1. Write the system in matrix form

The given system is

$$6\lambda x-3y+3z=4\lambda^2,$$ $$2x+6\lambda y+4z=1,$$ $$3x+2y+3\lambda z=\lambda.$$

So the coefficient matrix is

$$A=\begin{pmatrix} 6\lambda & -3 & 3\\ 2 & 6\lambda & 4\\ 3 & 2 & 3\lambda \end{pmatrix},$$

and the RHS vector is

$$B=\begin{pmatrix}4\lambda^2\\1\\\lambda\end{pmatrix}.$$

For the system to have no solution, we need:

• $\det(A)=0$ (so the system is singular), and
• the augmented system must be inconsistent.

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2. Find when $\det(A)=0$

Compute

$$\det(A)=\begin{vmatrix} 6\lambda & -3 & 3\\ 2 & 6\lambda & 4\\ 3 & 2 & 3\lambda \end{vmatrix}.$$

Expanding along the first row,

$$\det(A)=6\lambda\begin{vmatrix}6\lambda & 4\\2 & 3\lambda\end{vmatrix} -(-3)\begin{vmatrix}2 & 4\\3 & 3\lambda\end{vmatrix} +3\begin{vmatrix}2 & 6\lambda\\3 & 2\end{vmatrix}.$$

Now,

$$\begin{vmatrix}6\lambda & 4\\2 & 3\lambda\end{vmatrix}=18\lambda^2-8,$$ $$\begin{vmatrix}2 & 4\\3 & 3\lambda\end{vmatrix}=6\lambda-12,$$ $$\begin{vmatrix}2 & 6\lambda\\3 & 2\end{vmatrix}=4-18\lambda.$$

Hence,

$$\det(A)=6\lambda(18\lambda^2-8)+3(6\lambda-12)+3(4-18\lambda).$$

Simplify:

$$\det(A)=108\lambda^3-48\lambda+18\lambda-36+12-54\lambda$$ $$=108\lambda^3-84\lambda-24.$$

Factor out $12$:

$$\det(A)=12(9\lambda^3-7\lambda-2).$$

So we solve

$$9\lambda^3-7\lambda-2=0.$$

Try rational roots. For $\lambda=1$,

$$9-7-2=0,$$

so $(\lambda-1)$ is a factor.

Divide:

$$9\lambda^3-7\lambda-2=(\lambda-1)(9\lambda^2+9\lambda+2).$$

Now,

$$9\lambda^2+9\lambda+2=(3\lambda+1)(3\lambda+2).$$

Thus,

$$\det(A)=12(\lambda-1)(3\lambda+1)(3\lambda+2).$$

Therefore singular values are

$$\lambda=1,\,-\frac13,\,-\frac23.$$

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3. Check consistency for each singular value

We must find which of these make the system inconsistent.

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Case 1: $\lambda=1$

System becomes

$$6x-3y+3z=4,$$ $$2x+6y+4z=1,$$ $$3x+2y+3z=1.$$

From the first equation,

$$2x-y+z=\frac43.$$

Multiply by $3$:

$$6x-3y+3z=4.$$

Now check if row dependence creates contradiction. Observe coefficient rows satisfy

$$R_1=3R_3-2R_2?$$

Let's verify more directly through elimination.

Take

$$R_2-\frac13R_1: \quad (2x+6y+4z)-\frac13(6x-3y+3z)=1-\frac43$$ $$0x+7y+3z=-\frac13.$$

Take

$$R_3-\frac12R_1: \quad (3x+2y+3z)-\frac12(6x-3y+3z)=1-2$$ $$0x+\frac72 y+\frac32 z=-1.$$

Multiplying the second derived equation by $2$ gives

$$7y+3z=-2,$$

whereas from the first derived equation,

$$7y+3z=-\frac13.$$

Contradiction. Hence no solution for $\lambda=1$.

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Case 2: $\lambda=-\frac13$

System becomes

$$-2x-3y+3z=\frac49,$$ $$2x-2y+4z=1,$$ $$3x+2y-z=-\frac13.$$

Check consistency by solving two equations and testing the third.

Add first and second:

$$(-2x-3y+3z)+(2x-2y+4z)=\frac49+1$$ $$-5y+7z=\frac{13}{9}. \quad (1)$$

From third,

$$3x+2y-z=-\frac13. \quad (2)$$

From second,

$$2x-2y+4z=1. \quad (3)$$

Solve (2) and (3): from (3),

$$x=y-2z+\frac12.$$

Substitute into (2):

$$3\left(y-2z+\frac12\right)+2y-z=-\frac13$$ $$5y-7z+\frac32=-\frac13$$ $$5y-7z=-\frac{11}{6}.$$

Thus

$$-5y+7z=\frac{11}{6}.$$

But from (1),

$$-5y+7z=\frac{13}{9}.$$

Contradiction. Hence no solution for $\lambda=-\frac13$.

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Case 3: $\lambda=-\frac23$

System becomes

$$-4x-3y+3z=\frac{16}{9},$$ $$2x-4y+4z=1,$$ $$3x+2y-2z=-\frac23.$$

Multiply the second equation by $2$:

$$4x-8y+8z=2.$$

Add with the first equation:

$$(-4x-3y+3z)+(4x-8y+8z)=\frac{16}{9}+2$$ $$-11y+11z=\frac{34}{9}$$ $$y-z=-\frac{34}{99}. \quad (1)$$

From third equation,

$$3x+2y-2z=-\frac23. \quad (2)$$

From second,

$$2x-4y+4z=1. \quad (3)$$

Use (3):

$$x=2y-2z+\frac12.$$

Substitute into (2):

$$3\left(2y-2z+\frac12\right)+2y-2z=-\frac23$$ $$8y-8z+\frac32=-\frac23$$ $$y-z=-\frac{13}{48}. \quad (4)$$

(1) and (4) contradict. Hence no solution for $\lambda=-\frac23$.

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4. Form the set $S$

Thus,

$$S=\left\{1,-\frac13,-\frac23\right\}.$$

So,

$$\sum_{i\in S}|\lambda|=\left|1\right|+\left|-\frac13\right|+\left|-\frac23\right|=1+\frac13+\frac23=2.$$

Therefore,

$$12\sum_{i\in S}|\lambda|=12\cdot 2=24.$$

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5. Compare with stored answer

Derived answer = $24$.

Stored correct answer = $24$.

They agree.