3.- COLUMN SPACE & NULLSPACE. let be the Matrix 1 0 -4 A = -2 1 13 E M³×4 1 0 1 5 -1 and let the REF (Reduced Echelon form of A be 1 0 -4 0 0 1 0 0 5 1 Find: A base for C(A), the column space of A A...


The CU values
CU1=1
CU2=3
CU3=3
CU4=3
CU5=6
CU6=3


3.- COLUMN SPACE & NULLSPACE.<br>let be the Matrix<br>1<br>0 -4<br>A =<br>-2 1<br>13<br>E M³×4<br>1<br>0 1 5<br>-1<br>and let the REF (Reduced Echelon form of A be<br>1 0 -4 0<br>0 1<br>0 0<br>5<br>1<br>Find:<br>A base for C(A), the column space of A<br>A base for N(A), the Null space for A.<br>Verify the dimension theorem, that is, where<br>dim (C(A)) + dim (N (A)) = 4<br>

Extracted text: 3.- COLUMN SPACE & NULLSPACE. let be the Matrix 1 0 -4 A = -2 1 13 E M³×4 1 0 1 5 -1 and let the REF (Reduced Echelon form of A be 1 0 -4 0 0 1 0 0 5 1 Find: A base for C(A), the column space of A A base for N(A), the Null space for A. Verify the dimension theorem, that is, where dim (C(A)) + dim (N (A)) = 4

Jun 04, 2022
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