1. Problem statement: Given the matrix $\begin{bmatrix}a&b\\c&d\end{bmatrix}$ compute $F(-3)$ where $F(t)=\det(A - tI)$.\n2. Formula used: For a 2x2 matrix $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$ the characteristic polynomial is $F(t)=\det(A - tI)=(a-t)(d-t)-bc$.\n3. Important rule: To evaluate at $t=-3$ we compute $F(-3)=\det(A+3I)$ because $-(-3)=+3$.\n4. Compute determinant: $$F(-3)=\det\left(\begin{bmatrix}a&b\\c&d\end{bmatrix}+3I\right)=\det\begin{bmatrix}a+3&b\\c&d+3\end{bmatrix}=(a+3)(d+3)-bc$$\n5. Expand intermediate steps: Expand $(a+3)(d+3)-bc$ to get $ad+3a+3d+9-bc$.\n6. Simplify and present final answer: Therefore $F(-3)=ad-bc+3(a+d)+9$.\n