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# 数学代写|线性代数代写Linear algebra代考|Why would we want to achieve this sort of augmented matrix?

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## 数学代写|线性代数代写Linear algebra代考|Why would we want to achieve this sort of augmented matrix?

The Gaussian elimination process can be extended in the above example so that the first non-zero number in the bottom row of $\left({ }^*\right)$ is 1 , that is
How do we convert $45 / 4$ into 1 ?
Multiply the bottom row $\mathrm{R}_3^{\dagger \dagger}$ by $\frac{4}{45}$.
$$\mathrm{R}_3^{\prime}=\frac{4 \mathrm{R}_3^{\dagger \dagger}}{45}\left(\begin{array}{rrc|c} 1 & 3 & 2 & 13 \ 0 & -8 & -11 & -49 \ 0 & 0 & \frac{4}{45}\left(\frac{45}{4}\right) & \frac{4}{45}\left(\frac{135}{4}\right) \end{array}\right)$$
which simplifies to
\begin{aligned} & \mathrm{R}_1 \ & \mathrm{R}_2^{\dagger} \ & \mathrm{R}_3^{\prime} \end{aligned}\left(\begin{array}{rrr|r} 1 & 3 & 2 & 13 \ 0 & -8 & -11 & -49 \ 0 & 0 & 1 & 3 \end{array}\right)
The advantage of this is that we get the $z$ value directly. From the bottom row, $\mathrm{R}_3^{\prime}$, we have $z=3$. We can extend these row operations further and obtain the following matrix:
$$\left(\begin{array}{lll|l} 1 & 0 & 0 & * \ 0 & 1 & 0 & * \ 0 & 0 & 1 & * \end{array}\right)$$

## 数学代写|线性代数代写Linear algebra代考|Why would we want to achieve this sort of augmented matrix?

Because we can read off the $x, y$ and $z$ values directly from this augmented matrix. The only problem is in doing the arithmetic, because achieving this sort of matrix can be a laborious process.
This augmented matrix $\left({ }^*\right)$ is said to be in reduced row echelon form.

A matrix is in reduced row echelon form, normally abbreviated to rref, if it satisfies all the following conditions:

1. If there are any rows containing only zero entries then they are located in the bottom part of the matrix.
2. If a row contains non-zero entries then the first non-zero entry is a 1 . This 1 is called a leading 1.
3. The leading 1’s of two consecutive non-zero rows go strictly from top left to bottom right of the matrix.
4. The only non-zero entry in a column containing a leading 1 is the leading 1.

## 数学代写|线性代数代写Linear algebra代考|Why would we want to achieve this sort of augmented matrix?

$$\mathrm{R}_3^{\prime}=\frac{4 \mathrm{R}_3^{\dagger \dagger}}{45}\left(\begin{array}{rrc|c} 1 & 3 & 2 & 13 \ 0 & -8 & -11 & -49 \ 0 & 0 & \frac{4}{45}\left(\frac{45}{4}\right) & \frac{4}{45}\left(\frac{135}{4}\right) \end{array}\right)$$

\begin{aligned} & \mathrm{R}_1 \ & \mathrm{R}_2^{\dagger} \ & \mathrm{R}_3^{\prime} \end{aligned}\left(\begin{array}{rrr|r} 1 & 3 & 2 & 13 \ 0 & -8 & -11 & -49 \ 0 & 0 & 1 & 3 \end{array}\right)

$$\left(\begin{array}{lll|l} 1 & 0 & 0 & * \ 0 & 1 & 0 & * \ 0 & 0 & 1 & * \end{array}\right)$$

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