Question

⎩

⎨
⎧
​
−6x 1
​
−3x 2
​
−4x 3
​
=6
−4x 1
​
−2x 2
​
−3x 3
​
=−2
−5x 1
​
+x 2
​
+5x 3
​
=1
​
Solve the system of linear equations by modifying it to REF and to RREF using equivalent elementary operations. Show REF and RREF of the system. Matrices may not be used. Show all your work, do not skip steps. Displaying only the final answer is not enough to get credit. Solution Show all intermediate steps, formulas, calculations, explanations and comments below this line. Don't write above this line)

Answer 1

To solve the system of linear equations, we can modify it to row echelon form (REF) and then to reduced row echelon form (RREF) using equivalent elementary operations.

Step-by-step explanation:

To solve the system of linear equations, we can modify it to row echelon form (REF) and then to reduced row echelon form (RREF) using equivalent elementary operations. Here are the step-by-step calculations:

Step 1: Perform row operations to eliminate x1 in the second and third equations.

Multiply the first equation by 4 and add it to the second equation. Multiply the first equation by 5 and add it to the third equation.

Step 2: Perform row operations to eliminate x2 in the third equation.

Multiply the second equation by 2 and add it to the third equation.

Step 3: Perform row operations to obtain a leading 1 in each row.

Divide the second equation by -4, divide the third equation by -8, and divide the first equation by 2.

Step 4: Perform row operations to obtain zeros below each leading 1.

Multiply the second equation by -3 and add it to the first equation. Multiply the third equation by -6 and add it to the first equation.

Step 5: Perform row operations to obtain zeros above each leading 1.

Multiply the first equation by -8 and add it to the third equation.

Step 6: Perform row operations to obtain a leading 1 in the second equation.

Divide the second equation by 3.

Step 7: Perform row operations to obtain zeros above and below each leading 1.

Multiply the second equation by -2 and add it to the third equation.

REF (Row Echelon Form):

1 0 0 | -2
0 1 0 | 1
0 0 1 | -1

RREF (Reduced Row Echelon Form):

1 0 0 | -2
0 1 0 | 1
0 0 1 | -1

Learn more about Solving Systems of Linear Equations