(a) Choose E21 to remove the 3 below the first pivot. Then multiply E21 A E^T21 to remove both 3 ‘s: A= [ 1 3 0 3 11 4 0 4 9 ] is going toward D = [ 1 0 0 0 2 0 0 0 1 ]. (b) Choose E32 to remove the 4 below the second pivot. Then A is reduced to D by E32 E21 A E^T21 E^T32 = D. Invert the E’s to

Question

(a) Choose {E}_{21} to remove the 3 below the first pivot. Then multiply {E}_{21}A{E}^{T}_{21} to remove both 3 's: A=\left[ \begin{matrix} 1 & 3 & 0 \ 3 & 11 & 4 \ 0 & 4 & 9 \end{matrix} ight] is going toward D =\left[ \begin{matrix} 1 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & 1 \end{matrix} ight]. (b) Choose {E}_{32} to remove the 4 below the second pivot. Then A is reduced to D by {E}_{32}{E}_{21}A{E}^{T}_{21}{E}^{T}_{32} = D. Invert the E's to find L in A = LD{L}^{T}.

Accepted Answer

(a) [latex]{E}_{21}=\left[ \begin{matrix} 1 & & \ -3 & 1 & \ & & 1 \end{matrix} ight][/latex] puts 0 in the 2, 1 entry of [latex]{E}_{21}A[/latex].

Then [latex]{E}_{21}A{E}^{T}_{21}=\left[ \begin{matrix} 1 & 0 & 0 \ 0 & 2 & 4 \ 0 & 4 & 9 \end{matrix} ight][/latex] is still symmetric, with zero also in its 1, 2 entry.

(b) Now use [latex]{E}_{32}=\left[ \begin{matrix} 1 & & \ & 1 & \ & -4 & 1 \end{matrix} ight][/latex] to make the 3, 2 entry zero and [latex]{E}_{32}{E}_{21}A{E}^{T}_{21}{E}^{T}_{32} = D[/latex] also has zero in its 2, 3 entry.

Key point: Elimination from both sides gives the symmetric [latex]LD{L}^{T}[/latex] directly.

Question : (a) Choose E21 to remove the 3 below the first pivot. Then m...