1.Algebra Booster

7.64 Algebra Booster
a b c
b c a = 0
c a b
fi (a 3 + b 3 + c 3 – 3abc) = 0
fi (a + b + c)(a 2 + b 2 + c 2 – ab – bc – ca) = 0
fi (a + b + c)(a + bw + cw 2 )(a + bw 2 + cw) = 0
fi (a + b + c) = 0,
(a + bw + cw 2 ) = 0
and (a + bw 2 + cw) = 0
When (a + b + c) = 0
fi c = –(a + b)
Now, first two given equations can be written as
ax + by – (a + b)z = 0
bx – (a + b)y + az = 0
Solving, we get
fi
x y z
= =
2 2 2
ab -( a + b) - b( a + b) -a - a( a + b)
-b
x y z
= =
2 2 2 2 2 2
-( a + ab+ b ) -( a + ab+ b ) -( a + ab+
b )
fi
x y z
= = 1 1 1
fi x : y : z = 1 : 1 : 1
Hence, the result.
21. The given equations can be written as
x - ay + az = 0
bx + y - bz = 0
cx -cy - z = 0
For a non-trivial solutions,
D = 0
1 -a
a
b 1 - b = 0
c -c
-1
fi 1(–1 – bc) + a(–b + bc) + a(–bc – c) = 0
fi –1 – bc – ab + abc – abc – ac = 0
fi –1 – bc – ab – ac = 0
fi 1 + ab + bc + ca = 0
22. The given system of equations can be written as
1 + bc + qr = 0
1 + ca + rp = 0
1 + ab + pq = 0
Multiply Eq. (i) by ap, Eq. (ii) by bq and Eq. (iii) by cr,
we get
ap + (abc)p + (pqr)a = 0
bq + (abc)q + (pqr)b = 0
cr + (abc)r + (pqr)c = 0
Put abc = x and pqr = y, we get
ap + px + ay = 0
bq + qx + by = 0
cr + rx + cy = 0
The above system of equations will be consistent,
if D = 0
ap p a
fi bq q b = 0
cr c r
Hence, the result.
23. The given determinant is
24. We have,
cosec a 1 0
1 2coseca
1
0 1 2coseca
pa qb rc
qc ra pb
rb pc qa
2
= cosec a(4 cosec a -1) -2 cosec a
2
= cosec a(4 cosec a -3)
Ê
2
cos a ˆ
= cosec a Á4 + 1
Ë 2 ˜
sin a ¯
Ê
2 2
4 cos a + sin aˆ
= Á
Ë
3 ˜
sin a ¯
Ê
2
4–3sin a ˆ
= Á
Ë 3 ˜
sin a ¯
2Êaˆ 2Êaˆ
4 – 12 sin Á cos
Ë
˜
2¯ Á
Ë
˜
2¯
=
3Êaˆ 3Êaˆ
8sin Á cos
Ë
˜ Á ˜
2¯ Ë 2¯
Ê
2Êaˆ 2Êaˆˆ
1–3sin cos
= 1 Á Á
Ë
˜
2¯ Á
Ë
˜
2¯˜
¥ Á ˜
2 sin 3Êaˆ cos 3Êaˆ
Á Á ˜ Á ˜
Ë Ë 2¯
Ë 2¯
˜
¯
Ê 6Êaˆ 6Êaˆˆ
sin cos
1 Á Á +
Ë
˜
2¯ Á
Ë
˜
2¯˜
= ¥Á
2 3 a 3 a
˜
Á
Ê ˆ Ê ˆ
sin Á ˜cos
Á ˜ ˜
Ë Ë 2¯ Ë 2¯
¯
Ê 3Êaˆ 3Êaˆˆ
sin cos
1 Á Á
Ë
˜
2¯ Á
Ë
˜
2¯˜
= ¥ Á +
2 3 a 3 a
˜
Á
Ê ˆ Ê ˆ
cos Á ˜ sin Á ˜ ˜
Ë Ë 2¯ Ë 2¯
¯
1 Ê 3Êaˆ 3Êaˆˆ
= ¥ tan cot
2
Á Á ˜ + Á ˜
Ë Ë 2¯ Ë 2¯˜
¯
= pa(qra 2 – p 2 bc) – qb(q 2 ac – prb 2 ) + rc(pqc 2 – r 2 ab)
= pqr(a 3 + b 3 + c 3 ) – abc(p 3 – q 3 + r 3 )
= pqr(a 3 + b 3 + c 3 – 3abc) – abc(p 3 – q 3 + r 3 – 3pqr)
= pqr(a 3 + b 3 + c 3 – 3abc) – 0, since p + q + r = 0
= pqr(a 3 + b 3 + c 3 – 3abc)