605-Reply to a Discussion 3

T10† Suppose that A is a square matrix. Prove that the constant term of the characteristic polynomial of A is equal to the determinant of A.

$$A(x) = \sum_{0}^{n} a_{n}x^{n}$$ For $n=0, x = 0$ => $A(0) = a_{0} = constant$

Therefore $\lambda = 0$,

=> $A(0) = det(A - \lambda I_{n}) = det(A - 0 I_{n}) = det(A)$