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.