Answer
the system of equation is: $\quad \begin{array}{llll}
&x &- &3y &+ &2z & =-6\\
&2x &- &5y &+ &3z & =-4\\
&-3x &- &6y &+&4z & =\space 6
\end{array}$
Performing the indicated operations yield:
$\left[\begin{array}{ccc|c}
{1}&{-3}&{2}&{-6}\\
{0}&{1}&{-1}&{8}\\
{0}&{-15}&{10}&{-12}\end{array}\right]$
Work Step by Step
The standard form of a linear equation can be expressed as:
$$a_{i1}x_{1}+a_{i2}x_{2}+..........+a_{in}x_{n}=b_{i}\\
\text{where, the index $i$ indicates that it is the i-th equation of a system of equations.
}$$
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 lline.
4. The constants of the $B=[b_{i}]$ must be placed to the right of the line.
We can write the system as an augmented matrix $[A|B]$ as follows:
$\left\{\begin{array}{llll}
x & -3y & +2z & =-6\\
2x & -5y & +3z & =-4\\
-3x & -6y & +4z & =6
\end{array}\right. \rightarrow\left[\begin{array}{rrr|r}
{1}&{-3}&{2}&{-6}\\
{2}&{-5}&{3}&{-4}\\
{-3}&{-6}&{4}&{6}\end{array}\right]$
We perform the row operation as:
$R_{2}=-2r_{1}+r_{2}$
$R_{3}=3r_{1}+r_{3}$
$=\left[\begin{array}{ccc|c}
{1} &{-3} &{2} &{-6}\\
{(-2)(1)+2} &{(-2)(-3)-5} &{(-2)(2)+3}&{(-2)(-6)-4}\\
{3(1)-3}&{3(-3)-6}&{3(2)+4}&{3(-6)+6}\end{array}\right] \\=\left[\begin{array}{ccc|c}
{1}&{-3}&{2}&{-6}\\
{0}&{1}&{-1}&{8}\\
{0}&{-15}&{10}&{-12}\end{array}\right]$
