Solution:
For elementary row operation we have,
![\begin{array}{l} A=I A \\ \Rightarrow\left[\begin{array}{ccc} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \mathrm{A} \end{array}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0ba0ed5f8227333695ee96c77a0e6671_l3.png "Rendered by QuickLaTeX.com")
Applying 
![\Rightarrow\left[\begin{array}{ccc} 1 & 0 & -\frac{1}{2} \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{array}\right]=\left[\begin{array}{lll} \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-993175531bf21aa9e3f75cb91ce7abf9_l3.png "Rendered by QuickLaTeX.com")
Applying 
![\Rightarrow\left[\begin{array}{ccc} 1 & 0 & -\frac{1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 1 & 3 \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{2} & 0 & 0 \\ -\frac{5}{2} & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ae0f465ac732629e146568c16d0f4a0d_l3.png "Rendered by QuickLaTeX.com")
Applying 
![\Rightarrow\left[\begin{array}{ccc} 1 & 0 & -\frac{1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 0 & \frac{1}{2} \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{2} & 0 & 0 \\ -\frac{5}{2} & 1 & 0 \\ \frac{1}{2} & -1 & 1 \end{array}\right] \mathrm{A}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-78e661866d71eab67453e432ce07c53c_l3.png "Rendered by QuickLaTeX.com")
Applying 
![\Rightarrow\left[\begin{array}{ccc} 1 & 0 & -\frac{1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{2} & 0 & 0 \\ -\frac{5}{2} & 1 & 0 \\ 5 & -2 & 2 \end{array}\right] A](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e78737f808a80b925e69f692e1599fbb_l3.png "Rendered by QuickLaTeX.com")
Applying 
![\Rightarrow\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right] A](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-73ac2bb3eca297d9a563279796bd0df3_l3.png "Rendered by QuickLaTeX.com")
Therefore, as it is known that
Therefore
![\Rightarrow A^{-1}=\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f20dc16d86d23c6c4e160c8386a12eb9_l3.png "Rendered by QuickLaTeX.com")
Find the inverse of the following matrices by using elementary row transformations: ![\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3658077a1580407fff60d37d9541b6e4_l3.png "Rendered by QuickLaTeX.com")
Solution:
For elementary row operation we have,
Applying
Applying
Applying
Applying
Applying
Therefore, as it is known that
Therefore
