Why is there a reduction to REF to find the characteristic polynomial of a 4x4 matrix? I'm asked to find the characteristic polynomial of the matrix: $A = \small\begin{pmatrix} 
2 & 2 & 0 & 0\\
2 & 2 & 0 & 0\\
0 & 0 & 2 & 2\\
0 & 0 & 0 & 4\\
\end{pmatrix}$. I've gotten $$\begin{equation}
\begin{split}
p_A(t) &= \det (A-tI)\\ 
&= det \small\begin{pmatrix}
2-t & 2 & 0 & 0\\
2 & 2-t & 0 & 0\\
0 & 0 & 2-t & 2\\
0 & 0 & 0 & 4-t\\
\end{pmatrix}
\end{split}
\end{equation}$$
However, the mistake I seem to make comes from my evaluation of $det \small\begin{pmatrix}
2-t & 2 & 0 & 0\\
2 & 2-t & 0 & 0\\
0 & 0 & 2-t & 2\\
0 & 0 & 0 & 4-t\\
\end{pmatrix}$. 
My way of evaluating is:
$$\begin{equation} \begin{split}
det \small\begin{pmatrix}
2-t & 2 & 0 & 0\\
2 & 2-t & 0 & 0\\
0 & 0 & 2-t & 2\\
0 & 0 & 0 & 4-t\\
\end{pmatrix} &= (2-t)(2-t)(2-t)(4-t) - 0\\
&=(2-t)^3(4-t).
\end{split} \end{equation}$$
When I use an eigenvalue calculator to check my polynomial, I see $p_A(t) = (4-t)^2(2-t)t$. The step-by-step solution says to find the determinant of $A-tI$ you have to reduce $A-tI$ to REF. My question is: Why do we have to reduce $A-tI$ to its REF?

For an $n \times n$ matrix $A$ the best formula to use is $$\det(\lambda I-A))=\lambda^n+\sum_{i=1}^n \beta_i\lambda^{n-i} $$ where $\beta_i=(-1)^i$ sum of principal minors of order $i$. Note that since $n=4$, $$\det(A-\lambda I)=(-1)^4 \det(\lambda I-A)=\det(\lambda I-A)$$ $$\beta_1=-\text {trace} A=-(2+2+2+4)=-10,$$ $$\beta_2=0+4+8+4+8+8=32$$
$$\beta_3=-(16+16+0+0)=-32$$ $$\beta_4=\det A=\text { by generalized Laplace expansion on rows 1&2 }=0 \times 8=0.$$
$$\det(A-\lambda I=\det(\lambda I-A)=\lambda^4-10\lambda^3+32\lambda^2-32\lambda.$$
