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 $n=4$, $$\det(A-\lambda I)=(-1)^4 \det(\lambda I-A)=\det(\lambda I-A)$$ $$\beta_1=-\text {trace} A=-(2+2+2+4)=-10,$$ $$\beta_2=0+4+8+4+8+8=32$$ $$\beta_3=-(16+16+0+0)=-32$$ $$\beta_4=\det A=\text { by generalized Laplace expansion on rows 1&2 }=0 \times 8=0.$$ $$\det(A-\lambda I=\det(\lambda I-A)=\lambda^4-10\lambda^3+32\lambda^2-32\lambda.$$