Question
A system of equations was written as an augmented matrix, which was row reduced to: \( \left[\begin{array}{cccc}1 & 0 & 0 & -3 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & -1\end{array}\right] \) What is the solution to the original system of equations? \( x= \) \( y= \) \( z= \)
Ask by Ellis Mitchell. in the United States
Jan 23,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
The solution to the original system of equations is:
\[
x = -3, \quad y = 0, \quad z = -1
\]
Solution
The given augmented matrix after row reduction is:
\[
\begin{bmatrix}
1 & 0 & 0 & -3 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & -1
\end{bmatrix}
\]
This matrix corresponds to the following system of equations:
\[
\begin{cases}
1x + 0y + 0z = -3 \\
0x + 1y + 0z = 0 \\
0x + 0y + 1z = -1
\end{cases}
\]
Simplifying each equation:
1. \( x = -3 \)
2. \( y = 0 \)
3. \( z = -1 \)
**Solution:**
\[
\boxed{
\begin{aligned}
x &= -3, \\
y &= 0, \\
z &= -1
\end{aligned}
}
\]
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Simplify this solution The Deep Dive
The row reduced augmented matrix gives us the solutions directly. From the matrix, we can see that: - The first row corresponds to \(x = -3\), - The second row corresponds to \(y = 0\), and - The third row corresponds to \(z = -1\). Therefore, the solution to the original system of equations is: \( x = -3 \) \( y = 0 \) \( z = -1 \) This means that the point \((-3, 0, -1)\) is where all three planes represented by the equations intersect.
