Answer
The system of equations is: $\quad \quad$$\left\{\begin{array}{llll}
x & -3y & +4z & =3\\
3x & -5y & +6z & =6\\
-5x & +3y & +4z & =6
\end{array}\right.$
Performing the idicated operatiions yield:
$\left[\begin{array}{ccc|c}
{1}&{-3}&{4}&{3}\\
{0}&{4}&{-6}&{-3}\\
{0}&{-12}&{24}&{21}\end{array}\right]$
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.
The system of equations can be expressed as follows:
$\left\{\begin{array}{llll}
x & -3y & +4z & =3\\
3x & -5y & +6z & =6\\
-5x & +3y & +4z & =6
\end{array}\right.$
We can write the system as an augmented matrix $[A|B]$ as follows:
$\left[\begin{array}{ccc|c}
{1}&{-3}&{4}&{3}\\
{3}&{-5}&{6}&{6}\\
{-5}&{3}&{4}&{6}\end{array}\right] $
We perform the row operation as:
$\begin{array}{l}{R_{2}=-3 r_{1}+r_{2}} \\ {R_{3}=5 r_{1}+r_{3}}\end{array}$
$\left[\begin{array}{ccc|c}
{1}&{-3}&{4}&{3}\\
{3}&{-5}&{6}&{6}\\
{-5}&{3}&{4}&{6}\end{array}\right] =\left[\begin{array}{ccc|c}
{1} &{-3} &{4} &{3}\\
{-3(1)+3}&{-3(-3)-5}&{-3(4)+6}&{-3(3)+6}\\
{5(1)-5}&{5(-3)+3}&{5(4)+4}&{5(3)+6}\end{array}\right] \\=\left[\begin{array}{ccc|c}
{1}&{-3}&{4}&{3}\\
{0}&{4}&{-6}&{-3}\\
{0}&{-12}&{24}&{21}\end{array}\right]$
