Use matrix to solve the following system of equations.
`x+y+z=3`
`x+2y+3z=4`
`2x+3y+4z=7`

Answer

To solve the system of equations using matrices, we will follow these steps:

### Step 1: Write the system of equations in matrix form
The given equations are:
1. \( x + y + z = 3 \)
2. \( x + 2y + 3z = 4 \)
3. \( 2x + 3y + 4z = 7 \)

We can express this system in matrix form as \( A \mathbf{x} = \mathbf{b} \), where:
- \( A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 2 & 3 & 4 \end{pmatrix} \)
- \( \mathbf{x} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \)
- \( \mathbf{b} = \begin{pmatrix} 3 \\ 4 \\ 7 \end{pmatrix} \)

### Step 2: Check if matrix A is invertible
To determine if we can use the inverse of \( A \), we need to calculate the determinant of \( A \).

\[ \text{det}(A) = 1(2 \cdot 4 - 3 \cdot 3) - 1(1 \cdot 4 - 3 \cdot 2) + 1(1 \cdot 3 - 2 \cdot 2) \]
Calculating this gives:
\[ = 1(8 - 9) - 1(4 - 6) + 1(3 - 4) = 1(-1) - 1(-2) + 1(-1) = -1 + 2 - 1 = 0 \]

Since the determinant is 0, matrix \( A \) is not invertible.

### Step 3: Use Row Echelon Form
Since \( A \) is not invertible, we will use the row echelon form to solve the system. We will augment the matrix \( A \) with \( \mathbf{b} \):

\[ \begin{pmatrix} 1 & 1 & 1 & | & 3 \\ 1 & 2 & 3 & | & 4 \\ 2 & 3 & 4 & | & 7 \end{pmatrix} \]

Now, we will perform row operations to convert this into row echelon form.

1. Subtract Row 1 from Row 2:
\[ R_2 = R_2 - R_1 \implies \begin{pmatrix} 1 & 1 & 1 & | & 3 \\ 0 & 1 & 2 & | & 1 \\ 2 & 3 & 4 & | & 7 \end{pmatrix} \]

2. Subtract 2 times Row 1 from Row 3:
\[ R_3 = R_3 - 2R_1 \implies \begin{pmatrix} 1 & 1 & 1 & | & 3 \\ 0 & 1 & 2 & | & 1 \\ 0 & 1 & 2 & | & 1 \end{pmatrix} \]

3. Subtract Row 2 from Row 3:
\[ R_3 = R_3 - R_2 \implies \begin{pmatrix} 1 & 1 & 1 & | & 3 \\ 0 & 1 & 2 & | & 1 \\ 0 & 0 & 0 & | & 0 \end{pmatrix} \]

### Step 4: Back substitution
From the row echelon form, we can interpret the equations:
1. \( x + y + z = 3 \) (from Row 1)
2. \( y + 2z = 1 \) (from Row 2)
3. The last row \( 0 = 0 \) indicates that we have a dependent system.

From the second equation, we can express \( y \) in terms of \( z \):
\[ y = 1 - 2z \]

Substituting \( y \) into the first equation:
\[ x + (1 - 2z) + z = 3 \]
\[ x + 1 - z = 3 \implies x = 2 + z \]

### Step 5: General solution
Let \( z = k \) (where \( k \) is any real number), then:
- \( z = k \)
- \( y = 1 - 2k \)
- \( x = 2 + k \)

Thus, the solution set is:
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 2 + k \\ 1 - 2k \\ k \end{pmatrix}, \quad k \in \mathbb{R} \]