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# 数学代写|数值分析代写Numerical analysis代考|Partial pivoting

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## 数学代写数值分析代写Numerical analysis代考|Partial pivoting

At the start of classical Gaussian elimination of $n$ equations in $n$ unknowns, the first step is to use the diagonal element $a_{11}$ as a pivot to eliminate the first column. The partial pivoting protocol consists of comparing numbers before carrying out each elimination step. The largest entry of the first column is located, and its row is swapped with the pivot row, in this case the top row.

In other words, at the start of Gaussian elimination, partial pivoting asks that we select the $p$ th row, where
$$\left|a_{p 1}\right| \geq\left|a_{i 1}\right|$$
for all $1 \leq i \leq n$, and exchange rows 1 and $p$. Next, elimination of column 1 proceeds as usual, using the “new” version of $a_{11}$ as the pivot. The multiplier used to eliminate $a_{i 1}$ will be
$$m_{i 1}=\frac{a_{i 1}}{a_{11}}$$
and $\left|m_{i 1}\right| \leq 1$
The same check is applied to every choice of pivot during the algorithm. When deciding on the second pivot, we start with the current $a_{22}$ and check all entries directly below. We select the row $p$ such that
$$\left|a_{p 2}\right| \geq\left|a_{i 2}\right|$$
for all $2 \leq i \leq n$, and if $p \neq 2$, rows 2 and $p$ are exchanged. Row 1 is never involved in this step. If $\left|a_{22}\right|$ is already the largest, no row exchange is made.

## 数学代写|数值分析代写Numerical analysis代考|Permutation matrices

Before showing how row exchanges can be used with the $L U$ factorization approach to Gaussian elimination, we will discuss the fundamental properties of permutation matrices.

A permutation matrix is an $n \times n$ matrix consisting of all zeros, except for a single 1 in every row and column.

Equivalently, a permutation matrix $P$ is created by applying arbitrary row exchanges to the $n \times n$ identity matrix (or arbitrary column exchanges). For example,
$$\left[\begin{array}{ll} 1 & 0 \ 0 & 1 \end{array}\right],\left[\begin{array}{ll} 0 & 1 \ 1 & 0 \end{array}\right]$$
are the only $2 \times 2$ permutation matrices, and
\begin{aligned} & {\left[\begin{array}{lll} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{array}\right],\left[\begin{array}{lll} 0 & 1 & 0 \ 1 & 0 & 0 \ 0 & 0 & 1 \end{array}\right],\left[\begin{array}{lll} 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 \end{array}\right],} \ & {\left[\begin{array}{lll} 0 & 0 & 1 \ 0 & 1 & 0 \ 1 & 0 & 0 \end{array}\right],\left[\begin{array}{lll} 0 & 0 & 1 \ 1 & 0 & 0 \ 0 & 1 & 0 \end{array}\right],\left[\begin{array}{lll} 0 & 1 & 0 \ 0 & 0 & 1 \ 1 & 0 & 0 \end{array}\right]} \end{aligned}
are the $\operatorname{six} 3 \times 3$ permutation matrices.
The next theorem tells us at a glance what action a permutation matrix causes when multiplied on the left of another matrix.

## 数学代写数值分析代写Numerical analysis代考|Partial pivoting

$$\left|a_{p 1}\right| \geq\left|a_{i 1}\right|$$

$$m_{i 1}=\frac{a_{i 1}}{a_{11}}$$

$$\left|a_{p 2}\right| \geq\left|a_{i 2}\right|$$

## 数学代写|数值分析代写Numerical analysis代考|Permutation matrices

$$\left[\begin{array}{ll} 1 & 0 \ 0 & 1 \end{array}\right],\left[\begin{array}{ll} 0 & 1 \ 1 & 0 \end{array}\right]$$
are the only $2 \times 2$ permutation matrices, and
\begin{aligned} & {\left[\begin{array}{lll} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{array}\right],\left[\begin{array}{lll} 0 & 1 & 0 \ 1 & 0 & 0 \ 0 & 0 & 1 \end{array}\right],\left[\begin{array}{lll} 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 \end{array}\right],} \ & {\left[\begin{array}{lll} 0 & 0 & 1 \ 0 & 1 & 0 \ 1 & 0 & 0 \end{array}\right],\left[\begin{array}{lll} 0 & 0 & 1 \ 1 & 0 & 0 \ 0 & 1 & 0 \end{array}\right],\left[\begin{array}{lll} 0 & 1 & 0 \ 0 & 0 & 1 \ 1 & 0 & 0 \end{array}\right]} \end{aligned}

## MATLAB代写

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