Question

The system $x + y - 4z = 2$, $3x + y + 5z = 7$, $2x + 3y + z = 5$ has

• A. infinite number of solutions
• B. unique solution
• C. trivial solution
• D. no solution

Hint:

If determinant of coefficient matrix is non-zero, system has unique solution.

Answer 1

The Correct Option is B

To determine the type of solution for the given system of equations, we need to analyze the system:

1. The given system of equations is: \(x + y - 4z = 2 \quad \quad (1)\), \(3x + y + 5z = 7 \quad \quad (2)\), \(2x + 3y + z = 5 \quad \quad (3)\).
2. We can express this system in matrix form as \(A\vec{x} = \vec{b}\), where: \(A = \begin{bmatrix} 1 & 1 & -4 \\ 3 & 1 & 5 \\ 2 & 3 & 1 \end{bmatrix}, \quad \vec{x} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad \vec{b} = \begin{bmatrix} 2 \\ 7 \\ 5 \end{bmatrix}\).
3. The first step is to check whether the matrix \(A\) is invertible by calculating its determinant. \(\text{det}(A) = \begin{vmatrix} 1 & 1 & -4 \\ 3 & 1 & 5 \\ 2 & 3 & 1 \end{vmatrix}\).
4. Using the formula for a 3x3 determinant, we have: \(\text{det}(A) = 1(1 \cdot 1 - 5 \cdot 3) - 1(3 \cdot 1 - 5 \cdot 2) + (-4)(3 \cdot 3 - 2 \cdot 1)\).
5. Solving this, we get: \(\text{det}(A) = 1(-14) - 1(1) + (-4)(9 - 2) = -14 - 1 - 28 = -43\). Since \(\text{det}(A) \neq 0\), the matrix \(A\) is invertible.
6. Since \(\text{det}(A) \neq 0\), the system of equations has a unique solution.

Therefore, the correct answer is that the system has a unique solution.