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

Suggest Corrections

0

Similar questions
View More

People also searched for
View More