If the system of equations $$\begin{aligned}x + 2y - 3z &= 2, \\2x + \lambda y + 5z &= 5, \\14x + 3y + \mu z &= 33\end{aligned}$$ has infinitely many solutions, then $$\lambda + \mu \text{ is equal to:} $$

For a system of three linear equations to possess infinitely many solutions, the following two conditions must hold:
  • The determinant of the coefficient matrix must be zero (so its rank is < 3).
  • Every equation must be a linear combination of the others, i.e. the augmented matrix must have the same rank as the coefficient matrix.

Write the coefficient matrix $$A$$ and the constant column $$\mathbf{b}$$:

$$A = \begin{bmatrix}1 & 2 & -3 \\ 2 & \lambda & 5 \\ 14 & 3 & \mu\end{bmatrix},\qquad \mathbf{b} = \begin{bmatrix}2 \\ 5 \\ 33\end{bmatrix}.$$

Step 1 : Determinant of the coefficient matrix

The determinant is

$$\begin{vmatrix}1 & 2 & -3 \\ 2 & \lambda & 5 \\ 14 & 3 & \mu\end{vmatrix} = 1(\lambda\mu-5\!\cdot\!3)\;-\;2(2\mu-5\!\cdot\!14)\;+\;(-3)(2\!\cdot\!3-\lambda\!\cdot\!14).$$

Simplifying term by term:

$$\lambda\mu-15\;-\;2(2\mu-70)\;-\;3(6-14\lambda)$$

$$=\lambda\mu-15\;-\;4\mu+140\;-\,18+42\lambda$$

$$=\mu(\lambda-4)+42\lambda+107.$$

For infinitely many solutions, this determinant must vanish:

$$\mu(\lambda-4)+42\lambda+107=0\quad -(1).$$

Step 2 : Ensuring the third equation is a linear combination of the first two

Assume constants $$a$$ and $$b$$ exist such that

$$a(x+2y-3z)+b(2x+\lambda y+5z)=14x+3y+\mu z \quad\text{and}\quad a(2)+b(5)=33.$$(The left-hand side recreates the third equation’s coefficients and constant term.)

Matching the coefficients of $$x,\,y,\,z$$ and the constants gives

$$\begin{aligned} a+2b &= 14 \quad &(x\text{-coeff})\\ 2a+\lambda b &= 3 \quad &(y\text{-coeff})\\ -3a+5b &= \mu \quad &(z\text{-coeff})\\ 2a+5b &= 33 \quad &(\text{constant}) \end{aligned}$$

From $$a+2b=14$$ obtain

$$a = 14-2b \quad -(2).$$

Substitute $$a$$ from $$(2)$$ into the $$y$$-coefficient condition:

$$2(14-2b)+\lambda b = 3$$

$$28-4b+\lambda b = 3$$

$$(\lambda-4)b = -25$$

$$b = \frac{-25}{\lambda-4} \quad -(3).$$

Insert $$b$$ from $$(3)$$ into the constant condition $$2a+5b=33$$. First compute $$2a$$ using $$(2)$$:

$$2a = 2(14-2b) = 28-4b.$$

Hence

$$28-4b+5b = 33 \;\Longrightarrow\; 28 + b = 33$$

$$b = 5.$$

Equating this value of $$b$$ to the one in $$(3)$$ gives

$$5 = \frac{-25}{\lambda-4}$$

$$\lambda-4 = -5$$

$$\lambda = -1.$$

With $$\lambda=-1$$, use $$(3)$$ to confirm $$b=5$$ and $$(2)$$ to find $$a$$:

$$a = 14 - 2(5) = 4.$$

Finally, compute $$\mu$$ from the $$z$$-coefficient relation:

$$\mu = -3a + 5b = -3(4)+5(5) = -12+25 = 13.$$

Step 3 : Value of $$\lambda+\mu$$

$$\lambda+\mu = (-1)+13 = 12.$$

Therefore, $$\lambda+\mu = 12$$, which corresponds to Option C.