Question

How do I find the adjoint of this matrix?
(1 3 1)
(2 1 1)
(-2 2 -1)

Answer 1

cofactor transpose = adjoint

Answer 2

\[\begin{array}c
+\left|\begin{matrix}1&1\\2&-1
\end{matrix} \right|&
-\left|\begin{matrix}2&1\\-2&-1
\end{matrix} \right|&
+\left|\begin{matrix}2&1\\-2&2
\end{matrix} \right| \\\\
-\left|\begin{matrix}3&1\\2&-1
\end{matrix} \right|&
+\left|\begin{matrix}1&1\\-2&-1
\end{matrix} \right|&
-\left|\begin{matrix}1&3\\-2&2
\end{matrix} \right| \\\\
+\left|\begin{matrix}3&1\\1&1
\end{matrix} \right|&
-\left|\begin{matrix}1&1\\2&1
\end{matrix} \right|&
+\left|\begin{matrix}1&3\\2&1
\end{matrix} \right| \\\\
\end{array}\]

i just cant recall if we include the "entry" like in a determinant at the moment

Answer 3

nah, thats right

Answer 4

all those are little determinants; number them out and transpose this thing :)

Answer 5

-3 0 6
5 1 -8 ; chk my math to be sure, but then we transpose
2 1 -5


-3 5 2
0 1 1 ; would be the adjoint if I did it right :)
6 -8 -5

Answer 6

lol, the wolf likes it :)

http://www.wolframalpha.com/input/?i=adjoint+%5B%281%2C+3%2C+1%29%2C+%282%2C+1%2C+1%29%2C+%28-2%2C+2%2C+-1%29%5D