+26387 0 4
- 6 5
\(\boxed{~ A=\begin{pmatrix} a&b\\ c&d \end{pmatrix}\\ A^{-1}=\frac{1}{ad-bc}\begin{pmatrix} d&-b\\ -c&a \end{pmatrix} ~}\)
\(\begin{array}{rcl} A&=&\begin{pmatrix} 0&4\\ -6&5 \end{pmatrix}\\ A^{-1}&=&\frac{1}{0\cdot 5- 4\cdot (-6)}\begin{pmatrix} 5&-4\\ -(-6)&0 \end{pmatrix}\\\\ A^{-1}&=&\frac{1}{24}\begin{pmatrix} 5&-4\\ 6&0 \end{pmatrix}\\\\ A^{-1}&=&\begin{pmatrix} \frac{5}{24}&\frac{-4}{24}\\ \frac{6}{24}&\frac{0}{24} \end{pmatrix}\\\\ A^{-1}&=&\begin{pmatrix} \frac{5}{24}&\frac{-1}{6}\\ \frac{1}{4}&0 \end{pmatrix} \end{array}\)
+26387 0 4
- 6 5
\(\boxed{~ A=\begin{pmatrix} a&b\\ c&d \end{pmatrix}\\ A^{-1}=\frac{1}{ad-bc}\begin{pmatrix} d&-b\\ -c&a \end{pmatrix} ~}\)
\(\begin{array}{rcl} A&=&\begin{pmatrix} 0&4\\ -6&5 \end{pmatrix}\\ A^{-1}&=&\frac{1}{0\cdot 5- 4\cdot (-6)}\begin{pmatrix} 5&-4\\ -(-6)&0 \end{pmatrix}\\\\ A^{-1}&=&\frac{1}{24}\begin{pmatrix} 5&-4\\ 6&0 \end{pmatrix}\\\\ A^{-1}&=&\begin{pmatrix} \frac{5}{24}&\frac{-4}{24}\\ \frac{6}{24}&\frac{0}{24} \end{pmatrix}\\\\ A^{-1}&=&\begin{pmatrix} \frac{5}{24}&\frac{-1}{6}\\ \frac{1}{4}&0 \end{pmatrix} \end{array}\)
+6251 as an alternative to the method heureka has used we can use Gaussian elimination and operate on an identity matrix as well. At the end of the elimation the identity matrix will have been transformed into the inverse matrix.
\(\begin{pmatrix}0 &4 &| &1 &0\\ -6 & 5 &| &0 &1\end{pmatrix} \Rightarrow \\ \begin{pmatrix}-6 &5 &| &0 &1 \\ 0 &4 &| &1 &0\end{pmatrix} \Rightarrow \\ \begin{pmatrix}-6 &5 &| &0 &1 \\ 0 &1 &| &\frac 1 4 &0\end{pmatrix} \Rightarrow \\ \begin{pmatrix}-6 &0 &| &-\frac 5 4 &1 \\ 0 &1 &| &\frac 1 4 &0\end{pmatrix} \Rightarrow \\ \begin{pmatrix}1 &0 &| &\frac 5{24} &-\frac 1 6 \\ 0 &1 &| &\frac 1 4 &0\end{pmatrix} \Rightarrow \\\)
and we read off the inverse as
\(\begin{pmatrix}\frac 5{24} &-\frac 1 6 \\ \frac 1 4 &0\end{pmatrix}\)
