Finding Eigenvalue det(λI - A); 
I want to know if what I'm doing to derive equation (2) from (M2) is correct or not; usually, before moving onto the next row in Guass-Jordan elimination we turn a_11 into a leading one or whatever is needed to get leading ones, but I want to know if I can multiply a certain row with a value (even if there are no leading ones in the said row) and add it to another row, e.g. multiplying row 1 in (M2) by $\left(\frac{8}{λ-3}\right)$ and adding that to row 2 in the same matrix, to produce (M3). Is this correct? Because I cannot think of any other way to obtain equation (2).
 A: We wish to compute $\det(\lambda I-A)$ where
\begin{align*}
I&=\begin{bmatrix}1&0\\0&1\end{bmatrix} &
A&=\begin{bmatrix}3&0\\8&-1\end{bmatrix}
\end{align*}
Keeping the formula
$$
\det\begin{bmatrix}a&b\\ c&d\end{bmatrix}=ad-bc
$$
in mind, we may compute our determinant directly
\begin{align*}
\det(\lambda I-A)
&=\det\left(\lambda\begin{bmatrix}1&0\\0&1\end{bmatrix}-\begin{bmatrix}3&0\\8&-1\end{bmatrix}\right) \\
&=\det\left(\begin{bmatrix}\lambda&0\\0&\lambda\end{bmatrix}-\begin{bmatrix}3&0\\8&-1\end{bmatrix}\right) \\
&= \det\begin{bmatrix}\lambda-3& 0-0\\ 0-8&\lambda-(-1)\end{bmatrix}\\
&=\det\begin{bmatrix}\lambda-3& 0 \\ -8& \lambda+1\end{bmatrix} \\
&= (\lambda-3)(\lambda+1)-(-8)(0) \\
&= (\lambda-3)(\lambda+1)
\end{align*}
which gives the desired expression.
A: Notice, method is straight forward  $$\lambda I=\lambda\begin{bmatrix}1&0 \\0&1
\end{bmatrix}=\begin{bmatrix}\lambda&0 \\0&\lambda
\end{bmatrix}$$ & $$A=\begin{bmatrix}3&0 \\8&-1 \end{bmatrix}$$ $$\implies \lambda I-A=\begin{bmatrix}\lambda&0 \\0&\lambda
\end{bmatrix}-\begin{bmatrix}3&0 \\8&-1 \end{bmatrix}=\begin{bmatrix}\lambda-3&0 \\-8&\lambda+1 \end{bmatrix}$$ Now, for calculating eigen values we have $$|\lambda I-A|=0$$ $$\implies \left|\begin{matrix}\lambda-3&0 \\-8&\lambda+1 \end{matrix}\right|=0$$ $$\implies \color{blue}{(\lambda-3)(\lambda+1)=0}$$ Obviously your derivation of equation (2) from ($M_2$) is correct.  
