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Chapter 12: Problem 40

Solve each system of equations using matrices (row operations). If the system has no solution, say that it is inconsistent. $$ \left\\{\begin{array}{l} x+2 y=5 \\ x+y=3 \end{array}\right. $$

Short Answer

Expert verified
The solution is \( x = 1 \) and \( y = 1 \).

Step by step solution

01

Write the system as an augmented matrix

Convert the system of equations into an augmented matrix. The given system is: \[ \begin{cases} x + 2y = 5 \ x + y = 3 \end{cases} \]This can be written as the matrix: \[ \left[ \begin{array}{cc|c} 1 & 2 & 5 \ 1 & 1 & 3 \end{array} \right] \]
02

Perform row operations to get a row in form [1, 1, c]

Subtract the second row from the first row: \[ R_1 = R_1 - R_2 \] This gives: \[ \left[ \begin{array}{cc|c} 1 & 1 & 2 \ 1 & 1 & 3 \end{array} \right] \]
03

Eliminate the 1 in the second row first column

Subtract the first row from the second row to create a leading zero below the leading 1: \[ R_2 = R_2 - R_1 \] This changes the matrix to: \[ \left[ \begin{array}{cc|c} 1 & 1 & 2 \ 0 & 1 & 1 \end{array} \right] \]
04

Back-substitute to find the solution

Since the obtained matrix is in row echelon form, interpret the rows to solve for the variables. The matrix corresponds to the equations: \[ \begin{cases} x + y = 2 \ y = 1 \end{cases} \]From the second equation, we have:\[ y = 1 \]Substitute \( y = 1 \) into the first equation:\[ x + 1 = 2 \]Solving for \( x \):\[ x = 1 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Row Operations
Row operations are fundamental manipulations of the rows of a matrix. They help us solve systems of linear equations. There are three main types of row operations: swapping two rows, multiplying a row by a nonzero scalar, and adding or subtracting a multiple of one row to another row. These operations help in transforming the given matrix into forms that are easier to work with, such as row echelon form.
Augmented Matrix
An augmented matrix combines the coefficients of the variables and the constants from a system of linear equations into one matrix.
For a system of equations:
(1) \[ x + 2y = 5 \] (2) \[ x + y = 3 \],
the augmented matrix is written as:
\[ \begin{pmatrix} 1 & 2 & | & 5 \1 & 1 & | & 3 \end{pmatrix} \]
The vertical line separates the coefficients of the variables (on the left) from the constants (on the right). This matrix format is useful for applying row operations to solve the system.
Row Echelon Form
Row echelon form (REF) is a type of matrix form that makes solving systems of linear equations easier. A matrix is in row echelon form when:
  • All nonzero rows are above any rows of all zeros.
  • The leading entry of each nonzero row is 1.
  • The leading 1 in each nonzero row is to the right of the leading 1 in the row above it.
In our example, we performed row operations to convert the matrix:
\[ \begin{pmatrix} 1 & 2 & | & 5 \ 1 & 1 & | & 3 \end{pmatrix} \] to: \[ \begin{pmatrix} 1 & 1 & | & 2 \ 0 & 1 & | & 1 \end{pmatrix} \], which is in row echelon form. This format helps us easily identify the solution using back-substitution.
Back-Substitution
Back-substitution is the process of solving for variables in a system of equations starting from the last row and moving upwards. Once a matrix is in row echelon form, back-substitution allows us to find the values of the variables sequentially.
Using the matrix:
\[ \begin{pmatrix} 1 & 1 & | & 2 \ 0 & 1 & | & 1 \end{pmatrix} \]
we interpret the rows to get the equations:
  • \( x + y = 2 \)
  • \( y = 1 \)
We first solve \( y = 1 \). Then, we substitute \( y = 1 \) into the equation \( x + y = 2 \) to get \( x + 1 = 2 \). Solving for \( x \): we find \( x = 1 \). This gives us the solution \( x = 1 \) and \( y = 1 \).

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Solve each system of equations. If the system has no solution, state that it is inconsistent. $$ \left\\{\begin{array}{l} 3 x-2 y=0 \\ 5 x+10 y=4 \end{array}\right. $$
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