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Problem 188

# $\begin{array}{cccccccc} & /0 & 0 & 0 & 0 & 1 & 1 & 1 \mid \\ \text { Let } \mathrm{A}= & /0 & 2 & 6 & 2 & 0 & 0 & 4 \mid . \\ &/0 & 1 & 3 & 1 & 2 & 1 & 2 \mid\end{array}$ Reduce A to the Hermite normal form.

Expert verified
The Hermite normal form of matrix A is: $$HNF(A) = \begin{bmatrix} 0 & 0 & 0 & 0 & 4 & 2 & 0 \\ 0 & 1 & 3 & 1 & -2 & -1 & 2 \end{bmatrix}$$
See the step by step solution

## Step 1: Ensure positive diagonal values

In the given matrix, the first diagonal element is already positive, so we don't need to make any changes: $$A = \begin{bmatrix} 0 & 2 & 6 & 2 & 0 & 0 & 4 \\ 0 & 1 & 3 & 1 & 2 & 1 & 2 \end{bmatrix}$$

## Step 2: Create zeros below the first nonzero number in a nonzero row

To achieve this, we can perform the following row operation R2 = R2 - R1: $$\begin{bmatrix} 0 & 2 & 6 & 2 & 0 & 0 & 4 \\ 0 & -1 & -3 & -1 & 2 & 1 & -2 \end{bmatrix}$$

## Step 3: Make the first nonzero number in a nonzero row positive

For the second row, R2 = -R2: $$\begin{bmatrix} 0 & 2 & 6 & 2 & 0 & 0 & 4 \\ 0 & 1 & 3 & 1 & -2 & -1 & 2 \end{bmatrix}$$

## Step 4: Create zeros above the first nonzero number in a nonzero row

Perform the row operation R1 = R1 - 2 * R2: $$\begin{bmatrix} 0 & 0 & 0 & 0 & 4 & 2 & 0 \\ 0 & 1 & 3 & 1 & -2 & -1 & 2 \end{bmatrix}$$

## Step 5: Ensure that the first nonzero number in a nonzero row divides all other numbers in the row

The current matrix meets this requirement: $$\begin{bmatrix} 0 & 0 & 0 & 0 & 4 & 2 & 0 \\ 0 & 1 & 3 & 1 & -2 & -1 & 2 \end{bmatrix}$$ Now we have the Hermite normal form of matrix A: $$HNF(A) = \begin{bmatrix} 0 & 0 & 0 & 0 & 4 & 2 & 0 \\ 0 & 1 & 3 & 1 & -2 & -1 & 2 \end{bmatrix}$$

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