An n×n matrix B is define e^B by e^B=∑(B^j/j!) . Let P be the char polynomial of B.then the matrix e^P(B) is An $n \times n$ matrix $B$ defines $e^B$ by $e^B=\sum\left(B^j/j!\right)$. Let $P$ be the char polynomial of $B$. Then the matrix $e^{P(B)}$ is


$I(n \times n)$
$O(n \times n)$
$eI(n \times n)$
$πI(n \times n)$
By the Cayley-Hamilton theorem, every matrix satisfies it's own characteristic polynomial.  Thus

$P(B) = 0; \tag 1$

it follows that

$e^{P(B)} = e^0 = I. \tag 2$
By the Cayley-Hamilton theorem, every matrix satisfies it's own characteristic polynomial.  Thus

$P(B) = 0; \tag 1$

it follows that

$e^{P(B)} = e^0 = I. \tag 2$