, a3,
b1, b2, b3, c1, c2, c3
are 1 or –1
Now D
= a1(b2c3–b3c2) – a2(b1c3
– b3c1) + a3(b1c2 –
b2c1)
= (a1b2c3 +
a2b3c1 + a3b1c2)
– (a1b3c2 + a2b1c3
+ a3b2c1)
value of these six terms will be either –1or
1 \ D £
6 …(1)
Since we have to find the largest value of D, therefore first of all we consider
the case when a1b2c3 = a2b3c1=
a3b1 c2 = 1
&⇒
a1a2a3b1b2b3c1c2c3
= 1×1×1 = 1 …(2)
Let x = a1b3c2
y= a2b1c3,z = a3b2c1,
Then xyz = a1a2a3b1b2b3c1c2c3
=1 [form (2)]
But x = 1 or –1, y = 1 or –1, z = 1 or –1
\
all of x, y,z cannot be –1
Two of x,y,z are 1 and third is –1 is also
not possible
\
either x = y = z = 1 or two of x, y, z are – 1 and other is 1
when x = y = z = 1, D = 0
When two of x, y, z are –1 and third is 1, D = 3 – (–1 –1+1) = 4
This is possible Let a1 = 1, b3
= 1, c2 = –1, a2= 1, b1 = –1, a3 =
b2 = c1 =1
In this case D =
=
4 \ max value of D = 4.
Example.5
Prove
that
= x1 x2 x3
.
Solution
The
given determinant can be written in two determinants as under
![](http://www.quizsolver.com/radix/dth/notif/Determinanant%20SUB%204_files/image021.gif)
Now first can be expanded along C1 and by
taking common b1 from C1 in the second determinant and
applying C2 → C2 – b2C1
and
C3 → (C3 – b3 C1), we can
get the result. .