Chapter 8 Systems of Equations and Inequalities

47. Solve for x:
x 2 3
1 x 0 = 7
6 1 − 2
x
x
1
0
−2 − 2
1
6
0
−2
+ 3
1
6
x
1
= 7
x( −2x) −2( − 2) + 3( 1− 6x) = 7
2
− 2x + 4+ 3− 18x = 7
2
−2x − 18x = 0
− 2x x+
9 = 0
48. Solve for x:
( ) ( ) ( )
2
( )
x = 0 or x =− 9
x 1 2
1 x 3 =−4x
0 1 2
x
x
1
3
− 1
1
2 0
3
+ 2
1
2 0
x
1
=−4x
x 2x−3 − 1 2 + 2 1 =−4x
x y z
49. u v w = 4
1 2 3
2x−3x− 2+ 2=−4x 2
2x + x = 0
x 2x+ 1 = 0
( )
1
x = 0 or x =−
2
By Theorem (11), the value of a determinant
changes sign if any two rows are interchanged.
1 2 3
Thus, u v w =− 4 .
x y z
x y z
50. u v w = 4
1 2 3
By Theorem (14), if any row of a determinant is
multiplied by a nonzero number k, the value of
the determinant is also changed by a factor of k.
x y z x y z
Thus, u v w = 2 u v w = 2(4) = 8.
2 4 6 1 2 3
Section 8.3: Systems of Linear Equations: Determinants
823
x y z
51. Let u v w = 4 .
1 2 3
x y z x y z
−3−6− 9 =−312
3 [Theorem (14)]
u v w u v w
x y z
=−3( −1)
u v w [Theorem (11)]
1
= 3(4)
= 12
2 3
x y z
52. Let u v w = 4
1 2 3
1
x−u u
2
y−v v
3
z− w
w
1
= x
u
2
y
v
3
z
w
[Theorem (15)]
( R2 = r2 + r3)
x y z
= ( −1)
1 2 3 [Theorem (11)]
u v w
x y z
= ( −1)( −1)
u v w [Theorem (11)]
1 2 3
x y z
= u v w
1 2 3
x y z
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= 4
53. Let u v w = 4
1 2 3
1 2 3
x−3 y−6 z−9
2u 2v 2w
1 2 3
= 2 x−3y−6z−9 [Theorem (14)]
u v w
x−3 y−6 z−9
= 2( −1)
1 2 3 [Theorem (11)]
u v w