Question

Solve the following sets of simultaneous equations by reducing the matrix to row echelon form. $$ \left\\{\begin{array}{l} 2 x+3 z=0 \\ 4 x+2 y+5 z=0 \\ x-y+2 z=0 \end{array}\right. $$

Short answer

The solutions are x = -1.5t, y = 0.5t, z = t.

Step-by-step solution

- Write the augmented matrix

First, convert the given system of equations into an augmented matrix. The system of equations can be written as: \[ \begin{cases} 2x + 3z = 0 \ 4x + 2y + 5z = 0 \ x - y + 2z = 0 \end{cases} \]. The corresponding augmented matrix is: \[ \begin{pmatrix} 2 & 0 & 3 & | & 0 \ 4 & 2 & 5 & | & 0 \ 1 & -1 & 2 & | & 0 \end{pmatrix} \].

- Apply row operations to form zeros below the first pivot

We start with the augmented matrix from Step 1: \[ \begin{pmatrix} 2 & 0 & 3 & | & 0 \ 4 & 2 & 5 & | & 0 \ 1 & -1 & 2 & | & 0 \end{pmatrix} \]. For the first pivot, divide Row 1 (R1) by 2: \[ R1 \rightarrow \frac{1}{2}R1 = \begin{pmatrix} 1 & 0 & 1.5 & | & 0 \end{pmatrix} \].Next, make zeros below this pivot by performing row operations: \[ R2 \rightarrow R2 - 4R1 \] and \[ R3 \rightarrow R3 - R1 \].This gives:\[ \begin{pmatrix} 1 & 0 & 1.5 & | & 0 \ 0 & 2 & -1 & | & 0 \ 0 & -1 & 0.5 & | & 0 \end{pmatrix} \].

- Form zeros below the second pivot

Identify the second pivot which is 2 in the matrix. First, simplify Row 2 (R2) by dividing it by 2: \[ R2 \rightarrow \frac{1}{2}R2 = \begin{pmatrix} 0 & 1 & -0.5 & | & 0 \end{pmatrix} \].Now, eliminate the element below this pivot by performing the row operation on Row 3 (R3): \[ R3 \rightarrow R3 + R2 \].This gives the matrix:\[ \begin{pmatrix} 1 & 0 & 1.5 & | & 0 \ 0 & 1 & -0.5 & | & 0 \ 0 & 0 & 0 & | & 0 \end{pmatrix} \].

- Interpret the row echelon form

The matrix is now in row echelon form: \[ \begin{pmatrix} 1 & 0 & 1.5 & | & 0 \ 0 & 1 & -0.5 & | & 0 \ 0 & 0 & 0 & | & 0 \end{pmatrix} \]. The third row indicates there is no pivot in the third column, thus one of the variables is free (in this case, z). Assume z = t, where t is a parameter.

- Solve for the variables

From Step 4, express the other variables in terms of t. Substitute z = t into the second row equation: \[ y - \frac{1}{2}t = 0 \rightarrow y = \frac{1}{2}t \].Substitute z = t into the first row equation: \[ x + 1.5t = 0 \rightarrow x = -1.5t \].Thus, the solutions are: \[ x = -1.5t, \ y = 0.5t, \ z = t \].

Key concepts

Augmented Matrix

The first step in solving a system of simultaneous equations using matrices is converting the set of equations into an augmented matrix. An augmented matrix combines the coefficients of the variables with the constants from the equations. This helps in organizing and systematically applying row operations to solve the equations.

For example, consider the given system of equations:
$$ \begin{cases} 2x + 3z = 0 \ 4x + 2y + 5z = 0 \ x - y + 2z = 0 \text. \text \text \text \text \textend{cases} $$
This system translates to the following augmented matrix:
\text $$ \begin{pmatrix} 2 & 0 & 3 & | & 0 \ 4 & 2 & 5 & | & 0 \ 1 & -1 & 2 & | & 0 \textpmatrix \text. \text $$

• The entries before the vertical bar represent the coefficients of the variables \text (x,\text y,\text z \text). \text
• The entries after the vertical bar represent the constants on the right-hand side of the equations.

This matrix provides a structured form for applying row operations and finding the solution.

Row Operations

Row operations are the backbone of transforming the matrix to a simpler form, known as row echelon form. They include:

• Swapping rows.
• Multiplying a row by a non-zero scalar.
• Adding or subtracting a multiple of one row to another row.

For instance, starting with our augmented matrix:
$$ \begin{pmatrix} 2 & 0 & 3 & | & 0 \ 4 & 2 & 5 & | & 0 \ 1 & -1 & 2 & | & 0 \textpmatrix \textend align of point $$
We perform the following operations:

• Divide \text R1 \text by 2: \text R1 \rightarrow \frac{1}{2}R1. \text
• Subtract 4 times \text R1 \text from \text R2: \text R2 \rightarrow R2 - 4R1. \text
• Subtract \text R1 \text from \text R3: \text R3 \rightarrow R3 - R1. \text

Resulting in:
$$ \begin {pmatrix}{1 & 0 & 1.5 & | & 0 \ 0 & 2 & -1 & | & 0 \ 0 & -1 & 0.5 & | & 0 }\textpmatrix \text. \text $$ These steps simplify the matrix and clear the way for finding pivot elements and free variables.

Pivot Element

A pivot element is a non-zero element in the matrix that helps to simplify it further. It's the first non-zero element in a row, and its position determines where row operations should focus.

In our ongoing example, the pivot elements are identified as follows:

• The first pivot element is 1 in the first row after dividing by 2.
• The second pivot element is 2 in the second row, which we simplify by dividing the entire row by 2.

After those steps, the matrix looks like:
$$ \begin{pmatrix} 1 & 0 & 1.5 & | & 0 \ 0 & 1 & -0.5 & | & 0 \ 0 & 0 & 0 & | & 0 \text pmatrix $$ \text.

• Note how subsequent rows should have zeros in the columns corresponding to the previous pivot elements.
• The zeros streamline solving the system and identifying free variables.

This progressive zeroing out below pivots moves the matrix closer to row echelon form.

Free Variables

In your simplified matrix form, free variables are the ones not corresponding to a pivot column. They offer additional degrees of freedom in the solution and can take on any value.

Using our example, the matrix in row echelon form is:
$$ \begin{pmatrix} 1 & 0 & 1.5 & | & 0 \ 0 & 1 & -0.5 & | & 0 \ 0 & 0 & 0 & | & 0 \text*pmatrix .$$
The absence of a pivot in the third column (for variable \text z) signifies \text z is a free variable.

• We can choose \text z = t, where t is any parameter.
• The values of pivot variables (\text x and \text y) then depend on this free parameter.

This flexibility allows expressing the entire set of solutions in terms of the free variable.

Simultaneous Equations

Simultaneous equations are sets of equations that you solve together. They share common variables, and the solution set satisfies all equations.

The final step is solving our simultaneous equations by back-substitution into the row echelon matrix:
$$ \begin{pmatrix} 1 & 0 & 1.5 & | & 0 \ 0 & 1 & -0.5 & | & 0 \ 0 & 0 & 0 & | & 0 *pmatrix \text.$$

• We assumed \text z = t.
• From the second row, \text y - 0.5t = 0, so \text y = 0.5t.
• From the first row, \text x + 1.5t = 0, so \text x = -1.5t.

Hence, the solution is:
$$ x = -1.5t, y = 0.5t, z = t.$$ When dealing with real-world problems, always express solutions in clear forms and check if your free variable choices meet all equations in the system.