1. Key Concept: Conditions for a System of Linear Equations

For a system of linear equations of the form $AX = B$, where $A$ is the coefficient matrix, $X$ is the column vector of variables, and $B$ is the column vector of constants, we use Cramer's Rule to determine the nature of solutions.

Let:

• $\Delta = \det(A)$ be the determinant of the coefficient matrix.
• $\Delta_x, \Delta_y, \Delta_z$ be the determinants of the matrices formed by replacing the respective variable column in $A$ with the constant vector $B$.

The conditions for the nature of solutions are:

1. Unique Solution: If $\Delta \neq 0$.
2. No Solution (Inconsistent System): If $\Delta = 0$ AND at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero.
3. Infinitely Many Solutions (Consistent System): If $\Delta = 0$ AND $\Delta_x = \Delta_y = \Delta_z = 0$.

The problem asks for the set $S$ of values of $\lambda$ for which the system has no solution. This means we need to identify $\lambda$ values such that $\Delta = 0$ AND ( $\Delta_x \neq 0$ OR $\Delta_y \neq 0$ OR $\Delta_z \neq 0$).

2. Setting up the System and Calculating $\Delta$

The given system of equations is: $6 \lambda x-3 y+3 z=4 \lambda^{2}$ $2 x+6 \lambda y+4 z=1$ $3 x+2 y+3 \lambda z=\lambda$

The coefficient matrix $A$ and the constant vector $B$ are: $A = \begin{pmatrix} 6\lambda & -3 & 3 \\ 2 & 6\lambda & 4 \\ 3 & 2 & 3\lambda \end{pmatrix}, \quad B = \begin{pmatrix} 4\lambda^2 \\ 1 \\ \lambda \end{pmatrix}$

Now, we calculate the determinant of the coefficient matrix, $\Delta$: $\Delta = \det(A) = \begin{vmatrix} 6\lambda & -3 & 3 \\ 2 & 6\lambda & 4 \\ 3 & 2 & 3\lambda \end{vmatrix}$ We expand the determinant along the first row for calculation: $\Delta = 6\lambda \begin{vmatrix} 6\lambda & 4 \\ 2 & 3\lambda \end{vmatrix} - (-3) \begin{vmatrix} 2 & 4 \\ 3 & 3\lambda \end{vmatrix} + 3 \begin{vmatrix} 2 & 6\lambda \\ 3 & 2 \end{vmatrix}$ $= 6\lambda ((6\lambda)(3\lambda) - (4)(2)) + 3 ((2)(3\lambda) - (4)(3)) + 3 ((2)(2) - (6\lambda)(3))$ $= 6\lambda (18\lambda^2 - 8) + 3 (6\lambda - 12) + 3 (4 - 18\lambda)$ $= 108\lambda^3 - 48\lambda + 18\lambda - 36 + 12 - 54\lambda$ $= 108\lambda^3 - (48 - 18 + 54)\lambda - (36 - 12)$ $= 108\lambda^3 - 84\lambda - 24$

3. Finding Values of $\lambda$ for which $\Delta = 0$

For the system to have no solution or infinitely many solutions, $\Delta$ must be zero. So, we set our expression for $\Delta$ to zero: $108\lambda^3 - 84\lambda - 24 = 0$ To simplify, divide the entire equation by 12: $9\lambda^3 - 7\lambda - 2 = 0$ We look for integer roots by testing factors of the constant term (-2) divided by factors of the leading coefficient (9). Let's test $\lambda = 1$: $9(1)^3 - 7(1) - 2 = 9 - 7 - 2 = 0$. So, $\lambda = 1$ is a root. This means $(\lambda - 1)$ is a factor. We can use polynomial division or synthetic division to find the other factors. Using synthetic division:

