Question

Show that the following equations have solutions only if \(\eta=1\) or 2 , and find them in these cases: $$ \begin{aligned} x+y+z &=1 \\ x+2 y+4 z &=\eta \\ x+4 y+10 z &=\eta^{2} \end{aligned} $$

Short answer

The solutions exist only if \(\eta = 1\) or \(\eta = 2\). For \(\eta = 1\), \(x = 1 + 2z, y = -3z, z = z\); for \(\eta = 2\), \(x = 2z, y = 1 - 3z, z = z\).

Step-by-step solution

Write down the system of equations

The given system of equations is:\[ \begin{aligned} x + y + z &= 1 \ x + 2y + 4z &= \eta \ x + 4y + 10z &= \eta^2 \end{aligned} \]

Create an augmented matrix

Form the augmented matrix corresponding to the system of equations:\[ \left[ \begin{array}{ccc|c} 1 & 1 & 1 & 1 \ 1 & 2 & 4 & \eta \ 1 & 4 & 10 & \eta^2 \end{array} \right] \]

Perform row operations to Row Echelon Form

Subtract the first row from the second and third rows to simplify:\[ \left[ \begin{array}{ccc|c} 1 & 1 & 1 & 1 \ 0 & 1 & 3 & \eta-1 \ 0 & 3 & 9 & \eta^2 - 1 \end{array} \right] \]

Continue row operations

Subtract 3 times the second row from the third row to simplify further:\[ \left[ \begin{array}{ccc|c} 1 & 1 & 1 & 1 \ 0 & 1 & 3 & \eta-1 \ 0 & 0 & 0 & \eta^2 - 1 - 3(\eta - 1) \end{array} \right] \]This simplifies to:\[ \left[ \begin{array}{ccc|c} 1 & 1 & 1 & 1 \ 0 & 1 & 3 & \eta-1 \ 0 & 0 & 0 & \eta^2 - 3\eta + 2 \end{array} \right] \]

Find the condition for a solution

For the system to have a solution, the last entry (\(\eta^2 - 3\eta + 2\)) must be 0:\[ \eta^2 - 3\eta + 2 = 0 \]Solve the quadratic equation:\[ (\eta - 1)(\eta - 2) = 0 \]Thus, \(\eta = 1\) or \(\eta = 2\).

Solve the system for \(\eta = 1\)

Substitute \(\eta = 1\) into the original system:\[ \begin{aligned} x + y + z &= 1 \ x + 2y + 4z &= 1 \ x + 4y + 10z &= 1 \end{aligned} \]By subtracting the first equation from the second and third as before:\[ \left[ \begin{array}{ccc|c} 0 & 1 & 3 & 0 \ 0 & 3 & 9 & 0 \end{array} \right] \]This simplifies to:\[ \left[ \begin{array}{ccc|c} 1 & 1 & 1 & 1 \ 0 & 1 & 3 & 0 \end{array} \right] \]From the second row:\[ y + 3z = 0 \rightarrow y = -3z \]Substitute into the first equation:\[ x + (-3z) + z = 1 \rightarrow x - 2z = 1 \rightarrow x = 1 + 2z \]Thus:\[ x = 1 + 2z, y = -3z, z = z \]

Solve the system for \(\eta = 2\)

Substitute \(\eta = 2\) into the original system:\[ \begin{aligned} x + y + z &= 1 \ x + 2y + 4z &= 2 \ x + 4y + 10z &= 4 \end{aligned} \]By subtracting the first equation from the second and third as before:\[ \left[ \begin{array}{ccc|c} 0 & 1 & 3 & 1 \ 0 & 3 & 9 & 3 \end{array} \right] \]This simplifies to:\[ \left[ \begin{array}{ccc|c} 1 & 1 & 1 & 1 \ 0 & 1 & 3 & 1 \end{array} \right] \]From the second row:\[ y + 3z = 1 \rightarrow y = 1 - 3z \]Substitute into the first equation:\[ x + (1 - 3z) + z = 1 \rightarrow x - 2z = 0 \rightarrow x = 2z \]Therefore:\[ x = 2z, y = 1 - 3z, z = z \]

Key concepts

augmented matrix

In the context of solving linear systems, an augmented matrix helps to represent the system of equations in a compact form. Imagine you have a system of equations like these:
\[ \begin{aligned} x + y + z &= 1 \ x + 2y + 4z &= \theta \ x + 4y + 10z &= \theta^2 \ \right] \]
To create the augmented matrix, you write down the coefficients of the variables and the constants (on the right side of the equals sign) in a matrix format:
\[ \begin{aligned} \begin{array}{ccc|c} 1 & 1 & 1 & 1 \ 1 & 2 & 4 & \theta \ 1 & 4 & 10 & \theta^2 \ \right] \ \right] \]
The vertical bar divides the coefficients from the constants. This matrix form allows for streamlined operations to find solutions.

row operations

Row operations are essential techniques for simplifying matrices, especially when moving towards Row Echelon Form (REF). They include swapping rows, multiplying a row by a non-zero scalar, and adding or subtracting the multiple of one row from another.
For our system of equations, we started with the augmented matrix:
\[ \begin{aligned} \begin{array}{ccc|c} 1 & 1 & 1 & 1 \ 1 & 2 & 4 & \theta \ 1 & 4 & 10 & \theta^2 \ \right] \ \right] \ \]
The goal is to simplify this into a form where solving for the variables is easy.
First, subtract the first row from the second and third rows:
\[ \begin{aligned} \begin{array}{ccc|c} 1 & 1 & 1 & 1 \ 0 & 1 & 3 & \theta-1 \ 0 & 3 & 9 & \theta^2 - 1 \ \right] \ \right] \ \]
Next, subtract three times the second row from the third row:
\[ \begin{aligned} \begin{array}{ccc|c} 1 & 1 & 1 & 1 \ 0 & 1 & 3 & \theta-1 \ 0 & 0 & 0 & \theta^2 - 3\theta + 2 \ \right] \ \right] \ \]
These steps simplify the matrix and reveal conditions for solutions and dependencies among variables.

quadratic equation

A quadratic equation is a polynomial equation of the form:
\[ ax^2 + bx + c = 0 \ \]
Where \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable. To find the solutions to a quadratic equation, you can either factorize it, complete the square, or use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \ \]
In our problem, we ended up with a quadratic equation in terms of \(\theta\):
\[ \theta^2 - 3\theta + 2 = 0 \ \]
We can factorize it:
\[ (\theta - 1)(\theta - 2) = 0 \ \]
Thus, the solutions are \(\theta = 1\) and \(\theta = 2\). These values are crucial because they determine when the given system of equations has solutions.
Understanding quadratic equations provides foundational skills not just for solving algebraic problems but also for topics in calculus, physics, and engineering.