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Chapter 8: Problem 218

If the method of Gauss elimination corresponds in its final form to an echelon matrix, what is the matrix analogue of the Gauss-Jordan method for solving linear systems of equations? Explain by example.

Short Answer

Expert verified
The matrix analogue of the Gauss-Jordan method for solving linear systems of equations is the Reduced Row Echelon Form (RREF). This method transforms the augmented matrix [A|B] into RREF, making it easier to read off the solution by back-substituting the known values. The process consists of performing row operations, such as swapping rows, multiplying or dividing by constants, and adding rows to transform the matrix. The RREF is achieved when the matrix has a diagonal of 1s, 0s below the diagonal, and can help to obtain the values of each variable in the system.

Step by step solution

01

Understand the Gauss-Jordan method

The Gauss-Jordan method is an extension of the Gauss Elimination method used to obtain solutions of a linear system of equations. The goal of this method is to transform the augmented matrix [A|B] into the Reduced Row Echelon Form (RREF). Once the RREF is obtained, it becomes much easier to read off the solution by back-substituting the known values.
02

Provide an example linear system of equations

In order to illustrate the method, we will consider a linear system of equations with 3 variables (x, y, and z): x + y + z = 6 \\ 2y + 5z = -4 \\ 2x + 5y - z = 27
03

Convert the system into an augmented matrix [A|B]

We will now convert the linear system into an augmented matrix: \[ [A|B] = \begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 2 & 5 & | & -4 \\ 2 & 5 & -1 & | & 27 \end{bmatrix} \]
04

Perform row operations to reach RREF

To obtain the RREF, we perform the necessary row operations: 1. Swap R1 and R3: \\ \[ \begin{bmatrix} 2 & 5 & -1 & | & 27 \\ 0 & 2 & 5 & | & -4 \\ 1 & 1 & 1 & | & 6 \end{bmatrix} \] 2. Replace R1 by (1/2)R1: \\ \[ \begin{bmatrix} 1 & 5/2 & -1/2 & | & 27/2 \\ 0 & 2 & 5 & | & -4 \\ 1 & 1 & 1 & | & 6 \end{bmatrix} \] 3. Replace R3 by R3 - R1: \\ \[ \begin{bmatrix} 1 & 5/2 & -1/2 & | & 27/2 \\ 0 & 2 & 5 & | & -4 \\ 0 & -1/2 & 3/2 & | & -3/2 \end{bmatrix} \] 4. Replace R2 by (1/2)R2: \\ \[ \begin{bmatrix} 1 & 5/2 & -1/2 & | & 27/2 \\ 0 & 1 & 5/2 & | & -2 \\ 0 & -1/2 & 3/2 & | & -3/2 \end{bmatrix} \] 5. Replace R3 by R3 + (1/2)R2: \\ \[ \begin{bmatrix} 1 & 5/2 & -1/2 & | & 27/2 \\ 0 & 1 & 5/2 & | & -2 \\ 0 & 0 & 2 & | & -1 \end{bmatrix} \] 6. Replace R3 by (1/2)R3: \\ \[ \begin{bmatrix} 1 & 5/2 & -1/2 & | & 27/2 \\ 0 & 1 & 5/2 & | & -2 \\ 0 & 0 & 1 & | & 1/2 \end{bmatrix} \] 7. Replace R2 by R2 - (5/2)R3: \\ \[ \begin{bmatrix} 1 & 5/2 & -1/2 & | & 27/2 \\ 0 & 1 & 0 & | & -3 \\ 0 & 0 & 1 & | & 1/2 \end{bmatrix} \] 8. Replace R1 by R1 + (1/2)R3: \\ \[ \begin{bmatrix} 1 & 5/2 & 0 & | & 13 \\ 0 & 1 & 0 & | & -3 \\ 0 & 0 & 1 & | & 1/2 \end{bmatrix} \] 9. Replace R1 by R1 - (5/2)R2: \\ \[ \begin{bmatrix} 1 & 0 & 0 & | & 5 \\ 0 & 1 & 0 & | & -3 \\ 0 & 0 & 1 & | & 1/2 \end{bmatrix} \]
05

Extract the solution from the RREF

Now, we can easily read off the solution from the RREF: x = 5 \\ y = -3 \\ z = \dfrac{1}{2} So, this is how the Gauss-Jordan method can be applied to solve a linear system of equations in its matrix form.

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Most popular questions from this chapter

Find the solutions of the following systems and describe the solutions in geometric terms. (a) \(-2 \mathrm{x}+\mathrm{y}+3 \mathrm{z}=0\) \(2 x-y-3 z=0\) \(-6 x+3 y+9 z=0\) (b) \(2 x+5 y+z=0\) \(\mathrm{x}-2 \mathrm{y}+\mathrm{z} \quad=0\) \(3 x+3 y+2 z=0\) (c) \(x-y+z \quad=0\) \(2 x-y+z \quad=0\) \(\mathrm{x}+\mathrm{y}+\mathrm{z} \quad=0\)

By forming the augmented matrix and row reducing, determine the solutions of the following system \(2 x-y+3 z=4\) \(3 \mathrm{x}+2 \mathrm{z}=5\) \(-2 x+y+4 z=6\)

Show that the following system has more than one solution. $$ 3 x-y+7 z=0 $$ $$ \begin{aligned} &2 \mathrm{x}-\mathrm{y}+4 \mathrm{z}=(1 / 2) \\ &\mathrm{x}-\mathrm{y}+\mathrm{z}=1 \\ &6 \mathrm{x}-4 \mathrm{y}+10 \mathrm{z}=3 \end{aligned} $$

Reduce the following matrices to echelon form and then to row reduced echelon form. \(\begin{array}{rrrrr}\text { (a) } & 10 & 1 & 3 & -2 \mid \\ \mathrm{A}= & 12 & 1 & -4 & 3 \mid \\ & 12 & 3 & 2 & 11\end{array}\) (b) \(\begin{array}{rlr} & \mid 6 & 3 & -4 \mid \\ \mathrm{A}= & 1-4 & 1 & -61 \\\ & 1 & 1 & 2 & -5 \mid\end{array}\)

Solve the following linear system of equations: \(2 x+3 y-4 z=5\) \(-2 \mathrm{x}+\mathrm{z}=7\) \(3 x+2 y+2 z=3\)
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