This gives us the quadratic factor $9\lambda^2 + 9\lambda + 2$. So, the equation becomes: $(\lambda - 1)(9\lambda^2 + 9\lambda + 2) = 0$ Now, we find the roots of the quadratic equation $9\lambda^2 + 9\lambda + 2 = 0$ using the quadratic formula $\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $\lambda = \frac{-9 \pm \sqrt{9^2 - 4(9)(2)}}{2(9)}$ $\lambda = \frac{-9 \pm \sqrt{81 - 72}}{18}$ $\lambda = \frac{-9 \pm \sqrt{9}}{18}$ $\lambda = \frac{-9 \pm 3}{18}$ This yields two more roots: $\lambda_1 = \frac{-9 + 3}{18} = \frac{-6}{18} = -\frac{1}{3}$ $\lambda_2 = \frac{-9 - 3}{18} = \frac{-12}{18} = -\frac{2}{3}$ Thus, the values of $\lambda$ for which $\Delta = 0$ are $1, -\frac{1}{3}, -\frac{2}{3}$. These are the candidate values for which the system has no solution or infinitely many solutions.

4. Testing Candidate Values of $\lambda$ for "No Solution"

We now check each of these $\lambda$ values by calculating $\Delta_x, \Delta_y, \Delta_z$. If $\Delta = 0$ and at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero, then the system has no solution.

Case 1: $\lambda = 1$ The constant vector $B$ becomes $\begin{pmatrix} 4(1)^2 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \\ 1 \end{pmatrix}$. Let's calculate $\Delta_x$: $\Delta_x = \begin{vmatrix} 4 & -3 & 3 \\ 1 & 6 & 4 \\ 1 & 2 & 3 \end{vmatrix}$ Expand along the first row: $\Delta_x = 4(6 \times 3 - 4 \times 2) - (-3)(1 \times 3 - 4 \times 1) + 3(1 \times 2 - 6 \times 1)$ $= 4(18 - 8) + 3(3 - 4) + 3(2 - 6)$ $= 4(10) + 3(-1) + 3(-4)$ $= 40 - 3 - 12 = 25$ Since $\Delta_x = 25 \neq 0$ and $\Delta = 0$ for $\lambda = 1$, the system has no solution for $\lambda = 1$. Thus, $\lambda = 1 \in S$.

Case 2: $\lambda = -\frac{1}{3}$ The constant vector $B$ becomes $\begin{pmatrix} 4(-\frac{1}{3})^2 \\ 1 \\ -\frac{1}{3} \end{pmatrix} = \begin{pmatrix} \frac{4}{9} \\ 1 \\ -\frac{1}{3} \end{pmatrix}$. The coefficient matrix $A$ becomes $\begin{pmatrix} -2 & -3 & 3 \\ 2 & -2 & 4 \\ 3 & 2 & -1 \end{pmatrix}$. Let's calculate $\Delta_x$: $\Delta_x = \begin{vmatrix} \frac{4}{9} & -3 & 3 \\ 1 & -2 & 4 \\ -\frac{1}{3} & 2 & -1 \end{vmatrix}$ Expand along the first row: $\Delta_x = \frac{4}{9}((-2)(-1) - (4)(2)) - (-3)((1)(-1) - (4)(-\frac{1}{3})) + 3((1)(2) - (-2)(-\frac{1}{3}))$ $= \frac{4}{9}(2 - 8) + 3(-1 + \frac{4}{3}) + 3(2 - \frac{2}{3})$ $= \frac{4}{9}(-6) + 3(\frac{1}{3}) + 3(\frac{4}{3})$ $= -\frac{24}{9} + 1 + 4 = -\frac{8}{3} + 5 = \frac{-8 + 15}{3} = \frac{7}{3}$ Since $\Delta_x = \frac{7}{3} \neq 0$ and $\Delta = 0$ for $\lambda = -\frac{1}{3}$, the system has no solution for $\lambda = -\frac{1}{3}$. Thus, $\lambda = -\frac{1}{3} \in S$.

