Solution:
For row transformation we have
![\begin{array}{l} A=I A \\ \Rightarrow\left[\begin{array}{ll} 1 & 6 \\ -3 & 5 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \mathrm{A} \end{array}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2f954e10d3c383763796c8199ff09139_l3.png "Rendered by QuickLaTeX.com")
Applying 
![\Rightarrow\left[\begin{array}{cc} 1 & 6 \\ 0 & 23 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 3 & 1 \end{array}\right] \mathrm{A}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5dab5b8f6a0b987963a044e257d32783_l3.png "Rendered by QuickLaTeX.com")
Applying 
![\Rightarrow\left[\begin{array}{ll} 1 & 6 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ \frac{3}{23} & \frac{1}{23} \end{array}\right] \mathrm{A}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f05407e4210d461967cee899156801cf_l3.png "Rendered by QuickLaTeX.com")
Applying 
![\Rightarrow\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} \frac{5}{23} & \frac{-6}{23} \\ \frac{3}{23} & \frac{1}{23} \end{array}\right] A](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ab4986a4e7324e93d0489227a5252be6_l3.png "Rendered by QuickLaTeX.com")
Therefore, as it is known that
So
![\Rightarrow A^{-1}=\frac{1}{23}\left[\begin{array}{rr} 5 & -6 \\ 3 & 1 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-11659c723eca16617af438f5d4bd9dc2_l3.png "Rendered by QuickLaTeX.com")
Find the inverse of the following matrices by using elementary row transformations: ![\left[\begin{array}{cc}1 & 6 \\ -3 & 5\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a9b72dc8657a64ba9320d3064be5f4f5_l3.png "Rendered by QuickLaTeX.com")
Solution:
For row transformation we have
Applying
Applying
Applying
Therefore, as it is known that
So
