Use the Gauss-Jordan method to solve each system of...

Use the Gauss-Jordan method to solve each system of equations. For systems in two variables with infinitely many solutions, give the solution with y arbitrary; for systems in three variables with infinitely many solutions, give the solution with z arbitrary. See Examples $1-4$ $$\begin{array}{l} 2 x+y-z+3 w=0 \\ 3 x-2 y+z-4 w=-24 \\ x+y-z+w=2 \\ x-y+2 z-5 w=-16 \end{array}$$

Use the Gauss-Jordan method to solve each system of equations. For systems in two variables with infinitely many solutions, give the solution with y arbitrary; for systems in three variables with infinitely many solutions, give the solution with z arbitrary. See Examples $1-4$
$$\begin{array}{l}
2 x+y-z+3 w=0 \\
3 x-2 y+z-4 w=-24 \\
x+y-z+w=2 \\
x-y+2 z-5 w=-16
\end{array}$$

Key Concepts

Gauss-Jordan Elimination

Gauss-Jordan elimination is an algorithmic method to solve systems of linear equations by transforming the augmented matrix of the system into reduced row-echelon form through a series of elementary row operations. This method systematically eliminates variables, making the solution or the dependency among variables directly accessible.

Augmented Matrix

An augmented matrix is a compact representation of a system of linear equations where the coefficients of the variables and the constants are combined into one matrix. This format is essential for applying matrix operations like row reduction, which are used in methods such as Gauss-Jordan elimination.

Elementary Row Operations

Elementary row operations—including row swapping, multiplying a row by a nonzero constant, and adding a multiple of one row to another—are the tools used to manipulate the augmented matrix without changing the solution set. These operations facilitate the conversion of the matrix into a form where the solutions are evident.

Pivot Variables and Free Variables

In the process of Gauss-Jordan elimination, pivot variables are those corresponding to the leading ones in each row of the reduced row-echelon matrix, while any variables that lack such a pivot are termed free variables. Free variables can take arbitrary values, leading to infinite solution sets when they exist.

Infinite Solution Sets

An infinite solution set arises when a system of linear equations has free variables, meaning that not all variables are fixed by the equations. The general solution is then expressed in terms of the free variables, which can be assigned arbitrary values, resulting in a family of solutions.

Transcript

00:01 For this question, i've taken all the coefficients of x, y, z, and w, and put them into this matrix here.

00:09 And i've taken the right hand side and put it into this column vector here and then condensed the form.

00:18 So using the gauss -jordan method, i'm going to eliminate this matrix on left -hand side and convert it into a 4x4 identity matrix.

00:31 So we're able to find the values of each of our variables so to make it a little easier i'm going to swap row two and row three and swap row one and row four so that'll give me one negative one two negative five one and negative one negative one one negative 3, negative 2, 1, negative 4, 2, 1, negative 1, 3.

01:16 Swapping the right hand side 2, we get negative 16, 2, negative 24, and 0.

01:31 So i'm going to take the first pivot and eliminate the other three positions.

01:40 So that'll give me 1, negative 1, 2, negative 5, 0, 2, negative 3, 6, 0, 1, negative 5, 11, and 0, negative 3, negative 5, and 13.

02:08 And on the right hand side we get negative 16, oops, 18, 24, and 32.

02:25 So we have a 1 here, so i'm going to swap row 2 and 3 again to make it, to make elimination a little bit easier for your fractions.

02:36 So then we would have 1, 2, negative 1, 2, negative 5, 0, 1, 1 ,000, 0, 1 ,000, 5, 11, 0, 2, negative 3, 6, 0, 3, negative 5, 13.

03:00 And here we have negative 16, 24, 18, and 32.

03:09 So taking 1 as to pivot, we'll eliminate row 3 and row 4, and by doing so, we'll get 1, 2, negative 1, 2, negative 5, 0 -1 -5 -11 -0 -0 -07 negative 16 and 0 -0 -0 -10 sorry positive 10 negative 20 and here we would have negative 16 24 negative 30 and negative 40 so in the last low last row everything is a factor or it's a multiple of 10 so we can reduce it, and then we'll have 1 -1, 2, negative 5, negative 16, 0, 1, negative 5, 11, 24, 0 -07, negative 0 ,0, 0 ,0, 0, 0, 2, negative 4...

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