Answer to Question #278751 in Linear Algebra for Petra

Question

Compute the determinant, and use Gauss-Jordan elimination to find the inverse of the following matrix (if it exists).

218        0             0

0             218        2

0             1             1

Answer 1

$\begin{vmatrix} 218 & 0&0 \\ 0 & 218&2\\ 0&1&1 \end{vmatrix}=218\cdot 218\cdot1+0\cdot0\cdot0+\\+0\cdot2\cdot0 -0\cdot218\cdot0-0\cdot0\cdot1-\\ -1\cdot2\cdot218=47524-436=47088\neq0$

$A^{-1}\\ \begin{pmatrix} 218 &0&0 &|&1&0&0\\ 0 & 218&2&|&0&1&0\\ 0&1&1&|&0&0&1 \end{pmatrix}\sim\\ 1r\cdot \frac{1}{218}\\ 2r\leftrightarrow3r$

$\sim\begin{pmatrix} 1 &0&0 &|&\frac{1}{218}&0&0\\ 0&1&1&|&0&0&1\\ 0 & 218&2&|&0&1&0 \end{pmatrix}\sim\\ 3r+2r\cdot(-218)$

$\sim\begin{pmatrix} 1 &0&0 &|&\frac{1}{218}&0&0\\ 0&1&1&|&0&0&1\\ 0 & 0&-216&|&0&1&-218 \end{pmatrix}\sim\\ 3r\cdot\frac{-1}{216}$

$\sim\begin{pmatrix} 1 &0&0 &|&\frac{1}{218}&0&0\\ 0&1&1&|&0&0&1\\ 0 & 0&1&|&0&\frac{-1}{216}&\frac{109}{108} \end{pmatrix}\sim\\ 2r+3r\cdot(-1)$

$\sim\begin{pmatrix} 1 &0&0 &|&\frac{1}{218}&0&0\\ 0&1&0&|&0&\frac{1}{216}&\frac{-1}{108}\\ 0 & 0&1&|&0&\frac{-1}{216}&\frac{109}{108} \end{pmatrix}$

$A^{-1}=\begin{pmatrix} \frac{1}{218}&0&0\\\\ 0&\frac{1}{216}&\frac{-1}{108}\\\\ 0&\frac{-1}{216}&\frac{109}{108} \end{pmatrix}$