Question

Solve the following sets of simultaneous equations by reducing the matrix to row echelon form. $$ \begin{array}{r} \mid x-2 y+3 z=0 \\ x+4 y-6 z=0 \\ 2 x+2 y-3 z=0 \end{array} $$

Short answer

The solutions are \((x, y, z) = t(0, \frac{3}{2}, 1)\) for any real number \(t\).

Step-by-step solution

Write the augmented matrix

Begin by writing the given system of equations in augmented matrix form. \[ \begin{bmatrix} 1 & -2 & 3 & | & 0 \ 1 & 4 & -6 & | & 0 \ 2 & 2 & -3 & | & 0 \end{bmatrix} \]

Transform to upper triangular form - Step 1

Subtract the first row from the second row to create a zero in the first entry of the second row. \[ R2 = R2 - R1 \]\[ \begin{bmatrix} 1 & -2 & 3 & | & 0 \ 0 & 6 & -9 & | & 0 \ 2 & 2 & -3 & | & 0 \end{bmatrix} \]

Transform to upper triangular form - Step 2

Subtract 2 times the first row from the third row to create a zero in the first entry of the third row. \[ R3 = R3 - 2 \cdot R1 \]\[ \begin{bmatrix} 1 & -2 & 3 & | & 0 \ 0 & 6 & -9 & | & 0 \ 0 & 6 & -9 & | & 0 \end{bmatrix} \]

Transform to upper triangular form - Step 3

Subtract the second row from the third row to create a zero in the second entry of the third row. \[ R3 = R3 - R2 \]\[ \begin{bmatrix} 1 & -2 & 3 & | & 0 \ 0 & 6 & -9 & | & 0 \ 0 & 0 & 0 & | & 0 \end{bmatrix} \]

Convert to row echelon form

Divide the second row by 6 to normalize the second leading coefficient. \[ R2 = \frac{1}{6}R2 \]\[ \begin{bmatrix} 1 & -2 & 3 & | & 0 \ 0 & 1 & -\frac{3}{2} & | & 0 \ 0 & 0 & 0 & | & 0 \end{bmatrix} \]

Back substitution

Since the system is homogeneous and we have a row of zeros, we can deduce that there are free variables. Let \(z = t\), a parameter.For the second row, \(y - \frac{3}{2}t = 0\Rightarrow y = \frac{3}{2}t \).For the first row, solve for \(x\) using the already found \(y\) and \(z\). \(x - 2(\frac{3}{2}t) + 3t = 0 \Rightarrow x = 0\).So the general solution is: \[(x, y, z) = (0, \frac{3}{2}t, t) = t(0, \frac{3}{2}, 1) \]

Key concepts

augmented matrix

To solve simultaneous equations using matrices, we first need to represent the system of equations in an augmented matrix form. An augmented matrix combines the coefficients of the variables and the constants from the right-hand side of the equations into one matrix.

In this case, the system of equations is:
\(x - 2y + 3z = 0\)
\(x + 4y - 6z = 0\)
\(2x + 2y - 3z = 0\)

The corresponding augmented matrix is:
\[\begin{bmatrix} 1 & -2 & 3 & | & 0 \ 1 & 4 & -6 & | & 0 \ 2 & 2 & -3 & | & 0 \end{bmatrix} \]
where each row corresponds to one equation, and the vertical bar separates the coefficients from the constants.

simultaneous equations

Simultaneous equations are a set of equations with multiple variables that need to be solved together. The solutions are the values of the variables that satisfy all equations simultaneously.

In this example, we have three equations with three variables \(x, y,\text{ and } z\). Our goal is to find values for these variables that satisfy all three equations at the same time. By converting these equations to matrix form, we simplify the process of finding such solutions using systematic mathematical operations.

upper triangular form

The next step in solving the system of simultaneous equations is to convert the augmented matrix into upper triangular form. This form has all zeros below the main diagonal, simplifying the system to make it easier to solve.

To transform the given matrix:
1. Subtract the first row from the second row:
\(R2 = R2 - R1\)
\[\begin{bmatrix} 1 & -2 & 3 & | & 0 \ 0 & 6 & -9 & | & 0 \ 2 & 2 & -3 & | & 0 \end{bmatrix}\]
2. Subtract 2 times the first row from the third row:
\(R3 = R3 - 2 \times R1\)
\[\begin{bmatrix} 1 & -2 & 3 & | & 0 \ 0 & 6 & -9 & | & 0 \ 0 & 6 & -9 & | & 0 \end{bmatrix}\]
3. Subtract the second row from the third row:
\(R3 = R3 - R2\)
\[\begin{bmatrix} 1 & -2 & 3 & | & 0 \ 0 & 6 & -9 & | & 0 \ 0 & 0 & 0 & | & 0 \end{bmatrix}\]
Finally, divide the second row by 6 to normalize the leading coefficient:
\(R2 = \frac{1}{6}R2\)
\[\begin{bmatrix} 1 & -2 & 3 & | & 0 \ 0 & 1 & -\frac{3}{2} & | & 0 \ 0 & 0 & 0 & | & 0 \end{bmatrix}\]

back substitution

With the upper triangular form achieved, we can now perform back substitution to solve for the variables.

Considering the transformed matrix:
\[\begin{bmatrix} 1 & -2 & 3 & | & 0 \ 0 & 1 & -\frac{3}{2} & | & 0 \ 0 & 0 & 0 & | & 0 \end{bmatrix}\]
We observe a row of zeros, indicating free variables and a homogeneous system.

Set \(z = t\) (a parameter).
From the second row equation: \(y - \frac{3}{2}t = 0\)
\(y = \frac{3}{2}t\)
From the first row equation: \(x - 2(\frac{3}{2}t) + 3t = 0\)
\(x = 0\)
Thus, the general solution is:
\((x, y, z) = (0, \frac{3}{2}t, t) = t(0, \frac{3}{2}, 1)\)
This implies that any value of \(t\) will satisfy the original system, indicating an infinite number of solutions.