below you have a correct solution, which i don't understand. please explain it to me in great detail. a$)$Writea functionthat, by receiving theLU$-$factors ofA, $$ b andavector$\backslash$gamma as input,

returns the solution$$x to the systemB$$x$=\backslash$gamma using the identity $(1).$ Therefore, the function

shouldnotcomputethematricesA$1,$B$,$B$1$explicitly.Assumethatyoualreadyhaveavailable

afunctionthatreturnsthesolutiontoalinearsystembyreceivingtheLU$-$factorsofamatrixand

aright$-$hand$-$side.Youcanusevector$($notmatrix$)$operationssum$,$subtraction,dotproduct,and

multiplication$/$divisionbyascalar$.$Hint: ifyouhaveamatrixS$=$C$+$D@E$,$ thenyouhavethe

indentityS@g$=$C@g$+$D@$($E@g

b$)$ Computethenumberofoperationsof thealgorithmsolvingthe linear systemB$$x$=\backslash$gamma $.$

Assumethatsolvingthe linearsystemsassociatedwithLandUcostsn$2$ elementaryoperations

each.

c$))$If thecostofdoinganLUfactorizationofamatrixinRn$\backslash$times n is $(2/3)$n$3,$whichstrategy

wouldleadtoareducednumberofoperationstosolvethelinearsystemB$$xwhentheLUfactors of A and B still need to be computed? Strategy $($i$)$: to compute the full matrix B then compute

LU and then solve, or Strategy $($ii$)$: the one you developed using identity $(1)?$ What about if A

has a sparse structure?

Solutions: a$)$

deflin$\_$sys$($L$,$U$,$b$,$rhs$)$:

#SolvelinearsystemwithAandb

invAb$=$lin$\_$sys$\_$A$($L$,$U$,$ b$)$

#SolvelinearsystemwithAandrhs

invArhs$=$lin$\_$sys$\_$A$($L$,$U$,$ rhs$)$

#Scalarfactor

scalar$=$np$.$dot$($b$,$invArhs$)/(1+$np$.$dot$($b$,$invAb$))$

#Computeresult

result$=$invArhs$-$invAb$*$scalar

returnresult

b$)$

SolvelinearsystemforL,Uand$$ b: $2$n$2$ops.

$(0\mathrm{.}2$pts$)$SolvelinearsystemforL$,$Uand$\backslash$gamma : $2$n$2$ops.

$(0\mathrm{.}2$pts$)$Compute the scalar factor: twodot products $2$neach, nsumsbyone, n

divisions. So$4$nintotal.

$(0\mathrm{.}2$pts$)$Computetheresult:multiplicationofvectorbyscalarn,nsubstractions. So

$2$nintotal.

$(0\mathrm{.}2$pts$)$Addingalloperations: $4$n$2+6$n

c$)$Computing the full matrix B can be performed in n$2$ multiplications of the rank

$1$ term and then n$2$ additions, hence $2$n$2(0\mathrm{.}5$ pts$).$ Then the total cost of solving B$$x $=\backslash$gamma

would be $2$n$2+2/3$n$3+2$n$2=4$n$2+2/3$n$3.$ For the strategy above one obtain $4$n$2+6$n $+$

$2/3$n$3.$ All cubic and quadratic terms are the same, so both strategies lead to the same results

asymptotically $(0\mathrm{.}25$ pts$).$ If A is sparse, then the LU decomposition of A may take much

less operations than the one of B and therefore strategy $($ii$)$ may be more efficient