The problem asks us to find the inverse of the matrix $B = \begin{pmatrix} 2 & -1 & 0 \\ 1 & 3 & 1 \\ 0 & -2 & 4 \end{pmatrix}$ using Gaussian elimination with pivoting.

1. Problem Description

The problem asks us to find the inverse of the matrix $B = \begin{pmatrix} 2 & -1 & 0 \\ 1 & 3 & 1 \\ 0 & -2 & 4 \end{pmatrix}$ using Gaussian elimination with pivoting.

2. Solution Steps

We start by augmenting the matrix $B$ with the identity matrix $I_3$:

$[B | I_3] = \begin{pmatrix} 2 & -1 & 0 & | & 1 & 0 & 0 \\ 1 & 3 & 1 & | & 0 & 1 & 0 \\ 0 & -2 & 4 & | & 0 & 0 & 1 \end{pmatrix}$.

Since we want to use pivoting, we need to check if the absolute value of the first element of the first row is the largest among the elements in the first column. Here we have $|2| > |1| > |0|$. Thus, no row exchange is needed.

Divide the first row by 2:

$\begin{pmatrix} 1 & -1/2 & 0 & | & 1/2 & 0 & 0 \\ 1 & 3 & 1 & | & 0 & 1 & 0 \\ 0 & -2 & 4 & | & 0 & 0 & 1 \end{pmatrix}$.

Subtract the first row from the second row:

$\begin{pmatrix} 1 & -1/2 & 0 & | & 1/2 & 0 & 0 \\ 0 & 7/2 & 1 & | & -1/2 & 1 & 0 \\ 0 & -2 & 4 & | & 0 & 0 & 1 \end{pmatrix}$.

Multiply the second row by $2/7$:

$\begin{pmatrix} 1 & -1/2 & 0 & | & 1/2 & 0 & 0 \\ 0 & 1 & 2/7 & | & -1/7 & 2/7 & 0 \\ 0 & -2 & 4 & | & 0 & 0 & 1 \end{pmatrix}$.

Add 2 times the second row to the third row:

$\begin{pmatrix} 1 & -1/2 & 0 & | & 1/2 & 0 & 0 \\ 0 & 1 & 2/7 & | & -1/7 & 2/7 & 0 \\ 0 & 0 & 32/7 & | & -2/7 & 4/7 & 1 \end{pmatrix}$.

Multiply the third row by $7/32$:

$\begin{pmatrix} 1 & -1/2 & 0 & | & 1/2 & 0 & 0 \\ 0 & 1 & 2/7 & | & -1/7 & 2/7 & 0 \\ 0 & 0 & 1 & | & -1/16 & 1/8 & 7/32 \end{pmatrix}$.

Subtract $2/7$ times the third row from the second row:

$\begin{pmatrix} 1 & -1/2 & 0 & | & 1/2 & 0 & 0 \\ 0 & 1 & 0 & | & -1/7 + 2/7(1/16) & 2/7 - 2/7(1/8) & -2/7(7/32) \\ 0 & 0 & 1 & | & -1/16 & 1/8 & 7/32 \end{pmatrix}$.

Simplifying the second row gives:

$\begin{pmatrix} 1 & -1/2 & 0 & | & 1/2 & 0 & 0 \\ 0 & 1 & 0 & | & -1/7 + 1/56 & 2/7 - 1/28 & -1/16 \\ 0 & 0 & 1 & | & -1/16 & 1/8 & 7/32 \end{pmatrix}$.

$\begin{pmatrix} 1 & -1/2 & 0 & | & 1/2 & 0 & 0 \\ 0 & 1 & 0 & | & -7/56 & 7/28 & -1/16 \\ 0 & 0 & 1 & | & -1/16 & 1/8 & 7/32 \end{pmatrix}$.

$\begin{pmatrix} 1 & -1/2 & 0 & | & 1/2 & 0 & 0 \\ 0 & 1 & 0 & | & -1/8 & 1/4 & -1/16 \\ 0 & 0 & 1 & | & -1/16 & 1/8 & 7/32 \end{pmatrix}$.

Add $1/2$ times the second row to the first row:

$\begin{pmatrix} 1 & 0 & 0 & | & 1/2 - 1/16 & 1/8 & -1/32 \\ 0 & 1 & 0 & | & -1/8 & 1/4 & -1/16 \\ 0 & 0 & 1 & | & -1/16 & 1/8 & 7/32 \end{pmatrix}$.

$\begin{pmatrix} 1 & 0 & 0 & | & 7/16 & 1/8 & -1/32 \\ 0 & 1 & 0 & | & -1/8 & 1/4 & -1/16 \\ 0 & 0 & 1 & | & -1/16 & 1/8 & 7/32 \end{pmatrix}$.

3. Final Answer

$B^{-1} = \begin{pmatrix} 7/16 & 1/8 & -1/32 \\ -1/8 & 1/4 & -1/16 \\ -1/16 & 1/8 & 7/32 \end{pmatrix}$

or

$B^{-1} = \begin{pmatrix} 14/32 & 4/32 & -1/32 \\ -4/32 & 8/32 & -2/32 \\ -2/32 & 4/32 & 7/32 \end{pmatrix} = \frac{1}{32}\begin{pmatrix} 14 & 4 & -1 \\ -4 & 8 & -2 \\ -2 & 4 & 7 \end{pmatrix}$.