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Mathematics 36 Online
OpenStudy (anonymous):

Linear Algebra: Let A be a 4x3 matrix and suppose that the vectors z1=(1,1,2)^T and (1,0,-1)^T form a basis for N(A). If the b = a1 + 2a2 + a3 (where b is a vector and a1, a2, a3 are column vectors of A), find all solutions of the system Ax = b.

OpenStudy (anonymous):

There should be error.. if a=4*3 and dim(n(a)) =2 ,z1 shold be consist of 4 points...like z1=(a,s,0,t),z1=(a,s,f,0) if and only we can solve that problem is't it

OpenStudy (anonymous):

if it has two free variable ..

OpenStudy (anonymous):

where are you ...elliotc..

OpenStudy (anonymous):

Am i correct....

OpenStudy (anonymous):

please type something.....

OpenStudy (helder_edwin):

this miight help: let \(z_1=(1,1,2)^t\) and \(z_2=(1,0,-1)^t\) in \(\ker A=N(A)\). then \[ \large 0=Az_1=a_1+a_2+2a_3 \] and \[ \large 0=Az_2=a_1-a_3 \].

OpenStudy (anonymous):

Yaaa..something wrong with me...

OpenStudy (helder_edwin):

from the second equation \[ \large a_1=a_3 \]

OpenStudy (helder_edwin):

so the first equation becomes \[ \large 0=Az_1=3a_1+a_2 \] so \[ \large a_2=-3a_1 \] therefore \[ \large b=a_1+2a_2+a_3=a_1-6a_1+a_1=-4a_1 \]

OpenStudy (helder_edwin):

so the system \(Ax=b\) actually is \(Ax=-4a_1\) so \(x=(-4,0,0)^t\)

OpenStudy (helder_edwin):

i think that's it.

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