Question

Find inverse of the matrix 1%5C%5C4%263%262%5Cend%7Barray%7D%5Cright%5D%20" id="TexFormula1" title=" \left[\begin{array}{ccc}1&2&1\\3&2&1\\4&3&2\end{array}\right] " alt=" \left[\begin{array}{ccc}1&2&1\\3&2&1\\4&3&2\end{array}\right] " align="absmiddle" class="latex-formula"> using the inversion algorithm.

Answer 1

$\left[\begin{array}{ccc|ccc}1&2&1&1&0&0\\3&2&1&0&1&0\\4&3&2&0&0&1\end{array}\right]$

Subtract 3 times the first row from the second row, and replace the second row.
Subtract 4 times the first row from the third row, and replace the third row.
This gives

$\left[\begin{array}{ccc|ccc}1&2&1&1&0&0\\0&4&-2&-3&1&0\\0&-5&-2&-4&0&1\end{array}\right]$

Now add (-4) times the third row to 5 times the second row, and replace the third row.
This gives

$\left[\begin{array}{ccc|ccc}1&2&1&1&0&0\\0&4&-2&-3&1&0\\0&0&-2&1&5&-4\end{array}\right]$

Multiply the third row by -1/2.

$\left[\begin{array}{ccc|ccc}1&2&1&1&0&0\\0&4&-2&-3&1&0\\0&0&1&-\frac12&-\frac52&2\end{array}\right]$

Next, add 2 times the third row to the second row, and replace the second row.

$\left[\begin{array}{ccc|ccc}1&2&1&1&0&0\\0&-4&0&-4&-4&4\\0&0&1&-\frac12&-\frac52&2\end{array}\right]$

Multiply the second row by -1/4.

$\left[\begin{array}{ccc|ccc}1&2&1&1&0&0\\0&1&0&1&1&-1\\0&0&1&-\frac12&-\frac52&2\end{array}\right]$

Finally, add (-2) times the second row and (-1) times the third row to the first row, and replace the first row. You'll end up with

$\left[\begin{array}{ccc|ccc}1&0&0&-\frac12&\frac12&0\\0&1&0&1&1&-1\\0&0&1&-\frac12&-\frac52&2\end{array}\right]$

So the inverse is

$\begin{bmatrix}1&2&1\\3&2&1\\4&3&2\end{bmatrix}^{-1}=-\dfrac12\begin{bmatrix}1&-1&0\\-2&-2&2\\1&5&-4\end{bmatrix}$