Here is the question,
"Let A be an invertible Matrix. Suppose
$$ Au = 5e_1 + 6e_2 + 7e_3 + 8e_4 $$ $$ Av = 5e_1 + 6e_2 + 9e_3 + 8e_4 $$ Where $e_i \epsilon \Re^4 $ is the $i^{th}$ standard basis vector. In terms of u and v find the third column of $A^{-1}$
My first thought is that I can represent v and u as $$v = [5\;6\; 9\; 8\;]A^{-1} $$ $$u = [5\; 6\; 7\; 8\;]A^{-1} $$ Which gives me the Matrix in u and v forms. But I am a little confused as to where to go from here. I feel like each column in $A^{-1} v and u are both 4*1 vectors meaning A is a 4*4 matrix. This is derived from the space we are in. That means that this 4*1 vector for each equation would be equal to the Inverse times that vector. But where do I go from here with this logic??
There was a hint at the end that the once the logic is worked out there is essentially no calculations at all.
they are writing vectors as column vectors when using matrix multiplication. You should be thinking of $u$ as a column vector of four numbers (the coefficients in the standard basis), then $$ Au = \left( \begin{array}{c} 5 \\ 6 \\ 7 \\ 8 \\ \end{array} \right) $$ $$ Av = \left( \begin{array}{c} 5 \\ 6 \\ 9 \\ 8 \\ \end{array} \right) $$ So then, what column vector is $w$ if I say $$ \frac{1}{2} A(v-u) = w \; ? $$ What happens when you multiply both sides of that on the left by $A^{-1} \; ? \;$