Chapter 8 Systems of Equations and Inequalities

37.
38.
Dz
1 4 −8
= 3 −1
12
1 1 1
−1 12 3
= 1 − 4
1 1 1
12 3
+ ( −8)
1 1
−1
1
= 1( −1−12) −4(3−12) − 8(3 + 1)
=− 13 + 36 −32
=−9
Find the solutions by Cramer's Rule:
Dx
−243
x= = = 3
D −81 Dz
−9
1
z = = =
D −81
9
Dy
216 8
y = = =−
D −81
3
⎛ 8 1⎞
The solution is ⎜3, − , ⎟
⎝ 3 9⎠
.
⎧ x− 2y+ 3z = 1
⎪
⎨ 3x+ y− 2z = 0
⎪
⎩ 2x− 4y+ 6z = 2
1 − 2 3
D = 3 1 − 2
2 − 4 6
1 −2 3 −2
3
= 1 −( − 2) + 3
1
−4 6 2 6 2 −4
= 1(6− 8) + 2(18+ 4) + 3( −12−2) =− 2+ 44−42 = 0
Since D = 0 , Cramer's Rule does not apply.
⎧ x− y+ 2z = 5
⎪
⎨ 3x+ 2y = 4
⎪
⎩ − 2x+ 2y− 4z = −10
1 −1
2
D = 3 2 0
−2 2 −4
2 0 3 0 3
= 1 −( − 1) + 2
2
2 −4 −2 −4 −2
2
= 1( −8− 0) + 1( −12− 0) + 2(6 + 4)
=−8− 12+ 20
= 0
Since D = 0 , Cramer's Rule does not apply.
Section 8.3: Systems of Linear Equations: Determinants
821
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39.
40.
⎧ x+ 2y− z = 0
⎪
⎨ 2x− 4y+ z = 0
⎪
⎩−
2x+ 2y− 3z = 0
1 2 −1
D = 2 − 4 1
− 2 2 −3
Dx
Dy
= 1
− 4
2
1
− 2
2
−3 −2 1
+ ( −1)
2
−3
−2
− 4
2
= 1(12 −2) −2( − 6 + 2) −1(4−8) = 10 + 8 + 4
= 22
0 2 −1
= 0 − 4 1 = 0 [By Theorem (12)]
0 2 −3
1 0 −1
= 2 0 1 = 0 [By Theorem (12)]
− 2 0 −3
1 2 0
Dz = 2 − 4 0 = 0 [By Theorem (12)]
− 2 2 0
Find the solutions by Cramer's Rule:
D 0 D
x
y 0
x = = = 0 y = = = 0
D 22 D 22
Dz
0
z = = = 0
D 22
The solution is (0, 0, 0).
⎧ x+ 4y− 3z = 0
⎪
⎨3x−
y+ 3z = 0
⎪
⎩ x+ y+ 6z = 0
1 4 −3
D = 3 −1
3
1 1 6
D
x
= 1
−1 1
3 3
− 4
6 1
3 3
+ ( −3)
6 1
−1
1
= 1( −6−3) −4(18 −3) − 3(3 + 1)
=−9−60−12 =−81
0 4 −3
= 0 −1
3 = 0 [By Theorem (12)]
0 1 6