Question

Solve each system of equations using matrices (row operations). If the system has no solution, say that it is inconsistent. $$ \left\\{\begin{array}{l} x+y=8 \\ x-y=4 \end{array}\right. $$

Short answer

The solution is \( x = 6 \) and \( y = 2 \).

Step-by-step solution

Write the system as an augmented matrix

First, rewrite the system of equations as an augmented matrix. The given system is: \[\begin{cases} x + y = 8, \ x - y = 4 \end{cases}\]Convert this to: \[\left( \begin{array}{cc|c} 1 & 1 & 8 \ 1 & -1 & 4 \end{array} \right)\]

Perform row operations to obtain row echelon form

Next, perform row operations to simplify the augmented matrix to row echelon form. We can start by subtracting the second row from the first row: \[\left( \begin{array}{cc|c} 1 & 1 & 8 \ 1 & -1 & 4 \end{array} \right) \rightarrow \left( \begin{array}{cc|c} 1 & 1 & 8 \ 0 & -2 & -4 \end{array} \right)\]

Simplify the second row

Now, divide the second row by -2 to simplify it: \[\left( \begin{array}{cc|c} 1 & 1 & 8 \ 0 & -2 & -4 \end{array} \right) \rightarrow \left( \begin{array}{cc|c} 1 & 1 & 8 \ 0 & 1 & 2 \end{array} \right)\]

Use the second row to eliminate y in the first row

Use the second row to eliminate the y term in the first row by subtracting the second row from the first row: \[\left( \begin{array}{cc|c} 1 & 1 & 8 \ 0 & 1 & 2 \end{array} \right) \rightarrow \left( \begin{array}{cc|c} 1 & 0 & 6 \ 0 & 1 & 2 \end{array} \right)\]

Interpret the row-reduced matrix

Finally, interpret the resulting row-reduced matrix. The augmented matrix is now: \[\left( \begin{array}{cc|c} 1 & 0 & 6 \ 0 & 1 & 2 \end{array} \right)\]This corresponds to the system of equations: \[\begin{cases} x = 6, \ y = 2 \end{cases}\]

Key concepts

Augmented Matrix

When you have a system of equations, like the one given in the exercise: \[\begin{cases} x + y = 8 \ x - y = 4 \ \end{cases} \] You can represent it as an augmented matrix. An augmented matrix combines the coefficients of the variables and the constants from the equations into a single matrix. Here's how you form it: Write down only the coefficients of the variables in their respective order and place the constants in a separated column. For our example, it looks like this: \(\begin{array}{cc|c} 1 & 1 & 8 \ 1 & -1 & 4 \ \end{array} \) Now, we proceed with solving the system using this augmented matrix as our starting point.

Row Operations

Row operations are moves you can perform on the rows of your matrix to simplify it. The goal is to get the matrix in a simpler form, often aiming for the row echelon form. Here are the allowable row operations:

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

In our example, we subtract the second row from the first row: \(\begin{array}{cc|c} 1 & 1 & 8 \ 1 & -1 & 4 \ \end{array} \) becomes \(\begin{array}{cc|c} 1 & 1 & 8 \ 0 & -2 & -4 \ \end{array} \) These operations don't change the solution of the system but prepare it for the next steps.

Row Echelon Form

Row echelon form is a type of matrix where each leading coefficient (the first non-zero number from the left, in each row) is 1, and the leading coefficient of each row is to the right of the leading coefficient of the row directly above it. Additionally, rows with all zeros should be at the bottom of the matrix. To get to this form, continue with row operations. In our example, we do the following: 1. Simplify the second row by dividing it by -2: \(\begin{array}{cc|c} 1 & 1 & 8 \ 0 & -2 & -4 \ \end{array} \) becomes \(\begin{array}{cc|c} 1 & 1 & 8 \ 0 & 1 & 2 \ \end{array} \) 2. Use the second row to eliminate the y term in the first row: \(\begin{array}{cc|c} 1 & 1 & 8 \ 0 & 1 & 2 \ \end{array} \) becomes \(\begin{array}{cc|c} 1 & 0 & 6 \ 0 & 1 & 2 \ \end{array} \). Now, the matrix is in row echelon form, which is much easier to interpret.

Inconsistent System

An inconsistent system of equations is one that has no solution. This happens when the equations describe parallel lines that never intersect. In terms of augmented matrices, this situation will usually reveal itself when you end up with a row where all the variables are zero but the constant is non-zero, like so: \(\begin{array}{cc|c} 0 & 0 & 1 \ \end{array} \) This translates to an impossible statement like 0 = 1, showing that there's no common solution to satisfy all equations in the system. In our worked example, however, the system is consistent, as seen from the final row-reduced matrix: \(\begin{array}{cc|c} 1 & 0 & 6 \ 0 & 1 & 2 \ \end{array} \) It leads us to the solution x = 6 and y = 2, which are consistent values.