Using Gauss-Jordan elimination to solve the following system (if possible)

Question

(i) Using Gauss-Jordan elimination to solve the following system (if possible)

_{x1 + x2 + x3 = 7}

_{x1 - x2 + 2x3 = 7}

_{5x1 + x2 + x3 = 11}

Then, find the ranks of the matrix of coefficients and the augmented matrix and use Theorem 1 to verify that the number of solutions obtained in part(i) is correct.

Answer 1

There should only be one correct answer for this problem. I solved manually and on my calculator and I got the same answer as you did, (1,2,4) Here are my manual steps:

1 1 1 7

1 -1 2 7

5 1 1 11

-1 R1+R2

1 1 1 7

0 -2 1 0

5 1 1 11

-5 R1+R3

1 1 1 7

0 -2 1 0

0 -4 -4 -24

-1/2R3

1 1 1 7

0 -2 1 0

0 2 2 12

R2+R3

1 1 1 7

0 -2 1 0

0 0 3 12

Now to solve

By the 3rd equation

3x₃ = 12 Divide both sides by 3

x₃=4

By the 2nd equation

-2x₂+x₃ = 0

-2x₂+4 = 0 Subtract 4 from both sides

-2x₂ = -4 Divide both sides by -2

x₂ = 2

By the 1st equation

x₁+x₂+x₃ = 7

x₁+2+4 = 7 Combine like terms

x₁+6 = 7 Subtract 6 from both sides

x₁ = 1

(1,2,4)

Answer 2

The way to solve the equation using ti-84 calculator:

1. 2nd then Martix then click edit
2. use the coefficient in front of the variables given in the equations and it supposed to 3 x 4 because it is three columns and four rows: 3 x4 [ 1 1 1 7, 1 1 2 7, 5 1 1 11]
3. then press 2nd then Matrix, afterwards click on Math, scroll to B rref(
4. Click 2nd and Martix
5. Click on (A)

Your answer should be: rref( [A] [1 0 0 1]

[ 0 1 0 6]

[ 0 0 1 0 ]

The first x = 1 The second x=6 The third x= 0

Substitute the answers in the equation:

_{x1 + x2 + x3 = 7} 1+6+0 = 7

_{x1 - x2 + 2x3 = 7} 1+ 6+ 2(0) = 7

_{5x1 + x2 + x3 = 11} 5(1)+ 6+0= 11