Question

The system of equations

$x_1 + 2x_2 + 3x_3 = \lambda x_1$,

$3x_1 + x_2 + 2x_3 = \lambda x_2$,

$2x_1 + 3x_2 + x_3 = \lambda x_3$ can possess a non-trivial solution then $\lambda =...$

• A. 1
• B. 2
• C. 3
• D. 6

Answer 1

Answer: (D)

6

Answer 2

The given system of equations can be written in matrix form as $Ax = 0$, where $x = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}$ and $A$ is the coefficient matrix:

$A = \begin{pmatrix} 1-\lambda & 2 & 3 \\ 3 & 1-\lambda & 2 \\ 2 & 3 & 1-\lambda \end{pmatrix}$

A homogeneous system of linear equations $Ax = 0$ has a non-trivial solution if and only if the determinant of the coefficient matrix $A$ is zero, i.e., $\det(A) = 0$.

Calculating the determinant of $A$ and setting it to zero allows us to solve for $\lambda$.

$\det(A) = (1-\lambda) \begin{vmatrix} 1-\lambda & 2 \\ 3 & 1-\lambda \end{vmatrix} - 2 \begin{vmatrix} 3 & 2 \\ 2 & 1-\lambda \end{vmatrix} + 3 \begin{vmatrix} 3 & 1-\lambda \\ 2 & 3 \end{vmatrix}$

$\det(A) = (1-\lambda)[(1-\lambda)^2 - 6] - 2[3(1-\lambda) - 4] + 3[9 - 2(1-\lambda)]$

$\det(A) = (1-\lambda)(\lambda^2 - 2\lambda - 5) - 2(-1 - 3\lambda) + 3(7 + 2\lambda)$

$\det(A) = -\lambda^3 + 3\lambda^2 + 15\lambda + 18$

Setting $\det(A) = 0$, we get:

$-\lambda^3 + 3\lambda^2 + 15\lambda + 18 = 0$

$\lambda^3 - 3\lambda^2 - 15\lambda - 18 = 0$

By testing the given options, we find that $\lambda = 6$ is a root of this equation:

$6^3 - 3(6)^2 - 15(6) - 18 = 216 - 108 - 90 - 18 = 0$

Therefore, the system of equations has a non-trivial solution when $\lambda = 6$.