Let $A$ be set of $3\times 3$ symmetric matrices whose entries are $1,1,1,-1,-1,-1,0,0,0$.
$B$ is one of the matrix of set $A$ and $X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] ,U=\left[ \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] $ then find:
Number of matrices in set $A$, and
Number of matrices in the set $A$ such that $BX=U$ has infinite solutions.
My attempt : Number of matrices $=3!\times 3!=36$ as there will be three different elements in the diagonal and three different elements in the upper triangle. But I am not able to solve the second part. For infinite solutions $|B|$ must be zero. Do I have to manually count cases? Please Help!
I agree with you.
Using the Sarrus rule, all NW-SE products are zero, and all SW-NE products involve an element of the main diagonal and a squared element from a triangle (hence $0$ or $1$). The sum of these products is $(-1,0,1)\cdot(1,0,1)$, with all $6\cdot6$ permutations allowed*, of which $6\cdot2$ yield zero. (There are $6\cdot2$ times $+1$ and $6\cdot2$ times $-1$.)
*Though two elements are equal, they originate from distinct elements in the matrix and all permutations count.