To solve the system of linear equations corresponding to the given coefficient matrix, we first assume the right-hand side vector is [6,6,6] (since the sum of each row in the coefficient matrix equals 6, a common setup for such problems). Here's the step-by-step solution:

Step 1: Define the system

Let the variables be (x, y, z). The system is:
[ \begin{cases} 2x + 3y + z = 6 \ 3x + y + 2z = 6 \ x + 2y + 3z = 6 \end{cases} ]

Step 2: Augmented Matrix & Row Operations

Convert to augmented matrix and simplify:
[ \left[\begin{array}{ccc|c} 1 & 2 & 3 & 6 \ 0 & 1 & 5 & 6 \ 0 & 0 & 1 & 1 \end{array}\right] \quad (\text{after row swaps and reductions}) ]

Step 3: Back Substitution

• From the third row: (z = 1)
• Substitute (z=1) into the second row: (y + 5(1) = 6 \implies y=1)
• Substitute (y=1,z=1) into the first row: (x + 2(1) +3(1)=6 \implies x=1)

Answer: (x=1, y=1, z=1)

(\boxed{(1,1,1)})

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