Hello bartleI dont understand at question 1 how it goes from the matrix [ 1 1 1 ] to [0 1 2]at question 2 i dont quite understand the explanation my teacher gave, could you explain it in some easier words? PROBLEM 5.Consider a plane in R³ given by2+t |1+s31t, s € RX31. Does the point x = [1 1 1]' lie on the plane?2. Is the plane a subspace of R3?PROBLEM 5. Solution[1 1 1]' in the equation and simplifyingThe first problem is solved by inserting x =the expressiont1+.3+t |1 +s2This equation is then solved using the standard procedure of row reducing the augmentedmatrix1 0|00 1|00 0 12 -11 -1|13From the bottom row we see that the equations are inconsistent and the point x doesnot lie on the plane.For the plane to be a subspace of R³ it has to be closed under addition, closed undermultiplication and contain the zero vector. Due to the offset [1 0 -1]' the plane willnot pass through the origin unless this vector happens to be a linear combination of thetwo vectors [2 1 3]' and [-1 -1 2]. This is easily seen not to be the case and theplane is therefore not a subspace of R³.

Answered: Consider a plane in R³ given by [2 +t…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Hello bartle

I dont understand at question 1 how it goes from the matrix [ 1 1 1 ] to
[0 1 2]

at question 2 i dont quite understand the explanation my teacher gave, could you explain it in some easier words?

PROBLEM 5.
Consider a plane in R³ given by
2
+t |1+s
3
1
t, s € R
X3
1. Does the point x = [1 1 1]' lie on the plane?
2. Is the plane a subspace of R3?
PROBLEM 5. Solution
[1 1 1]' in the equation and simplifying
The first problem is solved by inserting x =
the expression
t1+.
3
+t |1 +s
2
This equation is then solved using the standard procedure of row reducing the augmented
matrix
1 0|0
0 1|0
0 0 1
2 -1
1 -1|1
3
From the bottom row we see that the equations are inconsistent and the point x does
not lie on the plane.
For the plane to be a subspace of R³ it has to be closed under addition, closed under
multiplication and contain the zero vector. Due to the offset [1 0 -1]' the plane will
not pass through the origin unless this vector happens to be a linear combination of the
two vectors [2 1 3]' and [-1 -1 2]. This is easily seen not to be the case and the
plane is therefore not a subspace of R³.

Expert Solution

Part 1.)

If we subtract the [ 1 0 -1 ] from [ 1 1 1 ] by taking it to the LHS we get

[ 1 1 1 ] - [ 1 0 -1 ] = [ 0 1 2 ].

The proof of vector x not lying on the plane is from the inconsistency of the system when we try to solve for t and s.

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