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a. Show that if $A$ is any $ n \times n$ matrix, the constant term in the characteristic polynomial of $A$ is $\operatorname{det}(A)$.

b. Deduce that $A$ is invertible if and only if $\lambda = 0$ is not an eigenvalue of $A$.

I know that the characteristic polynomial $P(λ) = \operatorname{det}( A - \lambda I_n )$ and $\lambda$ is an eigenvalue if and only if $P(λ) = \operatorname{det}( A - \lambda I_n )=0$ but I'm not sure where to go from there.

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    $\begingroup$ What happens if you set $\lambda$ to zero in the formula for the characteristic polynomial? $\endgroup$ – amd Apr 10 '17 at 20:55
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Hint:

From the definition we have: $$ P(\lambda)= \det \begin{bmatrix} a_{11}-\lambda & a_{12} & \cdots &a_{1n}\\ a_{21} & a_{22}-\lambda & \cdots &a_{2n}\\ \cdots\\ a_{n1} & a_{n2} & \cdots &a_{nn}-\lambda\\ \end{bmatrix}= (-1)^n(\lambda^n+c_1\lambda^{n-1}+c_2\lambda^{n-2}+\cdots c_n) $$

take $\lambda=0$

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