Word Count : 572

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Great — I'll do the elimination with partial pivoting by hand, record the multipliers, give the pivot (permutation) vectors, show the resulting $P A = L U$ factors, and finish with back-substitution. System: $$ \begin{cases} 3x+2y+4z=7\\[4pt] 2x+y+z=7\\[4pt] x+3y+5z=2 \end{cases} \qquad\text{(write as }A\mathbf{x}=\mathbf b\text{)} $$ Write rows $R_1,R_2,R_3$ as usual. --- ## Step 1 — column 1 pivoting Look at column 1 entries $|3|,|2|,|1|$. Largest is $|3|$ (row 1) — no swap. Multipliers to eliminate below pivot (pivot = 3): $$ m_{21}=\frac{2}{3},\qquad m_{31}=\frac{1}{3}. $$ Apply $$ R_2 \leftarrow R_2 - \frac{2}{3}R_1,\qquad R_3 \leftarrow R_3 - \frac{1}{3}R_1. $$ Compute the new rows: * $R_2:\;[\,0,\;1-\tfrac{4}{3}=-\tfrac{1}{3},\;1-\tfrac{8}{3}=-\tfrac{5}{3}\;|\;7-\tfrac{14}{3}=\tfrac{7}{3}\,]$. * $R_3:\;[\,0,\;3-\tfrac{2}{3}=\tfrac{7}{3},\;5-\tfrac{4}{3}=\tfrac{11}{3}\;|\;2-\tfrac{7}{3}=-\tfrac{1}{3}\,]$. After Step 1 the matrix is $$ \begin{bmatrix} 3 & 2 & 4 &|& 7\\[4pt] 0 & -\tfrac{1}{3} & -\tfrac{5}{3} &|& \tfrac{7}{3}\\[4pt] 0 & \tfrac{7}{3} & \tfrac{11}{3} &|& -\tfrac{1}{3} \end{bmatrix}. $$ Pivot vector after step 1: $p^{(1)}=[1,2,3]$ (no change). ___ _______ ____ ____ _____ __________ _____ ____ ___ ______.
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