Put this matrix in Reduced Row Echelon Form. Make sure that you list the Elementary Row Operations that you use and show every step in order to solve this.

[ 1 2 1
-2 -3 1
3 5 0]

To put the given matrix in Reduced Row Echelon Form (RREF), we need to perform a series of Elementary Row Operations. Here are the steps to solve this:

Step 1: Start with the given matrix:

[ 1 2 1
-2 -3 1
3 5 0]

Step 2: We want to create a leading 1 in the first row, first column. To achieve this, we divide the first row by 1:

R1 -> R1 / 1

This step is optional since the first element is already 1, but it can be useful to keep consistency in the operations.

[ 1 2 1
-2 -3 1
3 5 0]

Step 3: Use row operations to create zeros below the leading 1 in the first column:

R2 -> R2 + 2 * R1 (Multiply the first row by -2 and add it to the second row, preserving the first row)

[ 1 2 1
0 1 3
3 5 0]

R3 -> R3 - 3 * R1 (Multiply the first row by -3 and subtract it from the third row, preserving the first row)

[ 1 2 1
0 1 3
0 -1 -3]

Step 4: Use row operations to create zeros above the leading 1 in the second column:

R1 -> R1 - 2 * R2 (Multiply the second row by -2 and subtract it from the first row, preserving the second row)

[ 1 0 -5
0 1 3
0 -1 -3]

Step 5: Use row operations to create zeros above and below the leading 1 in the third column:

R3 -> R3 + R2 (Add the second row to the third row, preserving the second row)

[ 1 0 -5
0 1 3
0 0 0]

Step 6: The matrix is now in row echelon form (REF). To convert it to reduced row echelon form (RREF), create zeros above the leading 1 in the first column:

R1 -> R1 + 5 * R2 (Multiply the second row by -5 and add it to the first row, preserving the second row)

[ 1 0 0
0 1 3
0 0 0]

Step 7: The matrix is now in reduced row echelon form (RREF). In this form, each leading 1 has zeros above and below it, and each row with nonzero elements comes before any row with all zeros.

So, the matrix in reduced row echelon form is:

[ 1 0 0
0 1 3
0 0 0]