Solve the system of equations using matrices. Use Gaussian elimination with back-substitution.

x + y + z = -5
x - y + 3z = -1
4x + y + z = -2

I really need help..
1 1 1 -5
1 -1 3 -1
4 1 1 1 -2

I've done so many different methods to solve this and I get stuck..

agree with

1 1 1 -5
1 -1 3 -1
4 1 1 -2 -----> you had an extra 1 in there

1 1 1 -5
0 2 -2 -4 #1 - #2
0 3 3 -18

1 1 1 -5
0 1 -1 -2 #2 divided by 2
0 1 1 -6 #3 divided by 3

1 1 1 -5
0 1 -1 -2
0 0 -2 4 #2 - #3

1 1 1 -5
0 1 -1 -2
0 0 1 -2

from #3:
z = -2
in #2:
y - z = -2
y + 2 = -2
y = -4

in #1
x + y + z = -5
x - 4 - 2 = -5
x = 1

x = 1, y = -4, z = -2

thank you!

No worries! I'll walk you through the steps to solve this system of equations using Gaussian elimination with back-substitution.

Step 1: Set up the augmented matrix
To solve this system using matrices, we first need to set up the augmented matrix. The augmented matrix is a matrix that includes the coefficients of the variables and the constants on the right-hand side of each equation.

The augmented matrix for the given system of equations is:

1 1 1 -5
1 -1 3 -1
4 1 1 -2

Step 2: Perform row operations to transform the matrix into row-echelon form
Now, let's perform row operations on the augmented matrix to transform it into row-echelon form. The goal is to have a matrix where each leading coefficient is 1 and all entries below and above the leading coefficients are zeros.

Start with the first row and eliminate the coefficients below it by adding or subtracting multiple of the first row from the other rows. The goal is to make the entries below the leading coefficients in the first column equal to zero.

Add -1 times the first row to the second row, and subtract 4 times the first row from the third row. The resulting matrix should be:

1 1 1 -5
0 -2 2 4
0 -3 -3 18

Next, deal with the leading coefficient in the second row. Divide the second row by -2 to make the leading coefficient 1:

1 1 1 -5
0 1 -1 -2
0 -3 -3 18

Finally, eliminate the coefficients below the leading coefficient in the second row. Add 3 times the second row to the third row:

1 1 1 -5
0 1 -1 -2
0 0 0 12

Step 3: Convert the matrix to reduced row-echelon form
The matrix is now in row-echelon form. To convert it to reduced row-echelon form, we'll perform additional row operations.

Divide the third row by 12 to make the leading coefficient 1:

1 1 1 -5
0 1 -1 -2
0 0 0 1

Step 4: Perform back-substitution to find the values of the variables
Now that the matrix is in reduced row-echelon form, we can perform back-substitution to find the values of the variables.

From the third equation, we can see that 0z = 1, which is not possible. This means that the system of equations is inconsistent and has no solution.