Question

Solve the sets of homogeneous equations by row reducing the matrix. $$\left\\{\begin{aligned}2 x+2 z &=0 \\\4 x+2 y+5 z &=0 \\\x-y+2 z &=0\end{aligned}\right.$$

Short answer

The solution is x = 0, y = 0, and z = 0.

Step-by-step solution

Write the augmented matrix

First, write the coefficient matrix for the system of equations and append a column of zeros to form the augmented matrix:\[\begin{pmatrix} 2 & 0 & 2 & | & 0 \ 4 & 2 & 5 & | & 0 \ 1 & -1 & 2 & | & 0 \end{pmatrix}\]

Row operations to obtain a row echelon form

Perform row operations to transform the matrix into row echelon form. Start by making the first entry in the first row a leading 1. Divide row 1 by 2:\[\begin{pmatrix} 1 & 0 & 1 & | & 0 \ 4 & 2 & 5 & | & 0 \ 1 & -1 & 2 & | & 0 \end{pmatrix}\]becomes\[\begin{pmatrix} 1 & 0 & 1 & | & 0 \ 2 & 1 & 2.5 & | & 0 \ 1 & -1 & 2 & | & 0 \end{pmatrix}\]

Eliminate the x-term from rows below

Subtract 2 times the first row from the second row and 1 times the first row from the third row:\[\begin{pmatrix} 1 & 0 & 1 & | & 0 \ 0 & 1 & 0.5 & | & 0 \ 0 & -1 & 1 & | & 0 \end{pmatrix}\]

Eliminate the y-term from the last row

Add the second row to the third row:\[\begin{pmatrix} 1 & 0 & 1 & | & 0 \ 0 & 1 & 0.5 & | & 0 \ 0 & 0 & 1.5 & | & 0 \end{pmatrix}\]

Back-substitution

Now, solve for the variables starting from the bottom row. Since the last non-zero row reads:\[ z = 0 \]Substitute \( z = 0 \) into the second row:\[ y + 0.5z = 0 \]which gives:\[ y = 0 \].Finally, substitute \( y = 0 \) and \( z = 0 \) into the first row:\[ x + z = 0 \]which gives:\[ x = 0 \].

Key concepts

homogeneous equations

Homogeneous equations are systems of linear equations where each equation equals zero. They often look something like this:

• $$2x + 2z = 0$$
• $$4x + 2y + 5z = 0$$
• $$x - y + 2z = 0$$

These types of systems always have the trivial solution where all variables are zero, like $$(x, y, z) = (0, 0, 0)$$. But they can also have infinitely many solutions depending on the relationships between the equations. When we solve these systems using matrices, one of the most effective methods is through row reduction.

row echelon form

To solve systems of equations using matrices, we aim to transform the matrix into row echelon form (REF). This form simplifies the system and makes it easier to solve. A matrix is in REF when:

• All non-zero rows are above rows of all zeros.
• Each leading entry (the first non-zero number from the left, in a row) of a row is to the right of the leading entry of the row above it.
• All entries below each leading entry are zeros.

For example, the given matrix was reduced step by step. After performing appropriate row operations, we achieved the row echelon form:

back-substitution

Back-substitution is the step where we solve for the variables starting from the bottom row and working upwards once the matrix is in row echelon form. In the example provided, our reduced matrix becomes:

• $$ \begin{pmatrix} 1 & 0 & 1 & | & 0 \ 0 & 1 & 0.5 & | & 0 \ 0 & 0 & 1.5 & | & 0 \end{pmatrix} $$

We start at the bottom where the equation reads: $$1.5z = 0$$, so, $$z = 0$$. Next, substitute back into the previous row: $$ y + 0.5 \times 0 = 0$$, so $$y = 0$$. Finally, substitute $$ y = 0$$ and $$ z = 0$$ into the first row: $$ x + 0 \times 1 = 0$$, so $$x = 0$$. This gives us the solution $$(x, y, z) = (0, 0, 0)$$.

augmented matrix

An augmented matrix is one that includes both the coefficients of the variables and the constants from the equations all in one matrix form. For the given system of equations, the augmented matrix looks like:

• $$\begin{pmatrix} 2 & 0 & 2 & | & 0 \ 4 & 2 & 5 & | & 0 \ 1 & -1 & 2 & | & 0 \end{pmatrix}$$

Here, each row represents one equation, and each column represents a variable plus an additional column for the constants, separated by a vertical line. This setup helps to easily perform row operations and transform the system into row echelon form for solving.