Question

The system of equations \[ x + 3by + bz = 0, \quad x + 2ay + az = 0, \quad x + 4cy + cz = 0 \] has:

• A. only zero solution for any values of \( a, b, c \).
• B. non-zero solution for any values of \( a, b, c \).
• C. non-zero solution whenever \( b(a+c) = 2ac \).
• D. non-zero solution whenever \( a+c = 2b \).

Hint:

To check if a system of equations has a non-trivial solution, compute the determinant of the coefficient matrix. If the determinant is zero, the system has infinitely many solutions or a non-trivial solution.

Answer 1

The Correct Option is C

Step 1: Construct the Coefficient Matrix

The coefficient matrix \( A \) is:

\[ A = \begin{bmatrix} 1 & 3b & b \\ 1 & 2a & a \\ 1 & 4c & c \end{bmatrix} \]

For a non-zero solution to exist, the determinant of \( A \) must be zero:

\[ \begin{vmatrix} 1 & 3b & b \\ 1 & 2a & a \\ 1 & 4c & c \end{vmatrix} = 0. \]

Step 2: Compute the Determinant

Expanding along the first column:

\[ \begin{vmatrix} 1 & 3b & b \\ 1 & 2a & a \\ 1 & 4c & c \end{vmatrix} = 1 \begin{vmatrix} 2a & a \\ 4c & c \end{vmatrix} - 3b \begin{vmatrix} 1 & a \\ 1 & c \end{vmatrix} + b \begin{vmatrix} 1 & 2a \\ 1 & 4c \end{vmatrix}. \]

Computing Each Minor Determinant

\[ \begin{vmatrix} 2a & a \\ 4c & c \end{vmatrix} = 2ac - 4ac = -2ac. \]

\[ \begin{vmatrix} 1 & a \\ 1 & c \end{vmatrix} = c - a. \]

\[ \begin{vmatrix} 1 & 2a \\ 1 & 4c \end{vmatrix} = 4c - 2a. \]

Substituting:

\[ 1(-2ac) - 3b(c-a) + b(4c - 2a) = 0. \]

\[ -2ac - 3bc + 3ab + 4bc - 2ab = 0. \]

\[ -2ac + ab + bc = 0. \]

Step 3: Conclusion

For a non-trivial solution to exist, we require:

\[ b(a + c) = 2ac. \]

Thus, the correct answer is:

\[ (3) \text{ non-zero solution whenever } b(a+c) = 2ac. \]