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## Q. 9.5

Solve the following tridiagonal system:

$\begin{bmatrix}2.04& −1& &\\ −1& 2.04& −1& \\& −1& 2.04& −1\\ && −1& 2.04\end{bmatrix}\begin{Bmatrix}x_1\\ x_2 \\x_3\\ x_4\end{Bmatrix} = \begin{Bmatrix}40.8\\ 0.8 \\0.8\\ 200.8\end{Bmatrix}$

## Verified Solution

As with Gauss elimination, the first step involves transforming the matrix to upper triangular form. This is done by multiplying the first equation by the factor $e_2∕f_1$ and subtracting the result from the second equation. This creates a zero in place of $e_2$ and transforms the other coefficients to new values,
$f_2 = f_2 −\frac{e_2}{f_1}g_1 = 2.04 − \frac{−1}{2.04}(-1) = 1.550$
$r_2 = r_2 −\frac{e_2}{f_1}r_1 = 0.8 −\frac{-1}{2.04} (40.8) = 20.8$
Notice that $g_2$ is unmodified because the element above it in the first row is zero.
After performing a similar calculation for the third and fourth rows, the system is transformed to the upper triangular form
$\begin{bmatrix}2.04& −1& &\\ &1.550& −1& \\& & 1.395& −1\\ && & .323\end{bmatrix}\begin{Bmatrix}x_1\\ x_2 \\x_3\\ x_4\end{Bmatrix} = \begin{Bmatrix}40.8\\ 20.8 \\14.221\\ 210.996\end{Bmatrix}$
Now back substitution can be applied to generate the final solution:
$x_4 =\frac{r_4}{f_4} = \frac{210.996}{1.323} = 159.480$
$x_3 = \frac{r_3−g_3 x_4}{f_3} = \frac{14.221 − (−1) 159.480}{1.395} = 124.538$
$x_2 = \frac{r_2 − g_2 x_3}{f_2} = \frac{20.800 −(−1)124.538}{1.550} = 93.778$
$x_1 =\frac{r_1− g_1 x_2}{f_1} = \frac{40.800 −(−1)93.778}{2.040} = 65.970$