Solve the system of linear equations, using the Gauss-Jordan elimination method. (If there is no solution, enter NO SOLUTION. If there are infinitely many solutions involving one parameter, enter the solution using t for the last variable. If the solution involves two parameters, add s for the second to last variable.)

4x−8y−24z=−20
x−2y−6z=−5
3x−6y−18z=15

By inspection, we see that the first equation is a linear combination (4 times) the second equation. Similarly, the third equation can be obtained by multiplying equation 2 three times.

Therefore the equations represent 3 coincident lines:
x-2y-6z=-5
Let z = t, and y = s, then we can solve for x in terms of s and t:
x = -5 +2s + 6t
y = s
z = t

Solve the system of linear equations, using the Gauss-Jordan elimination method. (If there is no solution, enter NO SOLUTION. If there are infinitely many solutions, express your answer in terms of the parameters t and/or s.)

x + 2y + z = −2
−2x − 3y − z = 2
4x + 8y + 4z = −8
(x, y, z) =

Let's start by writing the augmented matrix for the system of linear equations:

[ 4 -8 -24 | -20 ]
[ 1 -2 -6 | -5 ]
[ 3 -6 -18 | 15 ]

To solve the system of equations using the Gauss-Jordan elimination method, we need to perform row operations to transform the augmented matrix into reduced row-echelon form.

Step 1: Divide the first row by 4 to make the leading coefficient of the first row equal to 1.
[ 1 -2 -6 | -5 ]
[ 1/4 -1 -3 | -5/4 ]
[ 3 -6 -18 | 15 ]

Step 2: Multiply the first row by -1/4 and add it to the second row, then multiply the first row by -3 and add it to the third row to eliminate the first column entries below the pivot element.
[ 1 -2 -6 | -5 ]
[ 0 1 0 | 1 ]
[ 0 0 0 | 30 ]

Step 3: To make the leading coefficient of the second row equal to 1, divide the second row by 1.
[ 1 -2 -6 | -5 ]
[ 0 1 0 | 1 ]
[ 0 0 0 | 30 ]

Step 4: Multiply the second row by 2 and add it to the first row to eliminate the second column entry above the pivot element.
[ 1 0 -6 | -3 ]
[ 0 1 0 | 1 ]
[ 0 0 0 | 30 ]

Step 5: Multiply the second row by 6 and add it to the first row to eliminate the second column entry below the pivot element.
[ 1 0 0 | -3 ]
[ 0 1 0 | 1 ]
[ 0 0 0 | 30 ]

From the reduced row-echelon form, we can see that the system of equations is inconsistent and has no solution because the last row of the augmented matrix becomes [0 0 0 | 30]. Therefore, the answer is NO SOLUTION.

To solve the system of linear equations using the Gauss-Jordan elimination method, we need to perform a series of row operations to obtain the solution.

Step 1: Write the augmented matrix
The first step is to write the augmented matrix of the system of equations. The augmented matrix is formed by taking the coefficients of the variables on the left-hand side of the equations and the constants on the right-hand side. In this case, the augmented matrix is:

[ 4 -8 -24 | -20 ]
[ 1 -2 -6 | -5 ]
[ 3 -6 -18 | 15 ]

Step 2: Perform row operations to get zeros in the first column below the first row
We want to create zeros in the first column below the first row. To do this, we can multiply the first row by a scalar and subtract it from the other rows.

-4R1 + R2 → R2
-4R1 + R3 → R3

The augmented matrix becomes:

[ 4 -8 -24 | -20 ]
[ 0 0 6 | 15 ]
[ 0 0 6 | 35 ]

Step 3: Perform row operations to get a leading 1 in the second row
We want to create a leading 1 in the second row. To do this, we can multiply the second row by a scalar.

(1/6)R2 → R2

The augmented matrix becomes:

[ 4 -8 -24 | -20 ]
[ 0 0 1 | 5 ]
[ 0 0 6 | 35 ]

Step 4: Perform row operations to get zeros in the second column above the second row
We want to create zeros in the second column above the second row. To do this, we can multiply the second row by a scalar and subtract it from the first and third rows.

-6R2 + R1 → R1
-6R2 + R3 → R3

The augmented matrix becomes:

[ 4 -8 0 | -50 ]
[ 0 0 1 | 5 ]
[ 0 0 0 | 5 ]

Step 5: Perform row operations to get a leading 1 in the third row
We want to create a leading 1 in the third row. To do this, we can multiply the third row by a scalar.

(1/5)R3 → R3

The augmented matrix becomes:

[ 4 -8 0 | -50 ]
[ 0 0 1 | 5 ]
[ 0 0 0 | 1 ]

Step 6: Perform row operations to get zeros in the third column
We want to create zeros in the third column. Since there is already a zero in the third column, we can skip this step.

Step 7: Back substitution and solution
Now that we have reduced the augmented matrix to its row echelon form, we can back substitute to find the values of the variables.

From the last row, we have:
0z = 1

Since this equation is not possible, it means that there is no solution to the system of linear equations. Hence, the answer is NO SOLUTION.