This question is in my assignment. We are not allowed to use any symbol to represent any elementary row and column operations used in the solution. We must solve it step-by-step. Please help me to check my solution word by word including my spelling and grammar.
Question:
Find the inverse of
$$A=\begin{pmatrix}2& 2& 3\\ 2& 5& 3\\ 1& 0& 8\end{pmatrix}$$
by using only elementary row operations.
Solution:
We begin by forming the matrix $\begin{pmatrix} A & | & I_3 \end{pmatrix}=\left(\begin{array}{ccc|ccc}2 & 2 & 3 & 1 & 0 & 0\\2 & 5 & 3 & 0 & 1 & 0\\1 & 0 & 8 & 0 & 0 & 1\end{array}\right)$. Interchanging the first and third rows of the matrix $\begin{pmatrix} A & | & I_3 \end{pmatrix}$, we obtain the matrix $\left(\begin{array}{ccc|ccc}1 & 0 & 8 & 0 & 0 & 1\\2 & 5 & 3 & 0 & 1 & 0\\2 & 2 & 3 & 1 & 0 & 0\end{array}\right)$. Adding $(-2)$ times the first row of the matrix $\left(\begin{array}{ccc|ccc}1 & 0 & 8 & 0 & 0 & 1\\2 & 5 & 3 & 0 & 1 & 0\\2 & 2 & 3 & 1 & 0 & 0\end{array}\right)$ to its second row, we obtain the matrix $\left(\begin{array}{ccc|ccc}1 & 0 & 8 & 0 & 0 & 1\\0 & 5 & -13 & 0 & 1 & -2\\2 & 2 & 3 & 1 & 0 & 0\end{array}\right)$. Multiplying the second row of the matrix $\left(\begin{array}{ccc|ccc}1 & 0 & 8 & 0 & 0 & 1\\0 & 5 & -13 & 0 & 1 & -2\\2 & 2 & 3 & 1 & 0 & 0\end{array}\right)$ by $\frac{1}{5}$, we obtain the matrix $\left(\begin{array}{ccc|ccc}1 & 0 & 8 & 0 & 0 & 1\\0 & 1 & -\frac{13}{5} & 0 & \frac{1}{5} & -\frac{2}{5}\\2 & 2 & 3 & 1 & 0 & 0\end{array}\right)$. Adding $(-2)$ times the first row of the matrix $\left(\begin{array}{ccc|ccc}1 & 0 & 8 & 0 & 0 & 1\\0 & 1 & -\frac{13}{5} & 0 & \frac{1}{5} & -\frac{2}{5}\\2 & 2 & 3 & 1 & 0 & 0\end{array}\right)$ to its third row, we obtain the matrix $\left(\begin{array}{ccc|ccc}1 & 0 & 8 & 0 & 0 & 1\\0 & 1 & -\frac{13}{5} & 0 & \frac{1}{5} & -\frac{2}{5}\\0 & 2 & -13 & 1 & 0 & -2\end{array}\right)$. Adding $(-2)$ times the second row of the matrix $\left(\begin{array}{ccc|ccc}1 & 0 & 8 & 0 & 0 & 1\\0 & 1 & -\frac{13}{5} & 0 & \frac{1}{5} & -\frac{2}{5}\\0 & 2 & -13 & 1 & 0 & -2\end{array}\right)$ to its third row, we obtain the matrix $\left(\begin{array}{ccc|ccc}1 & 0 & 8 & 0 & 0 & 1\\0 & 1 & -\frac{13}{5} & 0 & \frac{1}{5} & -\frac{2}{5}\\0 & 0 & -\frac{39}{5} & 1 & -\frac{2}{5} & -\frac{6}{5}\end{array}\right)$. Multiplying the third row of the matrix $\left(\begin{array}{ccc|ccc}1 & 0 & 8 & 0 & 0 & 1\\0 & 1 & -\frac{13}{5} & 0 & \frac{1}{5} & -\frac{2}{5}\\0 & 0 & -\frac{39}{5} & 1 & -\frac{2}{5} & -\frac{6}{5}\end{array}\right)$ by $(-\frac{5}{39})$, we obtain the matrix $\left(\begin{array}{ccc|ccc}1 & 0 & 8 & 0 & 0 & 1\\0 & 1 & -\frac{13}{5} & 0 & \frac{1}{5} & -\frac{2}{5}\\0 & 0 & 1 & -\frac{5}{39} & \frac{2}{39} & \frac{2}{13}\end{array}\right)$. Adding $(\frac{13}{5})$ times the third row of the matrix $\left(\begin{array}{ccc|ccc}1 & 0 & 8 & 0 & 0 & 1\\0 & 1 & -\frac{13}{5} & 0 & \frac{1}{5} & -\frac{2}{5}\\0 & 0 & 1 & -\frac{5}{39} & \frac{2}{39} & \frac{2}{13}\end{array}\right)$ to its second row, we obtain the matrix $\left(\begin{array}{ccc|ccc}1 & 0 & 8 & 0 & 0 & 1\\0 & 1 & 0 & -\frac{1}{3} & \frac{1}{3} & 0\\0 & 0 & 1 & -\frac{5}{39} & \frac{2}{39} & \frac{2}{13}\end{array}\right)$. Adding $(-8)$ times the third row of the matrix $\left(\begin{array}{ccc|ccc}1 & 0 & 8 & 0 & 0 & 1\\0 & 1 & 0 & -\frac{1}{3} & \frac{1}{3} & 0\\0 & 0 & 1 & -\frac{5}{39} & \frac{2}{39} & \frac{2}{13}\end{array}\right)$ to its first row, we obtain the matrix $\left(\begin{array}{ccc|ccc}1 & 0 & 0 & \frac{40}{39} & -\frac{16}{39} & -\frac{3}{13}\\0 & 1 & 0 & -\frac{1}{3} & \frac{1}{3} & 0\\0 & 0 & 1 & -\frac{5}{39} & \frac{2}{39} & \frac{2}{13}\end{array}\right)$. Thus, $A^{-1}=\begin{pmatrix}\frac{40}{39} & -\frac{16}{39} & -\frac{3}{13}\\ -\frac{1}{3} & \frac{1}{3} & 0\\ -\frac{5}{39} & \frac{2}{39} & \frac{2}{13}\end{pmatrix}$.
Bingo!
Well done. I even checked by multplying $AA^{-1}$ and $A^{-1}A$ and obtained $I_3$ each time. And this is confirmed here, as well.
Your solution is correct, very well-written, and easy to follow (albeit there's a lot to follow!).