61873 s140 141

141)

141)

' 2kxi +(k + l)x2 = 2
xi + x2 - x3 = 0	81.
, ^'2:1 + (2k — l)x2 = 1
f x + y + x = 1 < x + y + kz = 1 [ kx + y + kz = k	83. •
' x + 2y + z = 4
2x + 5y + (k + 2 )z = 12	85. -
2x + 4y + 2z = 8 - k

,    —x + (k² - l)z = 4

—Xi + (fc + 1)X2 = 1 - 2:₂ + (k - 2)x₃ = O . *1 + (k - l)x₂ + 2x₃ = 1

x + y + z = -1 kx — y + z = 3 5x + 5y + 3z = — 7 { 2x + (1 — k)y - z =    - 6

Xi - 2x₂ + 4x₃ = 1 x₂ + 2x₃ = -1 Xi — 3x2 + 2x₃ = k , 2xi — 4x2 + kx₃ = 2

86.

x_x + (3A: - l)x₂ + (k - 7)x₃ = -1 —x\ + (1 — k)x 2 + (1 — k)x₃ = 1 (k - l)x₂ + (k - 3)x₃ = -2

Xi + x₂ - x₃ = 3

2x\ + (k + 6)x₂ + (2A; + 6)23 = 4k + 4 —X\ - x₂ + 2kx₃ = 2k + 6 3xi + (k + 7)x₂ + (2 k + 5)13 = 4 k + 7

X\ + 2x2 + x₃ — O xi + (fc + l)x₃ = k + 1 -kx\ - 2kx₂ = k + 2 . kx 1 + (2k + k²)x₃ = k² + 2k - 1

' (1 - k)x 1 + (2 - 2k)x₂ + x₃ = k + 2 -xj - 4x2 + {k — 1)2:3 = k + 1

—2kx₂ + (k² + k)x 3 = k² + 2fc + 2 3A:xi + 4fcx₂ + (2k + k²)x₃ = k² - 5

x\ + x₂ x₃ — O

kx 2 + 2:3 + ^2x4 = k X3 "ł" kx4 — 1

X₂ 4- X3 + 2X4 = 1

' 2xi - x₂ + x₄ = O

kx i+x₃-2x₄=0    _gi<

x’i + 2:2 - 2:3 + 2:4 = O 2xi - x-2 + 5x₃ = O

HI

Znaleźć takie wartości parametru fc, dla których dany układ równań liniowych nic ma rozwiązania:

92.

kx 1 + T₂ + T₃ = k

< 2xi + kx2 — t₃ — 3

X2 + X3 = 0

Xi + X₂ — X3 = 1

93. x\ + 2x2 — kx3 = 2

kXy + X2 + X3 ⁼ k -f" 1

94.

x\ + 2x₂ + t₃ — 1

<    — X\ — X2 + kx3 = —4

-x₂ + 2x₃ = 3

96.

97.

	Ti + 2t2 + t3 — —2
	kx2 + t3 = 2
. —2tx -	(2fc + 4)t2 + (fc — 2)t3 = fc
	-Ti + 3t2 + t3 = -2
Ti + (fc	— 3)t2 + (2 fc — l)x3 = fc — 1

_k 2ti - (k + 6)t₂ — (k + 5)t₃ = 7

xi + 2t₂ - t₃ = 1 (k + 1)t₂ + (A: + 1)t₃ = k —Ti + (—2k — 4)t ₂ + 2t₃ = —k

.Tl + T₂ + 2t₃ kx\ + 2kx3 kx2 + (2 — k)x3 ,kx\ + 2kx2 + 4t₃

= 1 = 2

98.

, 2ti + (fc + 5)t₂ + (k — 1)t₃ = k + 2

99.

Ti + 2t₂ + 3t₃ = —2 Ti + kx₂ + {k + 1)t₃ = 2A; + 1 —Ti + (k - 4)x₂ + (—fc - 8)t₃ = fc + 7

. -Ti + (4fc - 10)x₂ + (-2fc - 20)t₃ = 5fc + 20

Ti + kx₃ = 3 2ti + kx2 + 2kx3 = k + 9 —Ti — kx2 + (6 + &)t₃ = —6 2xi + (4fc + 6)t₃ = fc + 6

100.