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Need to solve this in matlab. On problem 2, have to compute the 1st and 2nd colu

ID: 2962563 • Letter: N

Question

Need to solve this in matlab.



On problem 2, have to compute the 1st and 2nd column of A inverse without computing the 3rd column.



On problem 4, the weights are x1 ,..., x5   and not x1 ,x2, x2, ..., x4 .

use two methods to determine whether A is invertible. First compute its reduced echelon form. What do you conclude about A? What does it mean about the set of functions in B? Second, compute det(A). What do you conclude about A? What does it mean about the set of functions in B? The det(A) obtained at point c) might be very small. You become suspicious, wondering if what you see are in fact roundoff errors, while the true value of det(A) is zero. To do a check, you repeat the calculations at points a-c, for example taking values of t=0, 0.5, 1 etc What is your final conclusion about the set of functions in B?

Explanation / Answer

1)

S=[0 1 0 1 0;0 0 2 0 1;0 0 0 4 0;0 0 0 0 4;0 0 0 0 0 ]

S =

     0     1     0     1     0
     0     0     2     0     1
     0     0     0     4     0
     0     0     0     0     4
     0     0     0     0     0


>> B=S^2

B =

     0     0     2     0     5
     0     0     0     8     0
     0     0     0     0    16
     0     0     0     0     0
     0     0     0     0     0

>> B=S^3

B =

     0     0     0     8     0
     0     0     0     0    32
     0     0     0     0     0
     0     0     0     0     0
     0     0     0     0     0

>> B=S^4

B =

     0     0     0     0    32
     0     0     0     0     0
     0     0     0     0     0
     0     0     0     0     0
     0     0     0     0     0

>> B=S^5

B =

     0     0     0     0     0
     0     0     0     0     0
     0     0     0     0     0
     0     0     0     0     0
     0     0     0     0     0

>> B=S^6

B =

     0     0     0     0     0
     0     0     0     0     0
     0     0     0     0     0
     0     0     0     0     0
     0     0     0     0     0


Matrix S changes to zero matrix as power increases.



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