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(a)A={([x],[y],[z])inR^(3)∣xy+xz=0}" let "{:[x_(1)=([x_(1)],[y_(1)],[z_(1)])=([0],[1],[2])inR^(3)],[:'x_(1)y_(1)+x_(1)z_(1)=0.1+0.2=0],[:.x^(˙)_(1)=([x_(1)],[y_(1)],[z_(1)])in A]:}{:[X_(2)=([x_(2)],[y_(2)],[z_(2)])=([1],[1],[-1])inR^(3)quad:'x_(2)y_(2)+x_(2)z_(2)=1.1+1*(-1)],[=0],[:.x_(2)=([x_(2)],[y_(2)],[z_(2)])in A]:}but x_(1)+x_(2)=([0],[1],[2])+([1],[1],[-1])=([1],[2],[1])!in A quad because 1.2+1.1=3!=0.since x_(1),x_(2)in A but x_(1)+x_(2)!in A therefare A is not closed under vectar addition thus A is not a vecter subspace of R^(3).(b) B={f inF∣e^(x)f^('')(x)+xf(x)=0}.let f,g in B then e^(x)f^('')(x)+xf(x)= ... See the full answer