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Question

Find inverse, by elementary row operations (if possible), of the following matrices
$$\begin{bmatrix} 1 & 3 \\ -5 & 7 \end{bmatrix}$$


Solution

 To check if the inverse exist we find the determinant:
We have:
$$A=\begin{bmatrix}
 1 & 3 \\
 -5 & 7
\end{bmatrix}$$

So, $$\left|A\right|=1\times 7-(-5\times 3)$$

$$\Rightarrow \left|A\right|=7+15=22$$

Since, $$|A|\ne0$$, hence the inverse exists
 
Now, in order to use elementary row operations, we may write $$A=IA.$$
$$\therefore \begin{bmatrix}
 1 & 3 \\
 -5 & 7
\end{bmatrix}= \begin{bmatrix}
 1 & 0 \\
 0 & 1
\end{bmatrix}A$$

$$\Rightarrow \begin{bmatrix}
 1 & 3 \\
0 & 22
\end{bmatrix}= \begin{bmatrix}
 1 & 0 \\
 0 & 1
\end{bmatrix}A$$                           $$R_2 \rightarrow R_2+ 5\times R_1$$
    
$$\Rightarrow \begin{bmatrix}
 1 & 3 \\
0 & 1
\end{bmatrix}= \begin{bmatrix}
 1 & 0 \\
 \dfrac{5}{22} & \dfrac{1}{22}
\end{bmatrix}A$$                      $$R_2 \rightarrow \dfrac{1}{22} \times R_2$$
$$\Rightarrow \begin{bmatrix}
 1 & 0 \\
0 & 1
\end{bmatrix}= \begin{bmatrix}
 \dfrac{7 }{22} & -\dfrac{3}{22} \\
 \dfrac{5}{22} & \dfrac{1}{22}
\end{bmatrix}A$$                   $$R_1 \rightarrow R_1- 3\times R_2$$
$$\Rightarrow \begin{bmatrix}
 1 & 0 \\
0 & 1
\end{bmatrix}=\dfrac{1}{22} \begin{bmatrix}
 7 & -3 \\
5 & 1
\end{bmatrix}A$$
$$\Rightarrow I=BA$$, where $$B$$ is the inverse of $$A$$.
$$B=\dfrac{1}{22} \begin{bmatrix}
 7 & -3 \\
5 & 1
\end{bmatrix}$$

Mathematics

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