The characteristic equation for the matrix is \[ \begin{bmatrix} t & 3 \ 3 & -\epsilon \ 3 & \epsilon \end{bmatrix} = \lambda I \text{ where } I = \text{Identity matrix} \]

Steps to Solve

1. Identify the Matrix and the Values The given matrix appears to be: $$ A = \begin{bmatrix} t & 3 & 3 \ -e & 3 & -t \ s & 1 & e \end{bmatrix} $$ We need to find the characteristic polynomial, which is derived from the determinant of $A - \lambda I$, where $I$ is the identity matrix and $\lambda$ represents the eigenvalues.

2. Set Up the Characteristic Equation The characteristic equation is determined by: $$ \text{det}(A - \lambda I) = 0 $$ We represent $I$, the identity matrix, and adjust it according to $\lambda$: $$ A - \lambda I = \begin{bmatrix} t - \lambda & 3 & 3 \ -e & 3 - \lambda & -t \ s & 1 & e - \lambda \end{bmatrix} $$

3. Calculate the Determinant To find the determinant, we can use the formula for the determinant of a 3x3 matrix: If $$ B = \begin{bmatrix} a & b & c \ d & e & f \ g & h & i \end{bmatrix}, $$ then $$ \text{det}(B) = a(ei - fh) - b(di - fg) + c(dh - eg) $$

For our matrix $A - \lambda I$, we will expand the determinant using this formula.

4. Expanding the Determinant Let’s denote the elements of $A - \lambda I$ as:

• $a = t - \lambda$
• $b = 3$
• $c = 3$
• $d = -e$
• $e = 3 - \lambda$
• $f = -t$
• $g = s$
• $h = 1$
• $i = e - \lambda$

We can substitute these into the determinant formula: $$ \text{det}(A - \lambda I) = (t - \lambda)((3 - \lambda)(e - \lambda) + t) - 3(-e(e - \lambda) + st) + 3(s(3 - \lambda) + e) $$

5. Simplify and Solve Now we need to simplify the expression obtained from the determinant and set it to zero to find the characteristic polynomial: $$ \text{det}(A - \lambda I) = 0 $$

6. Resulting Characteristic Polynomial After simplifying, the final form of the characteristic polynomial will yield the eigenvalues of the matrix.