1.Algebra Booster

7.58 Algebra Booster
27. The given determinant is
13 + 3 2 5 5
15 + 26 5 10
3+
65 15 5
13 2 5 5 3 2 5 5
= 26 5 10 + 15 5 10
65 15 5 3 15 5
1 2 1
= 13 ◊ 5 ◊ 5 2 5 2
5 3 5
1 2 1
+ 3◊ 5 ◊ 5 5 5 2
3 3 5
1 2 1
= 0+ 3◊ 5 ◊ 5 5 5 2
3 3 5
1 2 1
= 5 3 0 - 5 2-
5
0 - 3 5-
3
1 2 1
= –5 3 0 5 2 - 5
0 3 5-
3
= –5 3[(5 - 15) -( 6 - 15)]
=-5 3(5-
6)
= 5 3( 6 -5)
28. We have
1 2 3
D = 2 4 1 = -20
3 2 9
6 2 3
D = 17 4 1 = -20
1
2 2 9
1 6 3
D2
= 2 17 1 = -80
3 2 9
1 2 6
D = 2 4 17 = 20
3
3 2 2
Thus,
D1
-20
x = = = 1
D -20
D2
-80
y = = = 4
D -20
D3
20
z = = = -1
D -20
29. We have,
2 p 6
D=
1 2 q
1 1 3
8 p 6
D1
= 5 2 q
4 1 3
2 8 6
D2
= 1 5 q
1 4 3
2 p 8
D3
= 1 2 5
1 1 4
(i) The given system of equations has no solution if D
= 0 and any one of D 1
, D 2
and D 3
is non-zero.
Now, D = 0 gives
2 p 6
1 2 q = 0
1 1 3
fi 2(6 – q) – p(3 – q) + 6(1 – 2) = 0
fi 12 – 2q – 3p + pq – 6 = 0
fi 6 – 2q – 3p + pq = 0
fi (3 – q)(2 – p) = 0
fi p = 2 , q = 3
8 p 6
Also, D1
= 5 2 q = (2- p)(15-4 q)
4 1 3
Thus, D 1
π 0 gives
(2 – p) (15 – 4q) π 0
15
fi p π 2, q π
4
Hence, the given system of equations has no solution,
if p π 2, q π 3.
(ii) The given system of equations has a unique solution,
if D π 0.
fi (p – 2)(q – 3) π 0
fi p π 2 and q π 3
(iii) The given system of equations has infinitely many
solutions, if D = 0 = D 1
+ D 2
= D 3