Show that two matrices are similar Show that $A_{n\times n}=[a_{jk}]$ is similar to $B_{n\times n}=[(-1)^{j+k}a_{jk}]$
I tried to find an $X$ such that $X^{-1}AX=B$. This is easy for a given 2x2 matrix. But how can I find an X for nxn matrices?
 A: Hint:
$$
b_{jk}= ( (-1)^j\delta_{jj}) \cdot a_{jk} \cdot ((-1)^k\delta_{kk})
= \sum_{i,l} ( (-1)^j\delta_{ji}) \cdot a_{il} \cdot ((-1)^k\delta_{lk})
$$
with $\delta_{ij}=0$ if $i\ne j$, $\delta_{ii}=1$.
A: Since similarity is an equivalence relation, try exploiting transitivity and using an induction argument. Informally, B is obtained from A by flipping the sign on (some) of the entries. Suppose $B_1$ were obtained from A by flipping only one of (those) entries, could you find the matrix X then (maybe by considering elementary matrix operations)? Then $B_2$ obtained from $B_1$ by flipping one more of (those) entries... etc. 
A: A hint that is almost an answer...
Let us take the case $n=3$, from which it is easy to establish the general case:
$$\pmatrix{1&&\\&-1&\\&&1}\pmatrix{a&b&c\\d&e&f\\g&h&i}\pmatrix{1&&\\&-1&\\&&1}=\left(\begin{array}{rrr}a&-b&c\\-d&e&-f\\g&-h&i\end{array}\right)$$
(note that this diagonal matrix is its own inverse)
Explanation: Left multiplication (resp. right multiplication) of $\pmatrix{a&b&c\\d&e&f\\g&h&i}$ by $\pmatrix{1&&\\&-1&\\&&1}$ gives a sign change on even numbered columns (resp. on even numbered rows). 
Underweaving the two actions provides the desired checkerboard pattern. 
Remark: See the use of a similar technique in my answer to (How to classify all nonsingular $n \times n$ matrices $A$ whose all entries in $A$ and $A^{-1}$ are non-negatives.)
A: $$det(X) = \sum_{k = 1}^{n!} ( \prod_{r=1}^n x_{r,p_k(r)})$$ where $p_k(r) runs through all the n! permutations of (1,2,3,...,n). 
Each of these products takes one element from each row and each column of the matrix X. 
Matrix B is the same as A with each entry entry multiplied by its position the "checkerboard" sign matrix
$$\pmatrix{+&-&+&-&\ldots& -\\-&+&-&+ &\ldots & +\\ +&-&+&-&\ldots& -\\-&+&-&+& \ldots&+\\ \ &\ & \  &\ddots\\-&+&-&+& \ldots & +\\}$$
This matrix illustrates concretely $sgn(-1)^{i+j}$ It is helpful in visualizing determinants by cofactors etc., and here to visualize how B is found from A.
We need to figure out the product of the parities $ (-1)^{r + p(r)}$ for A and B and show they are the same, or else the hypothesis is false.
I'll leave you with this much here and see if I can come back later.
