Question

The system of equations:
\[ x - y + 2z = 4 \] \[ 3x + y + 4z = 6 \] \[ x + y + z = 1 \] has:

• A. unique solution
• B. infinitely many solutions
• C. no solution
• D. two solutions

Hint:

If \( \Delta = 0 \) and all augmented determinants are also zero, the system has infinitely many solutions.

Answer 1

The Correct Option is B

Step 1: {Compute the determinant of the coefficient matrix}
\[ \Delta = \begin{vmatrix} 1 & -1 & 2 \\ 3 & 1 & 4 \\ 1 & 1 & 1 \end{vmatrix}. \] Expanding along the first row: \[ \Delta = 1(1 \times 1 - 4 \times 1) + (-1)(3 \times 1 - 4 \times 1) + 2(3 \times 1 - 1 \times 1). \] \[ = (1 - 4) + (-3 + 4) + 2(3 - 1). \] \[ = -3 + 1 + 4 = 0. \] Step 2: {Compute determinant of augmented matrix}
\[ \Delta_1 = \begin{vmatrix} 4 & -1 & 2 \\ 6 & 1 & 4 \\ 1 & 1 & 1 \end{vmatrix}. \] Expanding along the first row: \[ \Delta_1 = 4(1 \times 1 - 4 \times 1) + (-1)(6 \times 1 - 4 \times 1) + 2(6 \times 1 - 1 \times 1). \] \[ = 4(1 - 4) + (-6 + 4) + 2(6 - 1). \] \[ = -12 + 2 + 10 = 0. \] Step 3: {Conclusion}
Since \( \Delta = 0 \) and \( \Delta_1 = 0 \), the system has infinitely many solutions.

Answer 2

Step 1: Write the system of equations
We are given the system of linear equations: \[ x - y + 2z = 4 \quad \text{(1)} \] \[ 3x + y + 4z = 6 \quad \text{(2)} \] \[ x + y + z = 1 \quad \text{(3)}. \] Step 2: Write the augmented matrix
The system can be written as the augmented matrix: \[ \begin{pmatrix} 1 & -1 & 2 & | & 4 \\ 3 & 1 & 4 & | & 6 \\ 1 & 1 & 1 & | & 1 \end{pmatrix}. \] Step 3: Perform row operations to reduce the matrix
We will perform Gaussian elimination to reduce the augmented matrix to row echelon form. First, subtract 3 times row 1 from row 2: \[ R_2 \rightarrow R_2 - 3R_1. \] \[ \begin{pmatrix} 1 & -1 & 2 & | & 4 \\ 0 & 4 & -2 & | & -6 \\ 1 & 1 & 1 & | & 1 \end{pmatrix}. \] Next, subtract row 1 from row 3: \[ R_3 \rightarrow R_3 - R_1. \] \[ \begin{pmatrix} 1 & -1 & 2 & | & 4 \\ 0 & 4 & -2 & | & -6 \\ 0 & 2 & -1 & | & -3 \end{pmatrix}. \] Now, subtract \( \frac{1}{2} \) of row 2 from row 3: \[ R_3 \rightarrow R_3 - \frac{1}{2}R_2. \] \[ \begin{pmatrix} 1 & -1 & 2 & | & 4 \\ 0 & 4 & -2 & | & -6 \\ 0 & 0 & 0 & | & 0 \end{pmatrix}. \] Step 4: Analyze the system
The third row represents \( 0 = 0 \), which is a trivial equation, and does not provide any new information. This means that there are infinitely many solutions, as the system is consistent and dependent. 
Answer:
The system has infinitely many solutions.