GAUSS elimination ¶

Go back

You will use GAUSS like you always did, but you will work on two matrices. Given your matrix $A_n$, you will have to reduce it so that it becomes $I_n$. While, at the same time, repeating every step you did on $A$, on a matrix $I_n$, which will result in $A^{-1}$.

Example ¶

Solve $A^{-1}$ given the following invertible matrix $A$

$A=\begin{pmatrix} 3 & -2 & 4 \\ 2 & -4 & 5 \\1 & 8 & 2\end{pmatrix}$
• Step 0
$\begin{split}\begin{pmatrix}3 & -2 & 4 \\ 2 & -4 & 5 \\1 & 8 & 2\end{pmatrix} \begin{pmatrix}1 & 0 & 0 \\0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}\end{split}$
• Step 1, 2, 3
• $L3 \iff L1$
• $L2 \leftarrow L2 - 3 L_1$
• $L3 \leftarrow L2 - 2 L_1$
$\begin{split} \begin{pmatrix} 1 & 8 & 2 \\ 0 & -26 & -2 \\ 0 & -20 & 1 \\ \end{pmatrix} \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & -3 \\ 0 & 1 & -2 \\ \end{pmatrix} \end{split}$
• ... a lot of steps ...
$\begin{split}\begin{pmatrix}1 & 0 & 0 \\0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix} \begin{pmatrix}8/11 & -6/11 & -1/11 \\ -1/66 & -1/33 & 7/66 \\-10/33 & 13/33 & 4/33\end{pmatrix} = A^{-1}\end{split}$

The second matrix is $A^{-1}$.

Code in R ¶

library('MASS') # fractions

A <- matrix(c(3,2,1,-2,-4,8,4,5,2), nrow = 3, ncol = 3)

# solve and convert to fractions
fractions(solve(A))
#     [,1]   [,2]   [,3]
# [1,]   8/11  -6/11  -1/11
# [2,]  -1/66  -1/33   7/66
# [3,] -10/33  13/33   4/33