Question

Prove that , without expanding:

Answer 1

Huh ? x}

Answer 2

Prove that it's equal to
= AX + BY

Answer 3

we can apply the Gauss method

Answer 4

Out of my curriculum.

Answer 5

we can substitute the second row with the subsequent row:
\[\begin{gathered}
\left( {B,1,0} \right) - \frac{B}{A}\left( {A,0,1} \right) = \hfill \\
\hfill \\
= \left( {B - B,1, - \frac{B}{A}} \right) = \left( {0,1, - \frac{B}{A}} \right) \hfill \\
\end{gathered} \]

Answer 6

Please write in det style , this style confuses me.

Answer 7

so we get a matrix equivalent by row to the original matrix, and the obtained matrix, is:
\[\left( {\begin{array}{*{20}{c}}
A&0&1 \\
0&1&{ - B/A} \\
0&x&y
\end{array}} \right)\]

Answer 8

what do you think about my method?

Answer 9

next we can substitute the third row, with this row:
\[\begin{gathered}
\left( {0,x,y} \right) - x\left( {0,1, - \frac{B}{A}} \right) = \hfill \\
\hfill \\
= \left( {0,x,y} \right) - \left( {0,x, - \frac{B}{A}x} \right) = \left( {0,0,y + \frac{B}{A}x} \right) \hfill \\
\end{gathered} \]

Answer 10

Nice , but I can't write twice I have to relog to write once again , something is wrong over here.

Answer 11

so we get the subsequent matrix equivalent by row to the original matrix:
\[\left( {\begin{array}{*{20}{c}}
A&0&1 \\
0&1&{ - B/A} \\
0&0&{y + \frac{B}{A}x}
\end{array}} \right)\]

Answer 12

namely a triangular matrix, whose determinant is given by the product of its diagonal elements, so we have:
\[\Large {\text{determinant}} = A \times 1 \times \left( {y + \frac{B}{A}x} \right) = ...?\]

Answer 13

@TrojanPoem

Answer 14

Sorry my pc was lagging and made me invisible.
THANKS , you are the best !