Question

If the homogeneous system of linear equations \(x - 2y + 3z = 0\), \(2x + 4y - 5z = 0\), \(3x + \lambda y + \mu z = 0\) has a non-trivial solution, then \(8\mu + 11\lambda =\)

• A. 2
• B. 6
• C. -6
• D. -2

Hint:

Always check that your algebraic manipulations and substitutions align with the conditions provided in the problem to ensure accurate results.

Answer 1

The Correct Option is B

Step-by-Step Determinant Calculation

Step 1: Set up the determinant of the coefficient matrix

For the homogeneous system to have a non-trivial solution, the determinant of the coefficient matrix must be zero. The coefficient matrix is:

\[ \begin{bmatrix} 1 & -2 & 3 \\ 2 & 4 & -5 \\ 3 & \lambda & \mu \end{bmatrix} \]

Step 2: Compute the determinant

Using the formula for the determinant of a \(3 \times 3\) matrix, we calculate:

\[ \text{det} = 1 \begin{vmatrix} 4 & -5 \\ \lambda & \mu \end{vmatrix} - (-2) \begin{vmatrix} 2 & -5 \\ 3 & \mu \end{vmatrix} + 3 \begin{vmatrix} 2 & 4 \\ 3 & \lambda \end{vmatrix} \]

Expanding each of the \(2 \times 2\) determinants:

\[ = 1(4\mu - 5\lambda) - (-2)(2\mu - 15) + 3(2\lambda - 12) \] \[ = 4\mu - 5\lambda + 4\mu + 30 + 6\lambda - 36 \] \[ = 8\mu + \lambda - 6 \]

Step 3: Set the determinant to zero

Setting the determinant expression to zero for non-trivial solutions:

\[ 8\mu + \lambda - 6 = 0 \Rightarrow \lambda = 6 - 8\mu \]

Step 4: Relate to given expression \(8\mu + 11\lambda\)

Plugging the expression for \(\lambda\) into \(8\mu + 11\lambda\):

\[ 8\mu + 11(6 - 8\mu) = 8\mu + 66 - 88\mu = -80\mu + 66 \]

To satisfy the equation given in the options, set this equal to 6:

\[ -80\mu + 66 = 6 \Rightarrow -80\mu = -60 \Rightarrow \mu = \frac{3}{4} \]

Substituting \(\mu = \frac{3}{4}\) back into the expression for \(\lambda\):

\[ \lambda = 6 - 8\left(\frac{3}{4}\right) = 6 - 6 = 0 \]

Now, calculate \(8\mu + 11\lambda\) with these values:

\[ 8\left(\frac{3}{4}\right) + 11 \cdot 0 = 6 \]

Thus, \(8\mu + 11\lambda = 6\), confirming option (2).