Answer
$\left[\begin{array}{ccc|c}
{1}&{7}&{-11}&{2}\\
{2}&{-5}&{6}&{-2}\\
{0}&{-9}&{16}&{2}\end{array}\right]$
Work Step by Step
The standard form of a linear equation can be expressed as:
$a_{i1}x_{1}+a_{i2}x_{2}+...+a_{in}x_{n}=b_{i}$
where, the index $i$ indicates that it is the i-th equation of a system of equations.
In order to write the augmented matrix $[A|B]$ of a system of equations in standard form, we must follow some important points:
1. To express a system in matrix form, we must extract the coefficients of the variables and constants.
2. Draw a vertical line to separate the coefficient entries from the constants (essentially replacing the equal signs).
3. The entries of the coefficient matrix $A=[a_{ij}]$ must be placed to the left of the line.
4. The constants of $B=[b_{i}]$ must be placed to the right of the line.
We can write the system as an augmented matrix $[A|B]$ as follows:
$\left\{\begin{array}{llll}
5x & -3y & +z & =-2\\
2x & -5y & +6z & =-2\\
-4x & +y & +4z & =6
\end{array}\right. \rightarrow\left[\begin{array}{rrr|r}
{5}&{-3}&{1}&{-2}\\
{2}&{-5}&{6}&{-2}\\
{-4}&{1}&{4}&{6}\end{array}\right]$
We perform the row operation as:
$R_{1}=-2r_{2}+r_{1}$
$R_{3}=2r_{2}+r_{3}$
$=\left[\begin{array}{ccc|c}
{(-2)(2)+5} &{(-2)(-5)-3}&{(-2)(6)+1}&{(-2)(-2)-2}\\
{2} &{-5} &{6} &{-2}\\
{2(2)+(-4)}&{(2)(-5)+1}&{2(6)+4}&{(2)(-2)+6}\end{array}\right]$
$=\left[\begin{array}{ccc|c}
{1}&{7}&{-11}&{2}\\
{2}&{-5}&{6}&{-2}\\
{0}&{-9}&{16}&{2}\end{array}\right]$
