Hi
Can someone help me solve this matrix equation ? (It's my first matrix equation, and I don't know how to start)
OK, for starters, you should make sure there's enough space between the numbers to ensure that they can be read individually and not as single numbers like I read them.
Probably the easiest thing to do would be to multiply out the LHS and then use row operations to solve x1, x2, x3.
You have $\displaystyle \begin{bmatrix}x_1 & x_2 & x_3\end{bmatrix}\begin{bmatrix}1 & 3 & 2 & 4 \\ 3 & 7 & 2 & 8 \\ 2 & 4 & 0 & 4\end{bmatrix}= \begin{bmatrix}2 & 5 & 2 & 6\end{bmatrix}$
Do you at least know how to multiply matrices? Doing the multiplication on the left,
$\displaystyle \begin{bmatrix}x_1+ 3x_2+ 2x_3 & 3x_1+ 7x_2+ 4x_3 & 2x_1+ 2x_2+ 2x_3 & 4x_1+ 8x_2+ 4x_3\end{bmatrix}= \begin{bmatrix}2 & 5 & 2 & 6\end{bmatrix}$
Can you solve the equations $\displaystyle x_1+ 3x_2+ 2x_3= 2$, $\displaystyle 3x_1+ 7x_2+ 4x_3= 5$, $\displaystyle 2x1+ 2x_2+ 2x_3= 2$, and $\displaystyle 4x_1+ 8x_2+ 4x_3= 6$?
Since you have 4 equations in 3 unknowns, these equations may be "over determined". Solve three of them for $\displaystyle x_1$, $\displaystyle x_2$, $\displaystyle x_3$ and see if those value satisfy the other equation.
Yes, it is wrong. You are reducing rows on the left but the but the right has only one row. That method would work for "Ax= b" but your equation is "xA= b". If you want to use that method (personally, I would prefer to solve the equations resulting from the matrix equation that I gave you before) take the transpose of the matrix and row reduce:
$\displaystyle \begin{bmatrix}1 & 3 & 2 & 2 \\ 3 & 7 & 2 & 5 \\ 2 & 2 & 0 & 2 \\ 4 & 8 & 4 & 6\end{bmatrix}\begin{bmatrix}2 \\ 5 \\ 2 \\ 6\end{bmatrix}$
You should find one row that is all 0s so that there are only three independent equations allowing you to solve the other three for $\displaystyle x_1$, $\displaystyle x_2$, and $\displaystyle x_3$.