Solve the following system of equations (swap rows if needed to avoid zero pivots): 0x + y + z = 4 x + y + z = 6 2x - y + z = 3
Understand the Problem
The question asks us to solve a system of linear equations with three variables (x, y, z), using row swaps if needed to perform partial pivoting and avoid zero pivots during the solution process. This involves using techniques from linear algebra, particularly Gaussian elimination with partial pivoting, to find the values of x, y, and z that satisfy all three equations simultaneously.
Answer
$x = 2$, $y = \frac{5}{2}$, $z = \frac{3}{2}$
Answer for screen readers
$x = 2, y = \frac{5}{2}, z = \frac{3}{2}$
Steps to Solve
- Initial System
The given system of equations is: $\begin{cases} 0x + y + z = 4 \ x + y + z = 6 \ 2x - y + z = 3 \end{cases}$
- Swap Rows to Avoid Zero Pivot
Since the coefficient of $x$ in the first equation is 0, we swap the first and second rows to avoid a zero pivot. This gives us: $\begin{cases} x + y + z = 6 \ 0x + y + z = 4 \ 2x - y + z = 3 \end{cases}$
- Eliminate $x$ from the Third Equation
Subtract 2 times the first equation from the third equation: $R_3 \rightarrow R_3 - 2R_1$ $\begin{cases} x + y + z = 6 \ y + z = 4 \ 2x - y + z - 2(x + y + z) = 3 - 2(6) \end{cases}$
Simplifying the third equation: $2x - y + z - 2x - 2y - 2z = 3 - 12$ $-3y - z = -9$
The new system is: $\begin{cases} x + y + z = 6 \ y + z = 4 \ -3y - z = -9 \end{cases}$
- Eliminate $y$ from the Third Equation
Multiply the second equation by 3 and add it to the third equation: $R_3 \rightarrow R_3 + 3R_2$ $\begin{cases} x + y + z = 6 \ y + z = 4 \ -3y - z + 3(y + z) = -9 + 3(4) \end{cases}$
Simplifying the third equation: $-3y - z + 3y + 3z = -9 + 12$ $2z = 3$
The new system is: $\begin{cases} x + y + z = 6 \ y + z = 4 \ 2z = 3 \end{cases}$
- Solve for $z$
From the third equation, $2z = 3$, so $z = \frac{3}{2}$
- Solve for $y$
Substitute $z = \frac{3}{2}$ into the second equation: $y + \frac{3}{2} = 4$ $y = 4 - \frac{3}{2} = \frac{8}{2} - \frac{3}{2} = \frac{5}{2}$
- Solve for $x$
Substitute $y = \frac{5}{2}$ and $z = \frac{3}{2}$ into the first equation: $x + \frac{5}{2} + \frac{3}{2} = 6$ $x + \frac{8}{2} = 6$ $x + 4 = 6$ $x = 2$
$x = 2, y = \frac{5}{2}, z = \frac{3}{2}$
More Information
The solution to the system of equations is $x=2$, $y=2.5$, and $z=1.5$. This means that the point $(2, 2.5, 1.5)$ is the intersection of the three planes defined by the original equations.
Tips
A common mistake is not swapping rows when a zero pivot is encountered, which can lead to incorrect results. Another common mistake is making arithmetic errors during the elimination steps. Always double-check calculations, especially when dealing with fractions.
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