Answer
$\left[\begin{array}{ccc|c}
{1}&{-3}&{-4}&{-6}\\
{6}&{-5}&{6}&{-6}\\
{0}&{-2}&{0}&{0}\end{array}\right]$
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:
$\left\{\begin{array}{llll}
x & -3y & -4z & =-6\\
26x & -5y & +6z & =-6\\
-x & +y & +4z & =6
\end{array}\right.\rightarrow\left[\begin{array}{ccc|c}
{1}&{-3}&{-4}&{-6}\\
{6}&{-5}&{6}&{-6}\\
{-1}&{1}&{4}&{6}\end{array}\right]$
We perform the row operation as:
$R_{2}=-6r_{1}+r_{2}$
$R_{3}=r_{1}+r_{3}$
$=\left[\begin{array}{ccc|c}
{1} &{-3} &{-4} &{-6}\\
{-6(1)+6}&{-6(-3)-5}&{(-6)(-4)+6}&{-6(-6)-6}\\
{1-1} &{(-3)+1} &{(-4)+4} &{0}\end{array}\right]$ $=\left[\begin{array}{ccc|c}
{1}&{-3}&{-4}&{-6}\\
{6}&{-5}&{6}&{-6}\\
{0}&{-2}&{0}&{0}\end{array}\right]$
