Question

Solve the matrix equation. [2,1,4,3][x,y]=[10,-2]

Answer 1

This?\[\left[\begin{matrix}
2 & 1\\
4 & 3
\end{matrix}\right]
\left[\begin{matrix}
x\\
y
\end{matrix}\right]=\left[\begin{matrix}
10\\
-2
\end{matrix}\right]\]

Answer 2

There are a few ways to solve a matrix. Are they saying you need to use a specific method, or just to get the answer?

Answer 3

Cramer's Rule
\[x=\frac{|A_x|}{|A|}\quad x=\frac{|A_y|}{|A|}\]
Gauss
Row reduction until it is in upper triangular form. Then use back substitution.

Gauss-Jordan
\[Ax=b\implies Is=x\]where s is a solution found through reduced row echelon form.

If you still need help when you get back online, reply and I (or many others) can work with you on any of those.

Answer 4

oops... and that second x in Cramer's is supposed to be a y.

Answer 5

Yes that's the equation I meant

Answer 6

there are many methods, so you'd need to let me know which to use.

you are probably using either row reduction or using the inverse. lmk :)

Answer 7

\[\left[\begin{matrix}
2 & 1\\
4 & 3
\end{matrix}\right]
\left[\begin{matrix}
x\\
y
\end{matrix}\right]=\left[\begin{matrix}
10\\
-2
\end{matrix}\right]\implies

\left[\begin{matrix}
2 & 1&|&10\\
4 & 3&|&-2
\end{matrix}\right]
\]That is the setup for row reduction. If you are using that, that is the start. If you are being told to use Cramers:
\[A=\left[\begin{matrix}
2 & 1\\
4 & 3
\end{matrix}\right],\quad
A_x=\left[\begin{matrix}
10 & 1\\
-2 & 3
\end{matrix}\right],\quad
A_y=\left[\begin{matrix}
2 & 10\\
4 & -2
\end{matrix}\right]
\]Then solve:
\[x=\frac{|A_x|}{|A|}\quad y=\frac{|A_y|}{|A|}\]Where the |A| stuff means determinant.