15.3 verify that dim(NS(A)) + Rank(A) = 5

[karush]

nmh{780}

15.3 For the matrix
$$A=\begin{bmatrix}
1 & 0 &0 & 4 &5\\
0 & 1 & 0 & 3 &2\\
0 & 0 & 1 & 3 &2\\
0 & 0 & 0 & 0 &0
\end{bmatrix}$$
(a)find a basis for RS(A) and dim(RS(A)).
ok I am assuming that since this is already in row echelon form, its nonzero rows form a basis for RS(A) then
So...
$$RS(A)=(1,0,0,4,5),\quad(0,1,0,3,2),\quad(0,0,1,3,2)$$
also
dim(RS(A))= ??

(b)verify that dim(NS(A)) + Rank(A) = 5.

ok I am a little unsure what this means

 

[jvicens]

but I think it means that the dimension of the null space plus the rank of the matrix should equal the total number of columns in the matrix, which in this case is 5. So let's check:
dim(NS(A)) = number of free variables = 2 (since the last two columns have no pivots)
Rank(A) = number of nonzero rows = 3

So, 2 + 3 = 5. This checks out!