Stuck on the MATLAB question and really need help Consider the matrix A = (3 2 3
ID: 3028252 • Letter: S
Question
Stuck on the MATLAB question and really need help
Consider the matrix A = (3 2 3 -6 -4 -6 9 7 6 0 2 -6). Use rref () in MATLAB along with the procedure shown in class to find a basis for the column space of A. Use colspace(sym(A)) to find another basis for the column space of A (read the output from Matlab as collection of column vectors). Notice that your answers in parts (a) and (b) are different. How does colspaceO generate an answer, and why does it yield a valid basis? Using rref () in MATLAB or row-reduction by hand, give a basis for R(A^T). Using rref () in MATLAB or row-reduction by hand, give a basis for N(A). In addition, verify that each vector in the basis is orthogonal to each vector in part (d).Explanation / Answer
A=[3 -6 9 0; 2 -4 7 2; 3 -6 6 -6]
A =
3 -6 9 0
2 -4 7 2
3 -6 6 -6
[R,p]=rref(A), which defines R to be the reduced row echelon form of A and p to be the vector listing the columns of R which contain pivots.
>> [R,p]=rref(A)
R =
1 -2 0 -6
0 0 1 2
0 0 0 0
p =
1 3
Extracting the pivot columns of A gives a basis for the column space of A.
>> A(:,p)
ans =
3 9
2 7
3 6
>> colspace(sym(A))
ans =
[ 1, 0]
[ 0, 1]
[ 3, -3]
c)
B=transpose(A)
B =
3 2 3
-6 -4 -6
9 7 6
0 2 -6
>> rref(B)
ans =
1 0 3
0 1 -3
0 0 0
0 0 0
here basis of row space of A transopse = basis of colunm space
coulmspace A= transpose([ 1 0 3] ) ,transpose([ 0 1 -3])
d)
rref(B)
ans =
1 0 3
0 1 -3
0 0 0
0 0 0
here basis of row space of A transopse
[1 0 3], [0 1 -3]
e)
null(A)
ans =
-0.9302 0.2295
-0.3582 -0.7375
0.0713 -0.5681
-0.0357 0.2841