Question

The system of linear equations $\lambda x+2y+2z=5$ $2\lambda x+3y+5z=8$ $4x+\lambda y+6z=10$ has :

• A. no soiution when $\lambda = 2$
• B. infinitely many solutions when $\lambda = 2$
• C. no solution when $\lambda = 8$
• D. a unique solution when $\lambda = -8$

Answer 1

The Correct Option is A

To analyze the system of linear equations given below:

$$\lambda x + 2y + 2z = 5$$ $$2\lambda x + 3y + 5z = 8$$ $$4x + \lambda y + 6z = 10$$

we are asked to determine under which conditions (depending on the value of $\lambda$) the system has no solution, infinitely many solutions, or a unique solution.

Start by examining the coefficients matrix of the system:

$$ \begin{pmatrix} \lambda & 2 & 2 \\ 2\lambda & 3 & 5 \\ 4 & \lambda & 6 \\ \end{pmatrix} $$

and the augmented matrix of the system:

$$ \begin{pmatrix} \lambda & 2 & 2 & | & 5 \\ 2\lambda & 3 & 5 & | & 8 \\ 4 & \lambda & 6 & | & 10 \\ \end{pmatrix} $$

Now, we check the determinant of the coefficients matrix:

$$ \Delta = \begin{vmatrix} \lambda & 2 & 2 \\ 2\lambda & 3 & 5 \\ 4 & \lambda & 6 \\ \end{vmatrix} $$

Calculate the determinant:

$$ \Delta = \lambda(3 \cdot 6 - 5 \cdot \lambda) - 2(2\lambda \cdot 6 - 5 \cdot 4) + 2(2\lambda \cdot \lambda - 3 \cdot 4) $$

Expanding gives:

$$ \Delta = \lambda(18 - 5\lambda) - 2(12\lambda - 20) + 2(2\lambda^2 - 12) $$

This simplifies to:

$$ \Delta = 18\lambda - 5\lambda^2 - 24\lambda + 40 + 4\lambda^2 - 24 $$ $$ \Delta = -\lambda^2 - 6\lambda + 16 $$

Set $\Delta = 0$ to find when the system might be inconsistent or have infinitely many solutions:

$$ -\lambda^2 - 6\lambda + 16 = 0 $$

Solve the quadratic equation:

$$ \lambda^2 + 6\lambda - 16 = 0 $$ $$ \lambda = \frac{-6 \pm \sqrt{6^2 - 4 \times 1 \times (-16)}}{2 \times 1} $$ $$ \lambda = \frac{-6 \pm \sqrt{36 + 64}}{2} $$ $$ \lambda = \frac{-6 \pm \sqrt{100}}{2} $$ $$ \lambda = \frac{-6 \pm 10}{2} $$

Thus:

$$ \lambda = 2 \quad \text{or} \quad \lambda = -8 $$

For $\lambda = 2$, calculate the rank of the augmented matrix and the coefficient matrix. Note that when the ranks are unequal, the system has no solution. Substitute $\lambda = 2$:

$$ \begin{matrix} 2x + 2y + 2z = 5 \\ 4x + 3y + 5z = 8 \\ 4x + 2y + 6z = 10 \\ \end{matrix} $$

The first two equations lead to a contradiction, hence no solution is possible when $\lambda = 2$. Therefore, the system has no solution when $\lambda = 2$.

Conclusion: The correct answer is "no solution when $\lambda = 2$".