1 -2 1 37 00 00 Lo 0 00 X1 x2 x3 ХА [x₁ x2 x3 X4 From this matrix, we get the equation, x₁2x₂ + x3 + 3x4 = 0 which we can solve for the single pivot variable. x₁ = 2x₂ - x3 - 3x4 = x₂ x2 2 1 0 We can rewrite this as a linear combination. 0 ] 1 0 = [] 0 +x3 + x4 3 0 ģ 0
1 -2 1 37 00 00 Lo 0 00 X1 x2 x3 ХА [x₁ x2 x3 X4 From this matrix, we get the equation, x₁2x₂ + x3 + 3x4 = 0 which we can solve for the single pivot variable. x₁ = 2x₂ - x3 - 3x4 = x₂ x2 2 1 0 We can rewrite this as a linear combination. 0 ] 1 0 = [] 0 +x3 + x4 3 0 ģ 0
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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How do you get the linear combination? I don't get how for x2 you get [2100], for x3 you get [-1010] and for x4 you get [-3001]. Please explain that in detail.
![1
0
LO
-2 1 37
0
00
0 0 0]
X1
x2
x3
XA
From this matrix, we get the equation,
x₁ - 2x₂ + x3 + 3x4 = 0
which we can solve for the single pivot variable.
x₁ = 2x2-x3-3x4
x1
We can rewrite this as a linear combination.
[]
0
= x₂
x2
x3
2
1
0
0
X4
+x3
+ x4
0
0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbea0901b-9b71-43dc-90d5-d9e10ed3de89%2F93532d06-a7f2-4ccb-9480-ee0a7939ed23%2Fhtdtxi_processed.png&w=3840&q=75)
Transcribed Image Text:1
0
LO
-2 1 37
0
00
0 0 0]
X1
x2
x3
XA
From this matrix, we get the equation,
x₁ - 2x₂ + x3 + 3x4 = 0
which we can solve for the single pivot variable.
x₁ = 2x2-x3-3x4
x1
We can rewrite this as a linear combination.
[]
0
= x₂
x2
x3
2
1
0
0
X4
+x3
+ x4
0
0
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