Question

Solve each system of equations using matrices (row operations). If the system has no solution, say that it is inconsistent. $$ \left\\{\begin{array}{r} x+2 y+z=1 \\ 2 x-y+2 z=2 \\ 3 x+y+3 z=3 \end{array}\right. $$

Short answer

Infinite solutions, variable dependent.

Step-by-step solution

- Write the Augmented Matrix

First, write the system of equations in matrix form. The augmented matrix is: \[ \begin{bmatrix} 1 & 2 & 1 & | & 1 \ 2 & -1 & 2 & | & 2 \ 3 & 1 & 3 & | & 3 \ \end{bmatrix} \]

- Row Operations for Row Echelon Form (REF)

Start by making the element at (1,1) equal to 1 (which it already is). Then, make the rest of the first column values zero using row operations. To make the element in (2,1) zero, use: \ R_2 \leftarrow R_2 - 2R_1 \ \ \[ \begin{bmatrix} 1 & 2 & 1 & | & 1 \ 0 & -5 & 0 & | & 0 \ 3 & 1 & 3 & | & 3 \ \end{bmatrix} \] Next, make the element in (3,1) zero using: \ R_3 \leftarrow R_3 - 3R_1 \ \ \[ \begin{bmatrix} 1 & 2 & 1 & | & 1 \ 0 & -5 & 0 & | & 0 \ 0 & -5 & 0 & | & 0 \ \end{bmatrix} \]

- Further Simplify the Matrix

Next, make the element at (2,2) equal to 1 by dividing the second row by -5: \ R_2 \leftarrow \frac{R_2}{-5} \[ \begin{bmatrix} 1 & 2 & 1 & | & 1 \ 0 & 1 & 0 & | & 0 \ 0 & -5 & 0 & | & 0 \ \end{bmatrix} \] Then, make the element in (3,2) zero using: \ R_3 \leftarrow R_3 - (-5)R_2 \ \ \[ \begin{bmatrix} 1 & 2 & 1 & | & 1 \ 0 & 1 & 0 & | & 0 \ 0 & 0 & 0 & | & 0 \ \end{bmatrix} \]

- Identify the Solution or Inconsistency

The matrix has two non-zero rows and one row of zeros. Since we have a zero row, we need to check the remaining rows to identify the relationships All rows are consistent. Therefore, the system has infinite solutions depending on the free variable z Since we see for example z=0, row becomes row 1, x +2y = 1 The consistent solution is: Infinite solution of the form:= [ -2y+1, -z, Z]

Key concepts

Row Operations

Row operations are steps we can take to simplify a matrix to make it easier to solve. These operations include: swapping rows, multiplying a row by a scalar, and adding or subtracting rows from each other.

For example, in our exercise, to make the first column below the first row zero, we performed :

• Row 2 operation: \( R_2 \leftarrow R_2 - 2R_1 \)

After row operation, the augmented matrix was

\[ \begin{bmatrix} 1 & 2 & 1 & \vert & 1 \ \ 0 & -5 & 0 & \vert & 0 \ \ 3 & 1 & 3 & \vert & 3 \ \end{bmatrix} \]

Here, we subtracted twice the first row from the second row. Similarly, for the third row:

• Row 3 operation: \( R_3 \leftarrow R_3 - 3R_1 \)

giving us:

\[ \begin{bmatrix} 1 & 2 & 1 & \vert & 1 \ \ 0 & -5 & 0 & \vert & 0 \ \ 0 & -5 & 0 & \vert & 0 \ \end{bmatrix} \]

Row operations help in systematically transforming the matrix.

Augmented Matrix

An augmented matrix is a way to capture both the coefficients and constants from a system of linear equations into a single matrix.

For the given system of equations:

\(\begin{array}{r} x+2 y+z=1 \ \ 2 x-y+2 z=2 \ \ 3 x+y+3 z=3 \ \end{array}\right. \),

The augmented matrix is:

\[ \begin{bmatrix} 1 & 2 & 1 & \vert & 1 \ \ 2 & -1 & 2 & \vert & 2 \ \ 3 & 1 & 3 & \vert & 3 \ \end{bmatrix} \]

Here, the vertical line separates the coefficients of the variables from the constants. This form is useful for applying row operations to solve the system. By converting the system to this matrix representation, we can use techniques like row operations to find solutions more easily.

Infinite Solutions

A system of equations has infinite solutions when the equations describe the same plane or line in space. This means there is no unique solution, but rather a whole range of values that satisfy all equations.

In our exercise, after performing row operations, we reached an augmented matrix that indicated the presence of dependent equations. The final form was:

\[ \begin{bmatrix} 1 & 2 & 1 & \vert & 1 \ \ 0 & 1 & 0 & \vert & 0 \ \ 0 & 0 & 0 & \vert & 0 \ \end{bmatrix} \]

This shows one row of zeros, indicating a free variable and hence infinite solutions. We can express the solution set as:

• \ x = -2y + 1 \
• \ y = -z \
• \ z = z \

This means for every value of z, y and x change accordingly, resulting in infinite possible combinations of (x, y, z) that satisfy the system.