Example.1
Given that a = cos q + i sin q, b = cos 2q -i sin2q, c = cos 3q + i sin 3q and if
,
show that q = 2np, n ∈ Z.
Solution
D = ![](http://www.quizsolver.com/radix/dth/notif/Determinanant%20SUB%203_files/image002.gif)
&⇒
a + b + c = 0 or a = b = c
I f a + b + c = 0, we have
cos q
+ cos 2q + cos 3q = 0, sin q - sin 2q + sin 3q = 0
\
cos 2q (2cos q + 1) = 0
and sin 2q
(1 –2cos q) = 0 ....(1).
which is not possible as cos 2q = 0 gives sin 2q >
0, cos q > 1/2
and cosq
= -1/2 gives sin 2q > 0, cos q >
½
\
Equation (1) does not hold simultaneously.
\
a + b + c > 0
\
a = b = c or eiq = e-2iq = e3iq
which is satisfied only by eiq = 1
i.e. cos q
= 1, sin q = 0, so q = 2np, n ∈
Z.
Example.2
For what value(s) of
m does the system of equations 3x + m y = m and 2x - 5y = 20 has a solution
satisfying the conditions x > 0, y > 0.
Solution
By using Cramer's
Rule, the solution of the system is
![](http://www.quizsolver.com/radix/dth/notif/Determinanant%20SUB%203_files/image003.gif)
![](http://www.quizsolver.com/radix/dth/notif/Determinanant%20SUB%203_files/image004.gif)
![](http://www.quizsolver.com/radix/dth/notif/Determinanant%20SUB%203_files/image005.gif)
![](http://www.quizsolver.com/radix/dth/notif/Determinanant%20SUB%203_files/image006.gif)
Example.3
Let l and a be real. Find the set of all values of l for which the system of linear
equations lx + sina .y + cosa . z = 0, x+cosa . y+ sina . z = 0, -x+ sina . y - cosa . z = 0 has a non-trivial solution.
For l = 1, find all the
values of a. .
Solution
The given system has a non-trivial solution
if ![](http://www.quizsolver.com/radix/dth/notif/Determinanant%20SUB%203_files/image007.gif)
By expanding the
determinant we get l = sin 2 a + cos 2 a
Since -
£ sin2a +
cos2a £ ![](http://www.quizsolver.com/radix/dth/notif/Determinanant%20SUB%203_files/image008.gif)
&⇒ -
£ l £ ![](http://www.quizsolver.com/radix/dth/notif/Determinanant%20SUB%203_files/image008.gif)
For l =
1, sin 2a +
cos 2a =
1
&⇒ ![](http://www.quizsolver.com/radix/dth/notif/Determinanant%20SUB%203_files/image009.gif)
, n being an integer.
Example.4
For what values of p and q, the system of equation 2x +
py + 6z = 8, x + 2y + qz = 5, x + y + 3z = 4
has (i) no solution (ii) a unique solution (iii)
infinitely many solutions
Solution
D =
= 2(6 – q) – p(3 –
q) + 6(1 – 2)
= 12 – 2q – 3p + pq – 6 = pq – 2q – 3p + 6 = (p –2)(q
–3)
D1 =
=
8(6 –q) – p(15 – 4q) + 6(5 – 8)
= 48 – 8q – 15p + 4pq –18 = 4pq – 8q – 15p + 30
= 4q(p – 2) – 15(p –2) = (4q – 15)(p –2)
D2 =
=
2(15 – 4q) – 8(3 – q) + 6(4 – 5) = 0
D3 =
=
2(8 –5) – p(4 – 5) + 8(1 – 2) = p –2
Case –I
When D =
0
&⇒ p
= 2, q = 3 and atleast one of D1, D2, D3 is
not zero which is only possible if p > 2 and q = 3.
When D > 0
, i.e. p > 2, q > 3, given system of equation has
unique solution .
Case–II
When D = 0, i.e. p = 2, or q = 3 .
When p = 2, D = 0, D1 =
0, D2 =
0, D3 =
0
\ given system of equation has infinitely many solutions
Case–III
when q = 3, p > 2, D =
0, D1 > 0.
.
Given equation has no solution. .
Example.5
If
a > p,
b > q,
c > r
and
= 0 then find the
value of
.
Solution
By operating R1→ R1 – R2 and R2→ R2 – R3
0 = ![](http://www.quizsolver.com/radix/dth/notif/Determinanant%20SUB%203_files/image017.gif)
= ( p – a){r ( q – b) –b (
c – r)} + a ( b – q ) (c – r)
= (p – a) ( rq – rb) + a( b
– q) (c – r) + b(p-a) (r-c)
=![](http://www.quizsolver.com/radix/dth/notif/Determinanant%20SUB%203_files/image018.gif)
&⇒ ![](http://www.quizsolver.com/radix/dth/notif/Determinanant%20SUB%203_files/image019.gif)
Therefore
.