Answer
$AX = B$:
\begin{gather}
\begin{bmatrix}
1 & 2 \\ 3 & 4
\end{bmatrix}
\begin{bmatrix}
x \\ y
\end{bmatrix}
=
\begin{bmatrix}
0 \\ 2
\end{bmatrix}
\end{gather}
\begin{gather}
X =
\begin{bmatrix}
x \\ y
\end{bmatrix}
=
\begin{bmatrix}
2 \\ - 1
\end{bmatrix}
\end{gather}
Work Step by Step
1. Write the system as a matrix equation.
$A$ is the coefficient matrix, $X$ is the matrix that contains $x$ and $y$, the unknowns. And $B$ has the constants:
Thus:
$AX = B$
\begin{gather}
\begin{bmatrix}
1 & 2 \\ 3 & 4
\end{bmatrix}
\begin{bmatrix}
x \\ y
\end{bmatrix}
=
\begin{bmatrix}
0 \\ 2
\end{bmatrix}
\end{gather}
2. Invert $A$:
\begin{gather}
\begin{bmatrix}
1 & 2 & 1 & 0 \\ 3 & 4 & 0 & 1
\end{bmatrix}
\begin{matrix}
\\ R_2 - 3R_1
\end{matrix}
\end{gather}
\begin{gather}
\begin{bmatrix}
1 & 2 & 1 & 0\\ 0 & -2 & -3 & 1
\end{bmatrix}
\begin{matrix}
R_1 + R_2 \\ {}
\end{matrix}
\end{gather}
\begin{gather}
\begin{bmatrix}
1 & 0 & - 2 & 1 \\ 0 & -2 & -3 & 1
\end{bmatrix}
\begin{matrix}
\\ -\frac 12 R_2
\end{matrix}
\end{gather}
\begin{gather}
\begin{bmatrix}
1 & 0 & - 2 & 1 \\ 0 & 1 & \frac 32 & -\frac 12
\end{bmatrix}
\begin{matrix}
\end{matrix}
\end{gather}
\begin{gather}
A^{-1} =
\begin{bmatrix}
- 2 & 1 \\ \frac 32 & -\frac 12
\end{bmatrix}
\begin{matrix}
\end{matrix}
\end{gather}
3. Using $AX = B \longrightarrow X = A^{-1}B$, calculate $X$
\begin{gather}
X =
\begin{bmatrix}
- 2 & 1 \\ \frac 32 & -\frac 12
\end{bmatrix}
\begin{bmatrix}
0 \\ 2
\end{bmatrix}
=
\begin{bmatrix}
0 + 2 \\ 0 - 1
\end{bmatrix}
=
\begin{bmatrix}
2 \\ - 1
\end{bmatrix}
\end{gather}
- Therefore, x = 2 and y = -1
