![\left[\begin{array}{ccc} -1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f03cb7f5858669bd60950918832693a7_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $A=\left(\begin{array}{ccc}-1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right)$ To get the inverse we will proceed by augmented matrix with elementary...
![\left[\begin{array}{ccc} 3 & 0 & -1 \\ 2 & 3 & 0 \\ 0 & 4 & 1 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-25dedef54c4e0355d69496a2dc00aa7b_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $A=\left(\begin{array}{ccc}1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5\end{array}\right)$ To get the inverse we will proceed by augmented matrix with elementary...
![\left[\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9b986c3bb2233699ef7411fc62105e4b_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $A=\left(\begin{array}{ccc}1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5\end{array}\right)$. To get the inverse we will proceed by augmented matrix with elementary...
![\left[\begin{array}{ccc} 3 & -1 & -2 \\ 2 & 0 & -1 \\ 3 & -5 & 0 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6f98ac5a8bdc0a4c68b59ad5fff13f27_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $A=\left(\begin{array}{ccc}3 & -1 & -2 \\ 2 & 0 & -1 \\ 3 & -5 & 0\end{array}\right)$ To get the inverse we will proceed by augmented matrix with elementary...
![\left[\begin{array}{ccc} 1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a3c18e3e41c6018e29633c0543144480_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $A=\left(\begin{array}{ccc}1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4\end{array}\right)$ To get the inverse we will proceed by augmented matrix with elementary...
![\left[\begin{array}{lll} 3 & 0 & 2 \\ 1 & 5 & 9 \\ 6 & 4 & 7 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c2d4be16f7e50993b9d9ddb636e32da5_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $A=\left(\begin{array}{lll}3 & 0 & 2 \\ 1 & 5 & 9 \\ 6 & 4 & 7\end{array}\right)$ To get the inverse we will proceed by augmented matrix with elementary row...
![\left[\begin{array}{ccc} 2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-17264154c86fb2057da2439f6e8db1ee_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $A=\left(\begin{array}{ccc}2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2\end{array}\right)$ To get the inverse we will proceed by augmented matrix with elementary...
![\left[\begin{array}{ccc} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9e97cd7932741e795ff74dfcc62f2080_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $A=\left(\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right)$ To get the inverse we will proceed by augmented matrix with elementary row...
![\left[\begin{array}{ll} 6 & 7 \\ 8 & 9 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-720c2b2a18ed5a12137848fe6835b85c_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $A=\left(\begin{array}{ll}6 & 7 \\ 8 & 9\end{array}\right)$. To get the inverse we will proceed by augmented matrix with elementary row transformation process is as follow:...
![\left[\begin{array}{cc} 2 & -3 \\ 4 & 5 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6d086db9247d6fa2dd93801088be8f5f_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $A=\left(\begin{array}{cc}2 & -3 \\ 4 & 5\end{array}\right)$ To get the inverse we will proceed by augmented matrix with elementary row transformation process is as follow:...
![\left[\begin{array}{rr} 2 & 5 \\ -3 & 1 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4e7a3750c7e99f5875f5fe5e581bc3bc_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $A=\left(\begin{array}{cc}2 & 5 \\ -3 & 1\end{array}\right)$ To get the inverse we will proceed by augmented matrix with elementary row transformation process is as follow:...
![\left[\begin{array}{ll} 1 & 2 \\ 3 & 7 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6a1968e1a662e8c4373a2fec80dd99d1_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $A=\left(\begin{array}{ll}1 & 2 \\ 3 & 7\end{array}\right)$ To get the inverse we will proceed by augmented matrix with elementary row transformation process is as follow:...