Question

3- a) Assume S = {(x,y,z, 1) : 2x - 3y + z-t=0). Show why S is a subspace of R. b) Given a matrix A- 1 -1 2 1 show a basis of the null space: N(A). 01-11

Answer 1

Solution I → aj Giren set s={(,4,2,1); 2X-344z =+=0%, claim: To prove s is subspace of 1R 4 and Given Reasons. - - the given UCX,4,3, t) = 22 - 3y +zot 20 let. Vcx', 4';z', t') be any vector in s then, Voce jy z',t') = 24'-3y'+z'-+-0 Now 1st condition of subspace ů = (2+x', y ty', z+z', Ł+E') Now to show that utrens E = 2 * 344' Utv - 20+x') -3Lyty') +(2+z')-CEE ty - 2x +zx'- 34 $3y'+z z '-t-t' =(2x-34 +2 -- t) + (2x'=3y'+z'-t') but equation 0 and a uitv - o to utv =0 - then utvenS id for any Scalar K . to show that ku in subspace s ku = (kr, ky, kz, kt) in ins : 2RX-3ky + kz-k+20

But 2kx -3 kytkz.-kt = KC 2X-34 +2-t) ako ..(by Equation o) of subspace .. ku is ins.. set 5 satisfies two condition and s also non-empby. then s is subspace an IRT. b] =>: Given matrix Claim :- To find basis for null space. A first step is to find the reduced row echelon form of the matrix Consider Ti -1 2 17 R1 = R1 + R2 This is the reduced row echelon form matrix, How solve the matrix equation let x, X2, X₃, X4 ER Ax = b ro127 Loingid iki X2 X3 xy-

then 2, +33 +224 50 X2-23 +74 =0 The given system equation are two and variables are four, This system has no unique solution and two variables are free variable. let Xz=t : x4 = 5 eduction © becomes 267t +2s =0 26 --25-t eeluchon becomes 22- t + s = 0 x2 = t-s solution of giren linears system is X, = -25-t , H2= t-5, 73=t, xq=s 1-25-t X2 I t-s Is C541,42,43,44) = (-25-t, t-s, b, s) =t(r,!,!,0) +S(-2,-1,0,1) : The basis for the null space is { (-1,1,1,0), (-2,-1,0,1) ? hence gequireel solution