Solve the following systems of equations by Cramer's rule. (b) (а) fа1х, + а2х, — b; lа21х, + а22X2 %3D bz, 2x1 + х, — 3хз — 5 X1 – 2x2 + x3 = 10 ( 3х, + 4x, — 2хз 3D 0 where aja22 - a12a21 # 0. (c) 2х1 + X2 — Зxз — (d) X1 - X2 + 4x3 = – 4 - 8x1 + 3x2 + X3 = х, — 2х2 + хз — 8 Зx, + 4x, — 2хз 2х1 — X2 + Хз — X2 + 4x3 - 8x1 + 3x2 + X3 = (e) - 2 (f) 3x1 + X2 + X3 = 4 — 2х, — X2 12 2x1 – X2 + X3 = X1 + 2x2 + x3 = -8
Solve the following systems of equations by Cramer's rule. (b) (а) fа1х, + а2х, — b; lа21х, + а22X2 %3D bz, 2x1 + х, — 3хз — 5 X1 – 2x2 + x3 = 10 ( 3х, + 4x, — 2хз 3D 0 where aja22 - a12a21 # 0. (c) 2х1 + X2 — Зxз — (d) X1 - X2 + 4x3 = – 4 - 8x1 + 3x2 + X3 = х, — 2х2 + хз — 8 Зx, + 4x, — 2хз 2х1 — X2 + Хз — X2 + 4x3 - 8x1 + 3x2 + X3 = (e) - 2 (f) 3x1 + X2 + X3 = 4 — 2х, — X2 12 2x1 – X2 + X3 = X1 + 2x2 + x3 = -8
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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
Transcribed Image Text:Solve the following systems of equations by Cramer's rule.
(b)
(а) fа1х, + а2х, — b;
lа21х, + а22X2 %3D bz,
2x1 + х, — 3хз — 5
X1 – 2x2 + x3 = 10
( 3х, + 4x, — 2хз 3D 0
where aja22 - a12a21 # 0.
(c)
2х1 + X2 — Зxз —
(d)
X1 - X2 + 4x3 = – 4
- 8x1 + 3x2 + X3 =
х, — 2х2 + хз —
8
Зx, + 4x, — 2хз
2х1 — X2 + Хз —
X2 + 4x3
- 8x1 + 3x2 + X3 =
(e)
- 2
(f)
3x1 + X2 + X3 =
4
— 2х, — X2
12
2x1 –
X2 + X3 =
X1 + 2x2 + x3 = -8
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