Let $A = \begin{pmatrix} 1 & 4 \\ 2 & 3 \end{pmatrix}$. According to the Cayley–Hamilton theorem, the matrix $A^3 - 4A^2 - 5A + 7I$ can be reduced to $k I$. The value of $k$ is ________ (answer in integer).
Answer Key:
7
Step-by-Step Solution & Explanation
**Method 1: Cayley–Hamilton Annihilation**
The characteristic equation of $A$ is:
$\det(A - \lambda I) = \begin{vmatrix} 1 - \lambda & 4 \\ 2 & 3 - \lambda \end{vmatrix} = (1 - \lambda)(3 - \lambda) - 8 = \lambda^2 - 4\lambda - 5 = 0$
By the Cayley–Hamilton theorem, $A$ satisfies its own characteristic equation:
$A^2 - 4A - 5I = O$
Multiplying through by $A$:
$A^3 - 4A^2 - 5A = O$
Therefore:
$A^3 - 4A^2 - 5A + 7I = O + 7I = 7I$
Thus $k = 7$.