How to compute characteristic polynomial? 
Given a square matrix $\mathbf{X}$, the characteristic polynomial of $\mathbf{X}$ is defined as
  $$P_X(t) = \det (t \mathbf{M} - \mathbf{X}).$$(The matrix $\mathbf{M}$ is the identity matrix that has the same dimensions as $\mathbf{X}$. Note that when we expand this determinant, we obtain a polynomial in $t$.)
(a) Compute the characteristic polynomial of the matrix
  $$\mathbf{X} = \begin{pmatrix} x & y \\ w & z \end{pmatrix}.$$Express your answer in the form $p_2 t^2 + p_1 t + p_0$.
(b) For the polynomial in part (a), find
  $$p_2 \mathbf{X}^2 + p_1 \mathbf{X} + p_0 \mathbf{M}.$$

I am a bit lost where to start. Am I supposed to express $p_2$, $p_1$, and $p_0$ as values of $x$, $y$, $w$, and $z$? Does $t$ mean any constant? Even making these assumptions, I am not sure where to start / how to do. Any help would be appreciated!
 A: A:
We are given the the Polynomial:
$$P_X(t) = \det (t \mathbf{I} - \mathbf{X}).$$
Substituting $\mathbf{X}$ and the identity matrix $\mathbf{I}$ yields:
$$P_X(t) = \det (t \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} - \begin{pmatrix} x & y \\ w & z \end{pmatrix})$$ $$\Rightarrow P_X(t) = \det (\begin{pmatrix} t & 0 \\ 0 & t \end{pmatrix} - \begin{pmatrix} x & y \\ w & z \end{pmatrix})$$ $$\Rightarrow P_X(t) = \det (\begin{pmatrix} t-x & -y \\ -w & t-z \end{pmatrix})$$
Now we can solve the determinant:
$$\Rightarrow P_X(t) = \det (\begin{pmatrix} t-x & -y \\ -w & t-z \end{pmatrix})$$$$\Rightarrow P_X(t) = (t-x)(t-z) - yw$$$$\Rightarrow P_X(t) = t^2 -xt -zt +xz - yw$$$$\boxed{ P_X(t) = t^2 -(x+z)t + xz - yw \quad \text{where} \quad p_2 = 1 \quad p_1 = -(x+z) \quad p_0 = xz - yw}$$
B:
Here we need to use $p_2$, $p_1$, and $p_0$ from problem A.
We know $p_2 = 1$, $p_1 = -(x+z)$, and $p_0 = xz - yw$. Subsituting this into what we need to find, we yield:
$$p_2 \mathbf{X}^2 + p_1 \mathbf{X} + p_0 \mathbf{I}.$$ $$\Rightarrow 1 \mathbf{X}^2 -(x+z)\mathbf{X} + (xz - yw) \mathbf{I}$$$$\Rightarrow \begin{pmatrix} x & y \\ w & z \end{pmatrix}^2 -(x+z)\begin{pmatrix} x & y \\ w & z \end{pmatrix} + (xz - wy) \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$$$\Rightarrow \begin{pmatrix} x^2 + yw & yx+yz \\ wx+wz & z^2 +yw \end{pmatrix} + \begin{pmatrix} -xz-x^2 & -xy-yz \\ -xw-wz & -xz-z^2 \end{pmatrix} + \begin{pmatrix} xz - yw & 0 \\ 0 & xz - yw \end{pmatrix} $$ $$\Rightarrow \begin{pmatrix} x^2 + (-x - z) x + z x & x y + (-x - z) y + z y \\ x w + (-x - z) w + z w & z^2 + x z + (-x - z) z \end{pmatrix} $$ $$ \Rightarrow \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = \boxed{0}$$
A: If you actually calculate what the characteristic polynomial is, you would get:
$\begin{pmatrix} t-x & y \\ w & t-z \end{pmatrix}=(t-x)(t-z)-wy=t^2-(x+z)t-wy$
Then $p_2=1,p_1=-(x+z),p_0=-wy$. While $t$ is a simply a variable, the same as in a regular polynomial.
The second part is simply to calculate what $\mathbf{X}^2$ is and add the three according matrices.  
