Question

After solving the system

2x₁ + 4x₂ - 6x₃ = - 8

x₁ + 3x₂ + x₃ = 10 and

2x₁ - 4x₂ - 2x₃ = - 12

Using the Gauss-Jordan method, the values of x₁, x₂ and x₃ are

1. 1, 2, 3
2. 1, 3, 2
3. 3, 2, 1
4. 3, 1, 2

Answer 1

Correct Answer - Option 1 : 1, 2, 3

Concept:

Gauss-Jordan Method:

• This method is used to simplify an augmented matrix to its reduced-row echelon form.
• The following are the steps that must be followed while using the Gauss-Jordan Elimination method -

1. Write the system as an augmented matrix.
2. Look at the first entry in the first row. Make this entry into a 1 and all other entries in that column 0s. This is called pivoting the matrix about this element.
3. Once this is done, move down the diagonal to the second entry of the second row and pivot about this entry.
4. Continue until the whole matrix is in row-reduced form.
5. Solve for the values of variables.

Calculation:

Given,

2x₁ + 4x₂ - 6x₃ = - 8         ---(1)

x₁ + 3x₂ + x₃ = 10            ---(2)

2x₁ - 4x₂ - 2x₃ = - 12        ---(3)

This system of equations can be written in the form of matrices as follows -

\(\left[ {\begin{array}{*{20}{c}} 2&4&{ - 6}\\ 1&3&1\\ 2&{ - 4}&{ - 2} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 8}\\ {10}\\ { - 12} \end{array}} \right]\)

⇒ [A][X]=[D]        ---(4)

Here,

\([A] = \left[ {\begin{array}{*{20}{c}} 2&4&{ - 6}\\ 1&3&1\\ 2&{ - 4}&{ - 2} \end{array}} \right]\)

\([D] = \left[ {\begin{array}{*{20}{c}} { - 8}\\ {10}\\ { - 12} \end{array}} \right]\)

\([X] = \left[ {\begin{array}{*{20}{c}} {x_1}\\ {x_2}\\ {x_3} \end{array}} \right]\)

Step 1: Writing Augment Matrix

\(\therefore[A~\vdots ~D] = \left[ {\begin{array}{*{20}{c}} 2&4&{ - 6}\\ 1&3&1\\ 2&{ - 4}&{ - 2} \end{array}}{\begin{array}{*{20}{c}} { \vdots}\\ {\vdots}\\ { \vdots} \end{array}}{\begin{array}{*{20}{c}} { - 8}\\ {10}\\ { - 12} \end{array}} \right]\)

Step 2: Applying elementary transformations or pivoting of the augment matrix

a) R₁ → \(\frac{1}{2}\)R₁

\( = \left[ {\begin{array}{*{20}{c}} 1&2&{ -3}\\ 1&3&1\\ 2&{ - 4}&{ - 2} \end{array}}{\begin{array}{*{20}{c}} { \vdots}\\ {\vdots}\\ { \vdots} \end{array}}{\begin{array}{*{20}{c}} { - 4}\\ {10}\\ { - 12} \end{array}} \right]\)

b) R₂ → R₂ - R₁; R₃ → R₃ - 2R₁

\( = \left[ {\begin{array}{*{20}{c}} 1&2&{ -3}\\ 0&1&4\\ 0&{ - 8}&{ 4} \end{array}}{\begin{array}{*{20}{c}} { \vdots}\\ {\vdots}\\ { \vdots} \end{array}}{\begin{array}{*{20}{c}} { - 4}\\ {14}\\ { - 4} \end{array}} \right]\)

c) R₃ → R₃ + 8R₂

\( = \left[ {\begin{array}{*{20}{c}} 1&2&{ -3}\\ 0&1&4\\ 0&{ 0}&{ 36} \end{array}}{\begin{array}{*{20}{c}} { \vdots}\\ {\vdots}\\ { \vdots} \end{array}}{\begin{array}{*{20}{c}} { - 4}\\ {14}\\ { 108} \end{array}} \right]\)

Step 3: Solving for variables -

Now, the reduced equations can be written as -

x₁ + 2x₂ - 3x₃ = - 4      ---(5)

x₂ + 4x₃ = 14              ---(6)

36x₃ = 108                 ---(7)

Solving (5), (6), and (7)

x₃ = 3

x₂ = 2

x₁ = 1