Question

Solve each system of equations using matrices (row operations). If the system has no solution, say that it is inconsistent. $$ \left\\{\begin{array}{r} x+y=1 \\ 2 x-y+z=1 \\ x+2 y+z=\frac{8}{3} \end{array}\right. $$

Short answer

The solution to the system is \( x = \frac{1}{3}, y = \frac{2}{3}, z = 1 \).

Step-by-step solution

- Write the system in augmented matrix form

First, represent the given system of equations in an augmented matrix form. The system is: \[ \begin{cases} x + y = 1 \ 2x - y + z = 1 \ x + 2y + z = \frac{8}{3} \end{cases} \] The augmented matrix is: \[ \begin{bmatrix} 1 & 1 & 0 & | & 1 \ 2 & -1 & 1 & | & 1 \ 1 & 2 & 1 & | & \frac{8}{3} \end{bmatrix} \]

- Perform row operations to get a leading 1 in the first row first column

The first row already has a leading 1. The augmented matrix remains: \[ \begin{bmatrix} 1 & 1 & 0 & | & 1 \ 2 & -1 & 1 & | & 1 \ 1 & 2 & 1 & | & \frac{8}{3} \end{bmatrix} \]

- Eliminate the first element of rows 2 and 3 to transform into upper triangular form

Eliminate the first entry in the second row by replacing row 2 (R2) with R2 - 2R1: \[ R_2 = R_2 - 2R_1 \] This gives: \[ R_2 = \begin{bmatrix} 2 & -1 & 1 & | & 1 \end{bmatrix} - 2 \begin{bmatrix} 1 & 1 & 0 & | & 1 \end{bmatrix} = \begin{bmatrix} 0 & -3 & 1 & | & -1 \end{bmatrix} \] Next, eliminate the first entry in the third row by replacing row 3 (R3) with R3 - R1: \[ R_3 = R_3 - R_1 \] This gives: \[ R_3 = \begin{bmatrix} 1 & 2 & 1 & | & \frac{8}{3} \end{bmatrix} - \begin{bmatrix} 1 & 1 & 0 & | & 1 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 1 & | & \frac{5}{3} \end{bmatrix} \] The updated augmented matrix is now: \[ \begin{bmatrix} 1 & 1 & 0 & | & 1 \ 0 & -3 & 1 & | & -1 \ 0 & 1 & 1 & | & \frac{5}{3} \end{bmatrix} \]

- Make the second leading coefficient 1 by modifying the second row

Replace row 2 (R2) with \( R_2 / -3 \): \[ R_2 = \frac{1}{-3}R_2 \] This gives: \[ R_2 = \begin{bmatrix} 0 & -3 & 1 & | & -1 \end{bmatrix} \times \frac{1}{-3} = \begin{bmatrix} 0 & 1 & - \frac{1}{3} & | & \frac{1}{3} \end{bmatrix} \] The augmented matrix is now: \[ \begin{bmatrix} 1 & 1 & 0 & | & 1 \ 0 & 1 & - \frac{1}{3} & | & \frac{1}{3} \ 0 & 1 & 1 & | & \frac{5}{3} \end{bmatrix} \]

- Eliminate the second element of rows 1 and 3

Eliminate the element in the first row second column by replacing R1 with \( R_1 - R_2 \): \[ R_1 = R_1 - R_2 \] This gives: \[ R_1 = \begin{bmatrix} 1 & 1 & 0 & | & 1 \end{bmatrix} - \begin{bmatrix} 0 & 1 & - \frac{1}{3} & | & \frac{1}{3} \end{bmatrix} = \begin{bmatrix} 1 & 0 & \frac{1}{3} & | & \frac{2}{3} \end{bmatrix} \] Eliminate the element in the third row second column by replacing R3 with \( R_3 - R_2 \): \[ R_3 = R_3 - R_2 \] This gives: \[ R_3 = \begin{bmatrix} 0 & 1 & 1 & | & \frac{5}{3} \end{bmatrix} - \begin{bmatrix} 0 & 1 & - \frac{1}{3} & | & \frac{1}{3} \end{bmatrix} = \begin{bmatrix} 0 & 0 & \frac{4}{3} & | & \frac{4}{3} \end{bmatrix} \] The updated augmented matrix is: \[ \begin{bmatrix} 1 & 0 & \frac{1}{3} & | & \frac{2}{3} \ 0 & 1 & - \frac{1}{3} & | & \frac{1}{3} \ 0 & 0 & \frac{4}{3} & | & \frac{4}{3} \end{bmatrix} \]

