Given $A=\left[\begin{array}{ccc}4 & 1 & 0 \\ 2 & -1 & 1 \\ 0 & 2 & 1\end{array}\right]$ And $P(x)=x^{3}-3 x+4$
Calculate $| P ( A )+2 I |$
a) $1648$
b) $1624$
c) $1528$
d) $1728$
e) $1696$
My try
$P(A)+2I=A^{3}-3 A+4+2I$
$P(A)+2I=A^{3}-3 A+4I+2I$
$P(A)+2I=A^{3}-3 A+6I$
= ${\left[\begin{array}{ccc}4 & 1 & 0 \\ 2 & -1 & 1 \\ 0 & 2 & 1\end{array}\right] }^{3} - 3{ \left[\begin{array}{ccc}4 & 1 & 0 \\ 2 & -1 & 1 \\ 0 & 2 & 1\end{array}\right]}+6I$
$P(A)+2I={\left[\begin{array}{ccc}72 & 14 & 4 \\ 28 & 10 & 2 \\ 16 & 4 & 6\end{array}\right] }$
$| P ( A )+2 I |$= $ \text{Det}(\left( \begin{array}{ccc} 72 & 14 & 4 \\ 28 & 10 & 2 \\ 16 & 4 & 6 \end{array} \right) )=1648$
Is there a short method for solving this problem since "cubing" $A$ is a bit time consuming .... thank you
It's not a lot easier, but here is the calculation using the characteristic polynomial and eigenvalues of $A$. We calculate the characteristic polynomial $$ |xI - A| = x^3 - 4x^2 - 5x + 14 = (x-2)(x-6x + 7). $$ By the Cayley Hamilton theorem, $$ P(A) + 2I = A^3 - 3A + 6I = A^3 - 3A + 6I - 0 \\= (A^3 - 3A + 6I) - (A^3 - 4A^2 - 5A + 14 I) \\= 4A^2 + 2A - 8I = 2(2A^2 + A - 4I). $$ Thus, we want to calculate the determinant of $P(A) + 2I = 2(2A^2 + A - 4I)$. Let $q(x) = 2(2x^2 + x - 4)$.
We find that the eigenvalues (the zeros of the characteristic polynomial) are $\lambda_1=-2, \lambda_2 = 3 + \sqrt{2}, \lambda_3 = 3 - \sqrt{2}$. The eigenvalues of $q(A)$ are equal to $q(\lambda_j)$ for $j = 1,2,3$, i.e. $$ q(\lambda_1) = 4, \quad q(\lambda_2) = 42 + 26\sqrt{2}, \quad q(\lambda_3) = 42 - 26\sqrt{2}. $$ The determinant of $q(A)$ is the product of these eigenvalues of $q(A)$, namely $$ 2^2 (2\cdot 21 + 2 \cdot 13\sqrt{2})(2\cdot 21 - 2 \cdot 13\sqrt{2}) = 2^4(21^2 - 2\cdot 13^2) = 16 \cdot 103 = 1648. $$