TITLE: Finding a certain entry in a matrix
QUESTION [6 upvotes]: Can we find the entry $s_{23}$ of $S=H^3$ where $H=\pmatrix{2&-1&0\\3&1&2\\-1&1&1}
$ without finding $S$. I know that $s_{23}$ is given by multiplying the second row of $H^2$ and the third coloumn of $H$ but it is dull. Thank you for any hints!

REPLY [6 votes]: The characteristic polynomial of $H$ is $$X^3-4X^2+6X-3.$$
Therefore $$H^3=4H^2-6H+3I=H(4H-6I)+3I.$$
The element $(H^3)_{23}$ is then the product of the second row of $H$ times the third column of $4H-6I$ (plus nothing since $3I$ has element $(2,3)$ equal to $0$).
We need $18$ products and $12$ additions to get the polynomial, plus the expanding of terms. After that we need 1 addition (from the -6I), 3 multiplications (from the 4H), plus 3 multiplications and 2 additions from the second row-third column product.
But hey, this way is not dull!