- Make the third leading coefficient 1

Replace row 3 (R3) with \( R_3 \times \frac{3}{4} \): \[ R_3 = \frac{3}{4} R_3 \] This gives: \[ R_3 = \begin{bmatrix} 0 & 0 & \frac{4}{3} & | & \frac{4}{3} \end{bmatrix} \times \frac{3}{4} = \begin{bmatrix} 0 & 0 & 1 & | & 1 \end{bmatrix} \] The augmented matrix is now: \[ \begin{bmatrix} 1 & 0 & \frac{1}{3} & | & \frac{2}{3} \ 0 & 1 & -\frac{1}{3} & | & \frac{1}{3} \ 0 & 0 & 1 & | & 1 \end{bmatrix} \]

- Back-substitute to solve for the variables

Back-substitute to find the values of the variables: From the third row: \[ z = 1 \] From the second row: \[ y - \frac{1}{3}(z) = \frac{1}{3} \] Substitute \( z = 1 \) into the equation: \[ y - \frac{1}{3}(1) = \frac{1}{3} \] Therefore, \[ y - \frac{1}{3} = \frac{1}{3} \] Adding \( \frac{1}{3} \) to both sides gives \[ y = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \] From the first row: \[ x + \frac{1}{3}(z) = \frac{2}{3} \] Substitute \( z = 1 \) into the equation: \[ x + \frac{1}{3}(1) = \frac{2}{3} \] Therefore, \[ x + \frac{1}{3} = \frac{2}{3} \] Subtracting \( \frac{1}{3} \) from both sides gives: \[ x = \frac{2}{3} - \frac{1}{3} = \frac{1}{3} \]

Key concepts

Row Operations

Solving systems of equations using matrices involves various strategies, one of which is performing row operations. Row operations help us manipulate the rows of an augmented matrix to eventually reach a form that makes it easier to deduce the solutions. There are three main types of row operations:

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

These operations are essential for transforming an augmented matrix into row echelon form (REF) or reduced row echelon form (RREF), which facilitates easier back-substitution.
For example, when we have an augmented matrix:
\[\begin{bmatrix} 1 & 1 & 0 & | & 1 \ 2 & -1 & 1 & | & 1 \ 1 & 2 & 1 & | & \frac{8}{3} \ \end{bmatrix}\]
We'll first make row 2 zero in the first column by performing the row operation: \[ R_2 \rightarrow R_2 - 2R_1 \]
This helps us eliminate the coefficient of x in the second row, simplifying our matrix.
Mastery of row operations ultimately helps in systematically solving the given system of equations.

Augmented Matrix

An augmented matrix represents a system of linear equations in a combined matrix form. It includes the coefficients of the variables and the constants from the equations, separated by a vertical line. This unified form helps in efficiently applying matrix operations to solve the system.
Consider the system of equations:
\[\begin{cases} x + y = 1 \ 2x - y + z = 1 \ x + 2y + z = \frac{8}{3} \ \end{cases}\]
We can write it in augmented matrix form as:
\[\begin{bmatrix} 1 & 1 & 0 & | & 1 \ 2 & -1 & 1 & | & 1 \ 1 & 2 & 1 & | & \frac{8}{3} \ \end{bmatrix}\]
This matrix enables us to use row operations and other matrix techniques to solve the system. It’s crucial to carefully form the augmented matrix correctly to avoid errors in the subsequent steps.
The vertical line in the augmented matrix separates the coefficients of the variables from the constants, clearly delineating the two parts of each equation.

Inconsistent Systems

An inconsistent system of equations is one that has no solution. This happens when the equations in the system contradict each other, leading to a situation where there's no possible value for the variables to satisfy all equations simultaneously.
When solving systems with matrices, an inconsistent system often becomes evident during row operations. If we encounter a row in the augmented matrix that translates to an impossible statement, such as \[ 0x + 0y + 0z = k \] where k is a non-zero constant, we can conclude the system is inconsistent.
For example, if during the row reduction process, the augmented matrix converts to the form:
\[\begin{bmatrix} 1 & 1 & 0 & | & 1 \ 0 & 1 & -\frac{1}{3} & | & \frac{1}{3} \ 0 & 0 & 0 & | & c \ \end{bmatrix}\]
And if c is a non-zero value, it indicates a contradiction, suggesting the system has no solution.
Recognizing inconsistent systems early in the solving procedure helps avoid unnecessary computations and focus on verifying the correctness of the system’s formation.