Case 3: $\lambda = -\frac{2}{3}$ The constant vector $B$ becomes $\begin{pmatrix} 4(-\frac{2}{3})^2 \\ 1 \\ -\frac{2}{3} \end{pmatrix} = \begin{pmatrix} \frac{16}{9} \\ 1 \\ -\frac{2}{3} \end{pmatrix}$. The coefficient matrix $A$ becomes $\begin{pmatrix} -4 & -3 & 3 \\ 2 & -4 & 4 \\ 3 & 2 & -2 \end{pmatrix}$. Let's calculate $\Delta_x$: $\Delta_x = \begin{vmatrix} \frac{16}{9} & -3 & 3 \\ 1 & -4 & 4 \\ -\frac{2}{3} & 2 & -2 \end{vmatrix}$ Notice that the third column ($C_3$) is $-1$ times the second column ($C_2$). Specifically, $C_3 = \begin{pmatrix} 3 \\ 4 \\ -2 \end{pmatrix}$ and $C_2 = \begin{pmatrix} -3 \\ -4 \\ 2 \end{pmatrix}$, so $C_3 = -C_2$. If two columns of a determinant are proportional, the determinant is zero. Therefore, $\Delta_x = 0$ for $\lambda = -\frac{2}{3}$. Since $\Delta_x = 0$, we must check $\Delta_y$ or $\Delta_z$ to see if they are non-zero. If all are zero, it's infinitely many solutions; otherwise, it's no solution. Let's calculate $\Delta_y$: $\Delta_y = \begin{vmatrix} -4 & \frac{16}{9} & 3 \\ 2 & 1 & 4 \\ 3 & -\frac{2}{3} & -2 \end{vmatrix}$ Expand along the first column: $\Delta_y = -4((1)(-2) - (4)(-\frac{2}{3})) - 2((\frac{16}{9})(-2) - (3)(-\frac{2}{3})) + 3((\frac{16}{9})(4) - (3)(1))$ $= -4(-2 + \frac{8}{3}) - 2(-\frac{32}{9} + 2) + 3(\frac{64}{9} - 3)$ $= -4(\frac{2}{3}) - 2(\frac{-32+18}{9}) + 3(\frac{64-27}{9})$ $= -\frac{8}{3} - 2(-\frac{14}{9}) + 3(\frac{37}{9})$ $= -\frac{8}{3} + \frac{28}{9} + \frac{37}{3}$ $= \frac{-24 + 28 + 111}{9} = \frac{115}{9}$ Since $\Delta_y = \frac{115}{9} \neq 0$ (and $\Delta = 0$ for $\lambda = -\frac{2}{3}$), the system has no solution for $\lambda = -\frac{2}{3}$. Thus, $\lambda = -\frac{2}{3} \in S$.

Tip: When $\Delta_x=0$ for a particular $\lambda$, it is crucial to check $\Delta_y$ or $\Delta_z$. If even one of them is non-zero, the system has no solution. Only if ALL of $\Delta_x, \Delta_y, \Delta_z$ are zero, does it imply infinitely many solutions.

5. Identifying the Set $S$

From our analysis, all three values of $\lambda$ for which $\Delta = 0$ lead to the condition for "no solution". Therefore, the set $S = \{1, -\frac{1}{3}, -\frac{2}{3}\}$.

6. Final Calculation

We need to calculate 12 \sum_\limits{i \in S}|\lambda|. First, find the sum of the absolute values of the elements in $S$: \sum_\limits{i \in S}|\lambda| = |1| + |-\frac{1}{3}| + |-\frac{2}{3}| $= 1 + \frac{1}{3} + \frac{2}{3}$ $= 1 + \frac{1+2}{3} = 1 + \frac{3}{3} = 1 + 1 = 2$ Now, multiply this sum by 12: 12 \sum_\limits{i \in S}|\lambda| = 12 \times 2 = 24

The final answer is $\boxed{24}$.

Key Takeaway: For a system of linear equations to have no solution, the determinant of the coefficient matrix ($\Delta$) must be zero, AND at least one of the determinants obtained by replacing a column with the constant vector ($\Delta_x, \Delta_y, \Delta_z$) must be non-zero. Thorough calculation of all relevant determinants for each candidate $\lambda$ value is essential to correctly classify the system's behavior.