Show that det(A) = 0 without directly evaluating the determinant. A = [-4 1 1 1 1; 1 -4 1 1 1; 1 1 -4 1 1; 1 1 1 -4 1; 1 1 1 1 -
2 answers:
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Answer:
[a,b] divides n
Step-by-step explanation:
Let us denote the least common multiple of a and b [a,b]=m.
We want to prove that m divides n, where n is a multiple of a and b.
We suppose m does not divide n, then by the Division Theorem, there exists q and r integers such that:
(1) ... n=mq+r, where 0<r<m
As n is a multiple of a and b, there exists s and t integers such that:
sa=n and tb=n
Same thing happens to m as it is the least common multiple, there exists u and v such that:
ua=m and vb=m
So (1) has the following form:
n=mq+r ⇒ sa=uaq+r ⇒sa-uaq=r⇒(s-uq)a=r and
n=mq+r ⇒ tb=vbq+r ⇒ tb-vbq=r⇒ (t-vq)b=r
So r is a multiple of a and b, but r<m which is a contradiction as, m is the least common multiple of a and b. So this concludes the proof.
So this means that is and integer.
As m= vb, then is an integer, lets say
=v; and as m=ua, then
=u.
So v=
=a, so
divides a; on the other hand,
u=
=b, so
divides b. From this we can conclude that
is a common divisor of a and b.