1.Algebra Booster

Matrices and Determinants 7.77
fi (sin x + 2 cos x)(sin x – cos x) 2 = 0
fi (sin x + 2 cos x) = 0, (sin x – cos x) = 0
fi tan x = –2, tan x = 1
since
p p
- £ x £
4 4
fi –1 £ tan x £ 1
p
Hence, the number of solutions is 1 at x = .
4
40. The given determinant is
ax -by - c bx + ay cx + a
bx + ay - ax + by - c cy + b
cx + a cy + b -ax - by + c
Applying C 1
Æ aC 1
+ bC 2
+ cC 3
, we get
2 2 2
( )
1 (
2 2 2 )
a
2 2 2
a + b + c x bx + ay cx + a
a + b + c y - ax + by - c cy + b
( a + b + c ) cy + b -ax - by + c
x bx + ay cx + a
1
= y - ax + by - c cy + b
a
1 cy + b -ax - by + c
ÊC2ÆC2-bC1ˆ
Applying Á
ËC ÆC -cC
˜ , we get
¯
3 3 1
x ay a
1
y -ax-c b
a
1 cy -ax -by
Applying R 1
Æ (xR 1
+ yR 2
+ R 3
), we get
x
1
ax
2 2
+ y + 1 0 0
y -ax - c b
1 cy -ax -by
1 (
2 2 1)[( )
2
x y ax acx abxy bcy bcy ]
= + + + + + -
ax
1 (
2 2 1)[( )
2
= x + y + ax + acx + abxy ]
ax
= (x 2 + y 2 + 1)(ax + by + c)
If the given determinant is zero, then
(ax + by + c) = 0, [(x 2 + y 2 + 1) π 0]
Thus, (ax + by + c) = 0 represents a straight line.
41. The given determinant is
1 1 1
1
2
-1-w
2
w
1
2
w
4
w
1 1 1
= 1 w
2
w
1
2
w w
3 1 1
2
= 0 w w -1 ( C1Æ C1 + C2 + C3)
2
0 w w
= 3(w 2 – w 4 )
= 3(w 2 – w)
= 3w(w – 1)
42. Given system of equations are
(k + 1)x + 8y = 4k; kx + (k + 3)y = 3k – 1
has infinitely many solutions, if
( k + 1) 8 4k
= =
k ( k + 3) (3k
-1)
fi
( k + 1) 8
=
k ( k + 3)
fi (k + 1)(k + 3) = 8k
fi k 2 + 4k + 3 – 8k = 0
fi k 2 – 4k + 3 = 0
fi (k – 1)(k – 3) = 0
fi k = 1, 3
43. We have
2
2 Êa 0ˆ Êa 0ˆ
Ê a 0ˆ
A = A◊ A= Á ◊ =
Ë1 1 ˜
¯ Á
Ë1 1 ˜
¯ Á ˜
Ëa
+ 1 1¯
Given relation is
A 2 = B
Ê
2
a 0ˆ Ê1 0ˆ
fi Á ˜ = Á
a 1 1 Ë5 1˜
Ë + ¯ ¯
fi a 2 = 1, a + 1 = 5
fi a = ±1, a = 4
44. The given system of equations has infinitely many solutions,
if
1 a 0
a 0 1 = 0
0 1 a
fi 1(0 – 1) – a(a 2 – 0) = 0
fi –1 – a 3 = 0
fi a 3 = –1
fi a = –1
45. We have,
Êa b cˆÊa b cˆ
T
A A= Áb Á
c a˜Áb ˜Á
c a˜
˜
Ëc a b¯Ëc a b¯
Ê
2 2 2
a + b + c ab + bc + ca ab + bc + caˆ
Á
2 2 2
˜
= Áab + bc + ca a + b + c ab + bc + ca˜
Á
2 2 2˜
Ëab + bc + ac ab + bc + ca a + b + c ¯
Êa b bˆ
= Áb Á
a b˜
˜
Ëb b a¯