Question
Asked Oct 8, 2019

linear algebra

(4) (Section 5.2 # 8) The matrix below is in reduced row echelon form. Write the solution to the
corresponding homogeneous system
as a span of vectors
1
-1
0 5 0
0
1
20 2
0
0
0
1
2
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(4) (Section 5.2 # 8) The matrix below is in reduced row echelon form. Write the solution to the corresponding homogeneous system as a span of vectors 1 -1 0 5 0 0 1 20 2 0 0 0 1 2

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check_circleExpert Solution
Step 1

To describe the solution space corresponding to the row reduced echelon form given

Step 2

Determining the equations of the homogeneous system

Let (x y zuvwdenote a vector
in the solution space.(' denotes
transpose)
Then the given system is
v+2w 0
+2u2w 0
(1)
(2)
(3)
r-y+5u 0
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Let (x y zuvwdenote a vector in the solution space.(' denotes transpose) Then the given system is v+2w 0 +2u2w 0 (1) (2) (3) r-y+5u 0

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Step 3

From (4), (5) and (6), it follows that y, u and w are independent parameters. So the solution space is three dimensional ; we may a...

From (),(2),(3)
v -21w
(4)
-2u-2w
(5)
(6)
x=y-5u
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From (),(2),(3) v -21w (4) -2u-2w (5) (6) x=y-5u

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