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clope023
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Homework Statement
A = floor(10*rand(6)) (6x6 matrix with random numbers)
B = A'(transpose)
A(:,6) = -sum(B(1:5,' (sum row 1st through 5th row entries and place in the 6th column and then transpose and take the negative)
x = ones(6,1) (vector with 6 entries all equal to 1)
Ax = 0, x=/=0 implies A is a singular matrix since it is in contradiction to the fact that a nonsingular matrix implies x=0 is a unique solution of the equation Ax=0
[1:6] (vector with 6 entries from 1 to 6 respectively)
B = x*[1:6] (matrix column of 1's, column of 2's, ..., column of 6's)
The product AB should be the zero matrix. Why?
Homework Equations
det(A) = 0 => singular matrix
The Attempt at a Solution
set O = zero matrix
therefore AB = O take determinant of both sides det(AB) = det(O) => det(A)det(B) = det(O)
0det(B) = 0, therefore 0 = 0, I also calculated the determinant of matrix B with MATLAB and it seems it was singular as well so more easily I can calculate (0)(0) = det(O) => 0 = 0
I also attempted the question without the use of the determinant
assuming B is nonsingular
AB = O, ABinv(B) = Oinv(B), AI = O, A = O -> contradicts the original statement that A was not the zero matrix orignially this also implies there does not exist an inverse of B and thus B is also singular
I'm not sure if this is the correct to infer that the product of these two matrices is necessarily the zero matrix without using determinants from here any help would be appreciated thank you
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