To write the given system of equations as an augmented matrix, each row of the augmented matrix will correspond to one equation, with the last column representing the constants on the right side of the equations. Let's break down the given system of equations:
[tex]\[
\left\{\begin{array}{r}
n + 3z - 7b = 250 \\
n + 4z - 12b = 100 \\
b = 150
\end{array}\right.
\][/tex]
1. For the first equation [tex]\( n + 3z - 7b = 250 \)[/tex]:
- The coefficient of [tex]\( n \)[/tex] is [tex]\( 1 \)[/tex],
- The coefficient of [tex]\( z \)[/tex] is [tex]\( 3 \)[/tex],
- The coefficient of [tex]\( b \)[/tex] is [tex]\( -7 \)[/tex],
- The constant on the right side is [tex]\( 250 \)[/tex].
2. For the second equation [tex]\( n + 4z - 12b = 100 \)[/tex]:
- The coefficient of [tex]\( n \)[/tex] is [tex]\( 1 \)[/tex],
- The coefficient of [tex]\( z \)[/tex] is [tex]\( 4 \)[/tex],
- The coefficient of [tex]\( b \)[/tex] is [tex]\( -12 \)[/tex],
- The constant on the right side is [tex]\( 100 \)[/tex].
3. For the third equation [tex]\( b = 150 \)[/tex]:
- There is no [tex]\( n \)[/tex] or [tex]\( z \)[/tex], so their coefficients are [tex]\( 0 \)[/tex],
- The coefficient of [tex]\( b \)[/tex] is [tex]\( 1 \)[/tex],
- The constant on the right side is [tex]\( 150 \)[/tex].
Now, we write these coefficients and constants into the augmented matrix format:
[tex]\[
\begin{bmatrix}
1 & 3 & -7 & \vrule & 250 \\
1 & 4 & -12 & \vrule & 100 \\
0 & 0 & 1 & \vrule & 150
\end{bmatrix}
\][/tex]
After removing the vertical line, the augmented matrix is:
[tex]\[
\begin{bmatrix}
1 & 3 & -7 & 250 \\
1 & 4 & -12 & 100 \\
0 & 0 & 1 & 150
\end{bmatrix}
\][/tex]
Thus, the system of equations:
[tex]\[
\left\{\begin{array}{r}
n + 3z - 7b = 250 \\
n + 4z - 12b = 100 \\
b = 150
\end{array}\right.
\][/tex]
can be written as the augmented matrix:
[tex]\[
\begin{bmatrix}
1 & 3 & -7 & 250 \\
1 & 4 & -12 & 100 \\
0 & 0 & 1 & 150
\end{bmatrix}
\][/tex]
