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I use Khan Academy but I have not seen this video yet, thanks, I take a look.

I see Heureka is already answering - thanks Heureka.

But I was wondering if you have looked at this Khan Academy you tube?

https://www.khanacademy.org/math/algebra-home/alg-matrices/alg-determinants-and-inverses-of-large-matrices/v/inverting-matrices-part-3

Can someone teach me how to get the inverse of a 3x3 matrix using Gaussian elimination.

Here is an example:

\(\begin{array}{|rcll|} \hline \text{We have Matrix }A = \begin{pmatrix} 1&-2&2\\ 1&0&1\\-1&1&3\end{pmatrix}\\ \hline \end{array}\)

The inverse Matrix \(A^{-1} = \ ?\)

\(\begin{array}{|rcll|} \hline \text{The identity matrix ( unit Matrix ) } I = \begin{pmatrix} 1&0&0\\ 0&1&0\\0&0&1\end{pmatrix}\\ \hline \end{array}\)

Using Gaussian elimination:

\(\left( \begin{matrix} 1&-2&2\\ 1&0&1\\-1&1&3 \end{matrix} \left| \begin{matrix} 1&0&0\\ 0&1&0\\0&0&1 \end{matrix} \right. \right)\)

\( \overset{II-I=II^*}{\curvearrowright} \left( \begin{matrix} 1&-2&2\\ 0&2&-1\\-1&1&3 \end{matrix} \left| \begin{matrix} 1&0&0\\ -1&1&0\\0&0&1 \end{matrix} \right. \right)\)

\( \overset{III+I=III^*}{\curvearrowright} \left( \begin{matrix} 1&-2&2\\ 0&2&-1\\0&-1&5 \end{matrix} \left| \begin{matrix} 1&0&0\\ -1&1&0\\1&0&1 \end{matrix} \right. \right)\)

\( \overset{2\times III+II=III^*}{\curvearrowright} \left( \begin{matrix} 1&-2&2\\ 0&2&-1\\0&0&9 \end{matrix} \left| \begin{matrix} 1&0&0\\ -1&1&0\\1&1&2 \end{matrix} \right. \right)\)

\( \overset{9\times II+III=II^*}{\curvearrowright} \left( \begin{matrix} 1&-2&2\\ 0&18&0\\0&0&9 \end{matrix} \left| \begin{matrix} 1&0&0\\ -8&10&2\\1&1&2 \end{matrix} \right. \right)\)

\( \overset{\frac19 \times II+I=I^*}{\curvearrowright} \left( \begin{matrix} 1&0&2\\ 0&18&0\\0&0&9 \end{matrix} \left| \begin{matrix} \frac19&\frac{10}{9}&\frac{2}{9}\\ -8&10&2\\1&1&2 \end{matrix} \right. \right)\)

\( \overset{ -\frac29 \times III+I=I^*}{\curvearrowright} \left( \begin{matrix} 1&0&0\\ 0&18&0\\0&0&9 \end{matrix} \left| \begin{matrix} -\frac19&\frac{8}{9}&-\frac{2}{9}\\ -8&10&2\\1&1&2 \end{matrix} \right. \right)\)

\( \overset{ \frac{1}{18} \times II=II^*}{\curvearrowright} \left( \begin{matrix} 1&0&0\\ 0&1&0\\0&0&9 \end{matrix} \left| \begin{matrix} -\frac19&\frac{8}{9}&-\frac{2}{9}\\ -\frac{8}{18}&\frac{10}{18}&\frac{2}{18}\\1&1&2 \end{matrix} \right. \right) = \left( \begin{matrix} 1&0&0\\ 0&1&0\\0&0&9 \end{matrix} \left| \begin{matrix} -\frac19&\frac{8}{9}&-\frac{2}{9}\\ -\frac{4}{9}&\frac{5}{9}&\frac{1}{9}\\1&1&2 \end{matrix} \right. \right)\)

\( \overset{ \frac{1}{9} \times III=III^*}{\curvearrowright} \left( \begin{matrix} 1&0&0\\ 0&1&0\\0&0&1 \end{matrix} \left| \begin{matrix} -\frac19&\frac{8}{9}&-\frac{2}{9}\\ -\frac{4}{9}&\frac{5}{9}&\frac{1}{9}\\ \frac{1}{9}&\frac{1}{9}&\frac{2}{9} \end{matrix} \right. \right)\)

\(\begin{array}{|rcll|} \hline \text{The inverse Matrix } A ^{-1} = \begin{pmatrix} -\frac19&\frac{8}{9}&-\frac{2}{9}\\ -\frac{4}{9}&\frac{5}{9}&\frac{1}{9}\\ \frac{1}{9}&\frac{1}{9}&\frac{2}{9} \end{pmatrix} = \frac19 \times \begin{pmatrix} -1&8&-2\\ -4&5&1\\ 1&1&2 \end{pmatrix}\\ \hline \end{array} \)