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Gauss-Jordan Method

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Introduction

The Gauss-Jordan Method, an extension of Gaussian elimination, is a powerful technique for solving systems of linear equations and finding matrix inverses. Named after Carl Friedrich Gauss and Wilhelm Jordan, this method transforms a matrix into reduced row echelon form, providing a direct solution to the system.

Key Concepts

Augmented Matrix

Similar to Gaussian elimination, we start with an augmented matrix. For a system of equations:

a11x1+a12x2+⋯+a1nxn=b1a21x1+a22x2+⋯+a2nxn=b2⋮am1x1+am2x2+⋯+amnxn=bm \begin{align} a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n &= b_1 \\ a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n &= b_2 \\ \vdots \\ a_{m1}x_1 + a_{m2}x_2 + \cdots + a_{mn}x_n &= b_m \end{align}

The augmented matrix is:

[a11a12⋯a1nb1a21a22⋯a2nb2⋮⋮⋱⋮⋮am1am2⋯amnbm] \left[ \begin{array}{cccc|c} a_{11} & a_{12} & \cdots & a_{1n} & b_1 \\ a_{21} & a_{22} & \cdots & a_{2n} & b_2 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} & b_m \end{array} \right]

Reduced Row Echelon Form (RREF)

A matrix is in reduced row echelon form if:

  1. The first non-zero element in each row (leading coefficient) is 1.
  2. Each leading 1 is the only non-zero entry in its column.
  3. Each leading 1 is to the right of the leading 1's in the rows above it.

Steps of Gauss-Jordan Method

  1. Form the augmented matrix from the system of equations.
  2. Convert the augmented matrix to reduced row echelon form:
    • Start with the leftmost non-zero column.
    • Find a non-zero entry in this column (pivot).
    • Use row operations to make the pivot 1 and all other entries in its column 0.
    • Repeat for all columns.
  3. Read the solution directly from the rightmost column of the resulting matrix.

Detailed Example

Let's solve this system of equations using the Gauss-Jordan method:

2x+y−z=8−3x−y+2z=−11−2x+y+2z=−3 \begin{align} 2x + y - z &= 8 \\ -3x - y + 2z &= -11 \\ -2x + y + 2z &= -3 \end{align}

Step 1: Form the Augmented Matrix

[21−18−3−12−11−212−3] \left[ \begin{array}{ccc|c} 2 & 1 & -1 & 8 \\ -3 & -1 & 2 & -11 \\ -2 & 1 & 2 & -3 \end{array} \right]

Step 2: Convert to Reduced Row Echelon Form

2a. Make the first element 1:

R1=12R1R_1 = \frac{1}{2}R_1
[10.5−0.54−3−12−11−212−3] \left[ \begin{array}{ccc|c} 1 & 0.5 & -0.5 & 4 \\ -3 & -1 & 2 & -11 \\ -2 & 1 & 2 & -3 \end{array} \right]

2b. Make all other elements in the first column 0:

R2=R2+3R1R3=R3+2R1 \begin{align} R_2 &= R_2 + 3R_1 \\ R_3 &= R_3 + 2R_1 \end{align}
[10.5−0.5400.50.510215] \left[ \begin{array}{ccc|c} 1 & 0.5 & -0.5 & 4 \\ 0 & 0.5 & 0.5 & 1 \\ 0 & 2 & 1 & 5 \end{array} \right]

2c. Make the second pivot 1:

R2=2R2R_2 = 2R_2
[10.5−0.5401120215] \left[ \begin{array}{ccc|c} 1 & 0.5 & -0.5 & 4 \\ 0 & 1 & 1 & 2 \\ 0 & 2 & 1 & 5 \end{array} \right]

2d. Make all other elements in the second column 0:

R1=R1−0.5R2R3=R3−2R2 \begin{align} R_1 &= R_1 - 0.5R_2 \\ R_3 &= R_3 - 2R_2 \end{align}
[10−13011200−11] \left[ \begin{array}{ccc|c} 1 & 0 & -1 & 3 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & -1 & 1 \end{array} \right]

2e. Make the third pivot 1:

R3=−R3R_3 = -R_3
[10−130112001−1] \left[ \begin{array}{ccc|c} 1 & 0 & -1 & 3 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 1 & -1 \end{array} \right]

2f. Make all other elements in the third column 0:

R1=R1+R3R2=R2−R3 \begin{align} R_1 &= R_1 + R_3 \\ R_2 &= R_2 - R_3 \end{align}
[10020103001−1] \left[ \begin{array}{ccc|c} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & -1 \end{array} \right]

Step 3: Read the Solution

The solution can be read directly from the rightmost column:

x=2,quady=3,quadz=−1x = 2, quad y = 3, quad z = -1

Verification

Let's verify our solution by substituting these values back into the original equations:

2(2)+3−(−1)=4+3+1=8quad✓−3(2)−3+2(−1)=−6−3−2=−11quad✓−2(2)+3+2(−1)=−4+3−2=−3quad✓ \begin{align} 2(2) + 3 - (-1) &= 4 + 3 + 1 = 8 quad \checkmark \\ -3(2) - 3 + 2(-1) &= -6 - 3 - 2 = -11 quad \checkmark \\ -2(2) + 3 + 2(-1) &= -4 + 3 - 2 = -3 quad \checkmark \end{align}

All equations are satisfied, confirming our solution is correct.

Gauss-Jordan Method for Matrix Inversion

The Gauss-Jordan method can also be used to find the inverse of a matrix. To do this, we augment the original matrix with the identity matrix and perform row operations until the left side becomes the identity matrix. The right side will then be the inverse.

For a matrix A:

[A∣I]→row operations[I∣A−1] [A|I] \xrightarrow{\text{row operations}} [I|A^{-1}]

Conclusion

The Gauss-Jordan Method is a powerful technique for solving systems of linear equations and finding matrix inverses. By transforming matrices to reduced row echelon form, it provides direct solutions without the need for back-substitution. This method is particularly useful in computer implementations due to its straightforward algorithm and ability to solve multiple systems with the same coefficient matrix efficiently.

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