1.Algebra Booster

Matrices and Determinants 7.71
fi (l + 5)z = 0
fi z = 0
Thus, for all real l π –5, we have
4 9
x = , y = - , z = 0
7 7
Now, for l = –5, then
1 1
z = k, x = (4- 5 k), y = (13k
-9)
7 7
11. We have
1 a bc
D= 1 b ca
1 c ab
=
1
= 1
1
1
2
b b abc
abc
c c
2 abc
a
abc
= ¥ b
abc
c
=D¢
a
b
c
2
2
2
2
a a abc
a
b
c
a
b
c
2
2
2
12. Since the system of equations has a non-zero solution,
l 1 1
- 1 l 1 = 0
-1 -1
l
fi l(l 2 + 1) – (1 – l) + (1 + l) = 0
fi l 3 + l – 1 + l + 1 + l = 0
fi l 3 + 3l = 0
fi (l 3 + 3)l = 0
fi l = 0
13. Since a is a repeated root of the quadratic equation
f(x) = 0, we can write
f(x) = a(x Рl) 2 , a ΠR
Also, it is given that, F(x) is a polynomial of degree
atmost 5 and F(a) = 0.
Thus,
A¢ () x B¢ () x C¢
() x
F ¢(x) = A( a) B( a) C( a)
A¢ ( a) B¢ ( a) C¢
( a)
fi F ¢(a) = 0
Thus, l is a repeated root of the polynomial F(x).
Therefore, F(x) is divisible by f(x).
1
1
1
14. We have,
D=
=
x x x
Cr Cr+ 1 Cr+
2
y y y
Cr Cr+ 1 Cr+
2
z z z
Cr Cr+ 1 Cr+
2
x x x x x
Cr Cr + Cr+ 1 Cr+ 1+
Cr+
2
y y y y y
Cr Cr Cr+ 1 Cr+ 1 Cr+
2
z z z z z
Cr Cr + Cr+ 1 Cr+ 1+
Cr+
2
= + +
=
ÊC2Æ C2+
C1ˆ
Á
ËC Æ C + C ˜
¯
x x+ 1 x+
1
Cr Cr+ 1 Cr+
2
y y+ 1 y+
1
Cr Cr+ 1 Cr+
2
z z+ 1 z+
1
Cr Cr+ 1 Cr+
2
x x+ 1 x+ 1 x+
1
Cr Cr+ 1 Cr+ 1+
Cr+
2
y y+ 1 y+ 1 y+
1
Cr Cr+ 1 Cr+ 1 Cr+
2
z z+ 1 z+ 1 z+
1
Cr Cr+ 1 Cr+ 1+
Cr+
2
= +
x x+ 1 x+
2
Cr Cr+ 1 Cr+
2
y y+ 1 y+
2
Cr Cr+ 1 Cr+
2
z z+ 1 z+
2
Cr Cr+ 1 Cr+
2
Hence, the result.
15. We have,
x1 y1 1 a1 b1
1
x2 y2 1 = a2 b2
1
x3 y3 1 a3 b3
1
x1 y1 1 a1 b1
1
1 1
fi x2 y2 1 = a2 b2
1
2 2
x y 1 a b 1
3 3 3 3
3 3 2
(C 3
Æ C 3
+ C 2
)
fi D 1
= D 2
Areas of two triangles are the same but they may not be
congruent.
16. Since the system of equations has a non-trivial solution,
so
fi
sin (3 q) -1 1
cos (2 q) 4 3 = 0
2 7 7
sin (3 q) 0 1
cos (2 q) 7 3 = 0
2 14 7
(C 2
Æ C 2
+ C 3
)