# Let $A=\begin{bmatrix} 1 & 2\\ 3& 4 \end{bmatrix}$ then det$(A^3-6A^2+5A+3I)=3$

Let $$A=\begin{bmatrix} 1 & 2\\ 3& 4 \end{bmatrix}$$ then det$$(A^3-6A^2+5A+3I)=3$$

det$$(A^3-6A^2+5A+3I)=$$det$$((A^2-5A-2I)(A-I)+2A+I)=$$det$$(2A+I)=3$$, Since a matrix satisfies its characteristic polynomial. Is this right?

• @J.W.Tanner Yes, is there a mistake? – rhaldryn Jun 26 at 10:30

Yes, this looks fine. Since $$A$$ satisfies its own characteristic polynomial, you have: $$\color{blue}{A^2-5A-2I=O}$$ and so, as you wrote: $$A^3-6A^2+5A+3I=\underbrace{\left(\color{blue}{A^2-5A-2I}\right)}_{\color{blue}{O}}\left(A-I\right)+2A+I=2A+I$$ which leaves you with the (easier) $$\det\left(2A+I\right)$$ and that is indeed $$3$$.
Here is another method. Assume that $$\lambda_i\in\mathbb{C}$$, $$i=1,2$$ are the two igenvalues of $$A$$. Let $$f(x)\in \mathbb{C}[x]$$ be the characteristic polynomial of $$A$$. Then $$f(x)=\det(A-xI)=x^2-5x-2.$$ The matrix $$A$$ can be diagonalized as $$A=P\begin{bmatrix} \lambda_1 & 0\\ 0& \lambda_2 \end{bmatrix}P^{-1},$$ where $$P$$ is an invertible matrix over $$\mathbb{C}$$. Let $$g(x)=x^3-6x^2+5x+3$$. We have $$g(x)=f(x)(x-1)+2x+1$$. It follows that $$\det(A^3-6A^2+5A+3I)=\det(g(A))=\det\left(g\left(\begin{bmatrix} \lambda_1 & 0\\ 0& \lambda_2 \end{bmatrix}\right)\right)=\det\left(\begin{bmatrix} g(\lambda_1) & 0\\ 0& g(\lambda_2) \end{bmatrix}\right)=g(\lambda_1)g(\lambda_2).$$ Note that $$f(\lambda_i)=0$$, $$i=1,2$$. Thus $$g(\lambda_i)=f(\lambda_i)(\lambda_i-1)+2\lambda_i+1=2\lambda_i+1$$. So $$\det(A^3-6A^2+5A+3I)=(2\lambda_1+1)(2\lambda_2+1)=4\lambda_1\lambda_2+2(\lambda_1+\lambda_2)+1.$$ According to Vieta theorem, we have $$\lambda_1\lambda_2=-2, \lambda_1+\lambda_2=5.$$ Consequently, $$\det(A^3-6A^2+5A+3I)=4\times (-2)+2\times 5 +1=3.$$