MH1201 - SOLUTIONS TO PROBLEM SET 2Problem 1.•Solution to part (1).The vectorvlies in the span ofSif and only if there are scalarsa, b∈Rsuch that:(5,1,-5) =a(1,-2,-3) +b(-2,3,-4).This is the precise meaning of the statement “vlies in the span ofS”.Scalarsaandbsolve this equation if and only if they solve the system of three linearequations that we get from the components of these vectors:a-2b=5,-2a+ 3b=1,-3a-4b=-5.These equations can be expressed as an augmented matrix as follows:1-25-231-3-4-5Now we do the usual row manipulations to put this system in row echelon form, in orderto decide if there are any solutions.1-25-231-3-4-5-→1-250-1110-1010-→103001001-1This hasno solutionsbecause it contains the equation 0b= 10, which cannot besatisfied. Sovdoesnotlie in the span ofS.•Solution to part (2).In this case the equation we are solving is:-2x3-11x2+ 3x+ 2 =a(x3-2x2+ 3x-1) +b(2x3+x2+ 3x-2).This is equivalent to the system of 4 equations that are obtained by equating coefficientsof like powers ofx:a+ 2b=-2,-2a+b=-11,3a+ 3b=3,-a-2b=2.Solving this is the usual way:12-2-21-11333-1-22-→12-205-150-39000-→10400001-3000Thus the system is solved bya= 4 andb=-3. We can check that by writing out theapparent solution:-2x3-11x2+ 3x+ 2 = 4(x3-2x2+ 3x-1)-3(2x3+x2+ 3x-2).Thusvdoeslie in the span ofS.

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2•Solution to part (3).We want to understand the solutions to the equation:"2-113#=a"1111#+b"0111#+c"0011#.Equating entries in the same positions gives us a system of 4 equations in 3 unknowns:2=a,-1=a+b,1=a+b+c,3=a+b+c.This clearly hasno solutions.In fact, we should have been able to see this right away. If you look at the matricesinS, you will notice that in each of them the bottom two entries are the same. So anymatrix we can ‘build’ from these matrices is going to share that property too. But thematrix we were given does not have this property. So it cannot be in the span of thesetS.Problem 2.•Solution to part (1).