Answer
The system of equations is: $\quad \quad \begin{array}{llll}
&\quad x &- &3y &+ &3z & =-5\\
&-4x &- &5y &- &3z & =-5\\
&-3x &- &2y &+& 4z & = \space 6
\end{array}$
Performing theh indicated operations yield: $\quad \left[\begin{array}{ccc|c}
{1}&{-3}&{3}&{-5}\\
{0}&{-17}&{9}&{-25}\\
{0}&{-11}&{13}&{-9}\end{array}\right]$
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 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:
$\left\{\begin{array}{llll}
x & -3y & +3z & =-5\\
-4x & -5y & -3z & =-5\\
-3x & -2y & +4z & =6
\end{array}\right. \rightarrow\left[\begin{array}{rrr|r}
{1}&{-3}&{3}&{-5}\\
{-4}&{-5}&{-3}&{-5}\\
{-3}&{-2}&{4}&{6}\end{array}\right]$
We perform the row operations as:
$R_{2}=4r_{1}+r_{2}$
$R_{3}=3r_{1}+r_{3}$
$=\left[\begin{array}{ccc|c}
{1} &{-3} &{3} &{-5}\\
{4(1)-4}&{4(-3)-5}&{4(3)-3}&{4(-5)-5}\\
{3(1)-3}&{3(-3)-2}&{3(3)+4}&{3(-5)+6}\end{array}\right]$
$=\left[\begin{array}{ccc|c}
{1}&{-3}&{3}&{-5}\\
{0}&{-17}&{9}&{-25}\\
{0}&{-11}&{13}&{-9}\end{array}\right]$
