The system of linear equations `x_(1)+2x_(2)+x_(3)=3, 2x_(1)+3x_(2)+x_(3)=3 and 3x_(1)+5x_(2)+2x_(3)=1`, has

Options

A. infinite number of solutions
B. exactly three solutions
C. a unique solution
D. no solution

Answer

To solve the system of linear equations given by:

1. \( x_1 + 2x_2 + x_3 = 3 \)
2. \( 2x_1 + 3x_2 + x_3 = 3 \)
3. \( 3x_1 + 5x_2 + 2x_3 = 1 \)

we will use the method of determinants.

### Step 1: Write the coefficient matrix and the constant matrix

The coefficient matrix \( A \) is:

\[ A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 3 & 5 & 2 \end{bmatrix} \]

The constant matrix \( B \) is:

\[ B = \begin{bmatrix} 3 \\ 3 \\ 1 \end{bmatrix} \]

### Step 2: Calculate the determinant of matrix \( A \)

To find the determinant of \( A \):

\[ \text{det}(A) = \begin{vmatrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 3 & 5 & 2 \end{vmatrix} \]

Calculating this determinant using the formula for a 3x3 matrix:

\[ \text{det}(A) = 1 \cdot (3 \cdot 2 - 1 \cdot 5) - 2 \cdot (2 \cdot 2 - 1 \cdot 3) + 1 \cdot (2 \cdot 5 - 3 \cdot 3) \]

Calculating each term:

1. \( 1 \cdot (6 - 5) = 1 \)
2. \( -2 \cdot (4 - 3) = -2 \)
3. \( 1 \cdot (10 - 9) = 1 \)

Putting it all together:

\[ \text{det}(A) = 1 - 2 + 1 = 0 \]

### Step 3: Analyze the determinant

Since \( \text{det}(A) = 0 \), we need to check the conditions for the existence of solutions.

### Step 4: Calculate the determinant of \( A_1 \)

Now we form the matrix \( A_1 \) by replacing the first column of \( A \) with the constant matrix \( B \):

\[ A_1 = \begin{bmatrix} 3 & 2 & 1 \\ 3 & 3 & 1 \\ 1 & 5 & 2 \end{bmatrix} \]

Calculating the determinant of \( A_1 \):

\[ \text{det}(A_1) = \begin{vmatrix} 3 & 2 & 1 \\ 3 & 3 & 1 \\ 1 & 5 & 2 \end{vmatrix} \]

Using the same determinant formula:

\[ \text{det}(A_1) = 3 \cdot (3 \cdot 2 - 1 \cdot 5) - 2 \cdot (3 \cdot 2 - 1 \cdot 3) + 1 \cdot (3 \cdot 5 - 3 \cdot 1) \]

Calculating each term:

1. \( 3 \cdot (6 - 5) = 3 \)
2. \( -2 \cdot (6 - 3) = -6 \)
3. \( 1 \cdot (15 - 3) = 12 \)

Putting it all together:

\[ \text{det}(A_1) = 3 - 6 + 12 = 9 \neq 0 \]

### Step 5: Conclusion

Since \( \text{det}(A) = 0 \) and \( \text{det}(A_1) \neq 0 \), the system of equations has no solution.

Thus, the answer is that the system of equations has **no solution**.

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