Question

Solve the following sets of equations by reducing the matrix to row echelon form. \(\mid 2 x+5 y+z=2$$x+y+2 z=1$$x+5 z=3\)

Short answer

x = -2, y = 1, z = 1

Step-by-step solution

Set Up the Augmented Matrix

First, set up the augmented matrix from the given system of equations. The system is:\[ \begin{cases} 2x + 5y + z = 2 \ x + y + 2z = 1 \ x + 5z = 3 \end{cases} \]The augmented matrix is:\[ \begin{pmatrix} 2 & 5 & 1 & | & 2 \ 1 & 1 & 2 & | & 1 \ 1 & 0 & 5 & | & 3 \end{pmatrix} \]

Make the first element of the first column a 1

Swap the first and second rows to bring the leading 1 to the first row:\[ R_1 \rightleftharpoons R_2 \]New matrix:\[ \begin{pmatrix} 1 & 1 & 2 & | & 1 \ 2 & 5 & 1 & | & 2 \ 1 & 0 & 5 & | & 3 \end{pmatrix} \]

Make the first column below the leading 1 a 0

Use row operations to make the elements below the leading 1 in the first row zeros.Subtract 2 times the first row from the second row:\[ R_2 = R_2 - 2R_1 \]Subtract the first row from the third row:\[ R_3 = R_3 - R_1 \]New matrix:\[ \begin{pmatrix} 1 & 1 & 2 & | & 1 \ 0 & 3 & -3 & | & 0 \ 0 & -1 & 3 & | & 2 \end{pmatrix} \]

Make the second element of the second row a 1

Divide the second row by 3 to make the leading coefficient 1:\[ R_2 = \frac{R_2}{3} \]New matrix:\[ \begin{pmatrix} 1 & 1 & 2 & | & 1 \ 0 & 1 & -1 & | & 0 \ 0 & -1 & 3 & | & 2 \end{pmatrix} \]

Make the second column below the leading 1 a 0

Add the second row to the third row:\[ R_3 = R_3 + R_2 \]New matrix:\[ \begin{pmatrix} 1 & 1 & 2 & | & 1 \ 0 & 1 & -1 & | & 0 \ 0 & 0 & 2 & | & 2 \end{pmatrix} \]

Make the third element of the third row a 1

Divide the third row by 2:\[ R_3 = \frac{R_3}{2} \]New matrix:\[ \begin{pmatrix} 1 & 1 & 2 & | & 1 \ 0 & 1 & -1 & | & 0 \ 0 & 0 & 1 & | & 1 \end{pmatrix} \]

Make the elements above the leading 1 in the third column 0

Subtract 2 times the third row from the first row:\[ R_1 = R_1 - 2R_3 \]Add the third row to the second row:\[ R_2 = R_2 + R_3 \]New matrix:\[ \begin{pmatrix} 1 & 1 & 0 & | & -1 \ 0 & 1 & 0 & | & 1 \ 0 & 0 & 1 & | & 1 \end{pmatrix} \]

Make the second element of the first row 0

Subtract the second row from the first row:\[ R_1 = R_1 - R_2 \]Final row echelon form matrix:\[ \begin{pmatrix} 1 & 0 & 0 & | & -2 \ 0 & 1 & 0 & | & 1 \ 0 & 0 & 1 & | & 1 \end{pmatrix} \]

Interpret the Results

The system of equations given by the matrix is:\[ \begin{cases} x = -2 \ y = 1 \ z = 1 \end{cases} \]

Key concepts

Augmented Matrix

To solve a system of linear equations using the Gauss-Jordan elimination method, we start by forming what is known as an augmented matrix. An augmented matrix includes coefficients of all the variables and the constants from each equation in the system. For example, the system has equations: \(2x + 5y + z = 2\) \(x + y + 2z = 1\) \(x + 5z = 3\).We transform these equations into matrix form, appending the constants as an additional column: \[ \begin{pmatrix} 2 & 5 & 1 & | & 2 \ 1 & 1 & 2 & | & 1 \ 1 & 0 & 5 & | & 3 \end{pmatrix} \]This structure aids in using row operations for simplifying the system step-by-step.

Row Operations

Row operations are special manipulations that we perform on the rows of the augmented matrix to simplify it into a form from which the solutions can easily be derived. There are three main types of row operations:

• Row swap: swapping two rows
• Row scaling: multiplying a row by a non-zero scalar
• Row addition: adding or subtracting the multiple of one row from another row

These operations are essential because they preserve the equivalence of the system while simplifying it. For instance, converting a leading coefficient to 1 or making the elements below a pivot zeroes are typical uses of row operations.

Row Echelon Form

The goal of Gauss-Jordan elimination is to transform the matrix into a row echelon form or even further into a reduced row echelon form. In row echelon form, we have:

• All zero rows, if any, are at the bottom
• The leading entry (pivot) of each non-zero row is 1
• The leading 1 of each row is to the right of the leading 1 of the row above it

For our example, after a series of row operations, we reach the row echelon form: \[ \begin{pmatrix} 1 & 1 & 2 & | & 1 \ 0 & 1 & -1 & | & 0 \ 0 & 0 & 1 & | & 1 \end{pmatrix} \].This form allows us to interpret the results directly.

System of Linear Equations

A system of linear equations consists of multiple equations with the same set of variables. The solution to this system is the set of values that satisfies all equations simultaneously. In representing such a system, we make use of the augmented matrix to simplify and solve via Gauss-Jordan elimination. The goal is to reach a matrix from which we can backtrack to find the values of the variables. For instance, the final form of the example system becomes \[ \begin{cases} x = -2 \ y = 1 \ z = 1 \end{cases} \].These values are obtained by interpreting the simplified matrix, allowing us to solve the original set of equations efficiently.