1.Algebra Booster

Matrices and Determinants 7.55
a b c
2
= b c a
c a b
a b c a b c
= b c a ¥ b c a
c a b c a b
a b c a – c b
= b c a ¥ b – a c
c a b c – b a
a b c a b c
= b c a ¥ – c – a – b
c a b b c a
2 2 2
a c 2ac - b
2 2 2
= 2ab - c b a
Hence, the result.
a
17. Let D= a
0
2 2 2
b 2bc - a c
0 c
b 0
b c
The determinant of the co-factors of D is
bc -ca ab
bc ca -ab
-bc ca ab
3-1 2
=D =D
a 0
c
= a b 0
0
a
b
0
c
c a 0 c
= a b 0 ¥ a b 0
0 b c 0 b c
2
2 2 2 2
c + a a c
2 2 2 2
= a a + b b
2 2 2 2
c b b + c
a b c
18. Let D= b c a
c a b
The determinant of the co-factors of D is
2 2 2
bc - a ca -b ab -c
2 2 2
ca -b ab -c bc - a
2 2 2
ab - c bc - a ac -b
3-1 2
=D =D
a b c
= b c a
c a b
a b c a b c
= b c a ¥ b c a
c a b c a b
2
2 2 2
a + b + c ab + bc + ca ab + bc + ca
2 2 2
2 2 2
2 2 2
2 2 2
= ab + bc + ca a + b + c ab + bc + ca
=
2 2 2
ab + bc + ca ab + bc + ca a + b + c
u v v
v u v
v v u
2 2 2
19. Let
x = X, y = Y and z = Z.
2 2 2
a b c
The given system of equations reduces to
X + Y - Z = 1
X - Y + Z = 1
X + Y + Z = 1
It can be written in matrix form as
Ê1 1 -1ˆÊX
ˆ Ê1ˆ
Á1 - 1 1 ˜ÁY
˜ = Á1˜
Á ˜Á ˜ Á ˜
Ë1 1 1 ¯ËZ
¯ Ë1¯
AX¢ = B, where
Ê1 1 -1ˆ ÊX
ˆ Ê1ˆ
A= Á1 - 1 1 ˜, X¢
= ÁY˜, B = Á1˜
Á ˜ Á ˜ Á ˜
Ë1 1 1 ¯ ËZ
¯ Ë1¯
1 1 -1
Now, | A| = 1 - 1 1 = -4π0
1 1 1
Thus, the system of equations have a unique solution.
20. Given
Ê1 2ˆ Êa
bˆ
A= Á and B=
Ë3 4
˜
¯
Á
Ëc
d
˜
¯
Ê1 2ˆÊa
bˆ
Now, AB = Á
Ë3 4 ˜Á
¯Ëc
d ˜
¯
Ê a+ 2c b+
2d
ˆ
= Á
Ë3a + 4c 3b+
4d
˜
¯
Also,
Êa
bˆÊ1 2ˆ
BA = Á
Ëc
d ˜Á
¯Ë3 4 ˜
¯
Êa + 3b 2a + 4bˆ
= Á
Ëc+ 3d 2c+
4d
˜
¯