To solve this problem, we need to understand the conditions under which a system of linear equations has infinitely many solutions. This is a crucial concept in Matrices and Determinants, often tested in JEE.

Key Concept: Conditions for Solutions of a System of Linear Equations

Consider a system of linear equations represented in matrix form as $AX=B$, where $A$ is the coefficient matrix, $X$ is the variable matrix, and $B$ is the constant matrix. For a system of $n$ linear equations in $n$ variables, the nature of solutions depends on the determinant of $A$, denoted as $|A|$, and the product $(\text{adj } A)B$:

1. Unique Solution: If $|A| \neq 0$, the system has a unique solution given by $X = A^{-1}B$.
2. No Solution (Inconsistent System): If $|A|=0$ AND $(\text{adj } A)B \neq 0$ (i.e., at least one element of the resulting column matrix is non-zero), the system has no solution.
3. Infinitely Many Solutions (Consistent System): If $|A|=0$ AND $(\text{adj } A)B = 0$ (i.e., all elements of the resulting column matrix are zero), the system has infinitely many solutions. This is the condition relevant to our problem.

Step 1: Represent the System in Matrix Form

First, let's write the given system of equations in the standard matrix form $AX=B$: $x+4 y-z=\lambda$ $7 x+9 y+\mu z=-3$ $5 x+y+2 z=-1$

Here, the coefficient matrix $A$, the variable matrix $X$, and the constant matrix $B$ are: $A = \begin{bmatrix} 1 & 4 & -1 \\ 7 & 9 & \mu \\ 5 & 1 & 2 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} \lambda \\ -3 \\ -1 \end{bmatrix}$

Step 2: Apply the Condition $|A|=0$ to Find $\mu$

For the system to have infinitely many solutions, the first necessary condition is that the determinant of the coefficient matrix $A$ must be zero. This condition ensures that the matrix $A$ is singular, meaning it does not have an inverse, which is a prerequisite for either no solution or infinitely many solutions.

Let's calculate the determinant of $A$: $|A| = \begin{vmatrix} 1 & 4 & -1 \\ 7 & 9 & \mu \\ 5 & 1 & 2 \end{vmatrix}$ Expanding the determinant along the first row for ease of calculation: $|A| = 1 \cdot (9 \cdot 2 - \mu \cdot 1) - 4 \cdot (7 \cdot 2 - \mu \cdot 5) + (-1) \cdot (7 \cdot 1 - 9 \cdot 5)$ $|A| = 1 \cdot (18 - \mu) - 4 \cdot (14 - 5\mu) - 1 \cdot (7 - 45)$ $|A| = (18 - \mu) - (56 - 20\mu) - 1 \cdot (-38)$ $|A| = 18 - \mu - 56 + 20\mu + 38$ Now, combine the constant terms and the terms involving $\mu$: $|A| = (18 - 56 + 38) + (-\mu + 20\mu)$ $|A| = (56 - 56) + 19\mu$ $|A| = 0 + 19\mu$ Since for infinitely many solutions, we must have $|A|=0$: $19\mu = 0$ $\Rightarrow \mu = 0$ Thus, we have found the value of $\mu$.

Step 3: Apply the Condition $(\text{adj } A)B = 0$ to Find $\lambda$

Now that we have $\mu=0$, we can update our coefficient matrix $A$: $A = \begin{bmatrix} 1 & 4 & -1 \\ 7 & 9 & 0 \\ 5 & 1 & 2 \end{bmatrix}$ The second condition for infinitely many solutions is $(\text{adj } A)B = 0$. This condition ensures that the system is consistent even when $|A|=0$. If this condition is not met, the system would have no solution.

To apply this condition, we first need to find the adjoint of matrix $A$. The adjoint matrix is the transpose of the cofactor matrix. Let's calculate the cofactors $C_{ij}$ for each element $a_{ij}$ of $A$:

• $C_{11} = \begin{vmatrix} 9 & 0 \\ 1 & 2 \end{vmatrix} = (9)(2) - (0)(1) = 18 - 0 = 18$
• $C_{12} = -\begin{vmatrix} 7 & 0 \\ 5 & 2 \end{vmatrix} = -((7)(2) - (0)(5)) = -(14 - 0) = -14$
• $C_{13} = \begin{vmatrix} 7 & 9 \\ 5 & 1 \end{vmatrix} = (7)(1) - (9)(5) = 7 - 45 = -38$
• $C_{21} = -\begin{vmatrix} 4 & -1 \\ 1 & 2 \end{vmatrix} = -((4)(2) - (-1)(1)) = -(8 + 1) = -9$
• $C_{22} = \begin{vmatrix} 1 & -1 \\ 5 & 2 \end{vmatrix} = (1)(2) - (-1)(5) = 2 + 5 = 7$
• $C_{23} = -\begin{vmatrix} 1 & 4 \\ 5 & 1 \end{vmatrix} = -((1)(1) - (4)(5)) = -(1 - 20) = -(-19) = 19$
• $C_{31} = \begin{vmatrix} 4 & -1 \\ 9 & 0 \end{vmatrix} = (4)(0) - (-1)(9) = 0 + 9 = 9$
• $C_{32} = -\begin{vmatrix} 1 & -1 \\ 7 & 0 \end{vmatrix} = -((1)(0) - (-1)(7)) = -(0 + 7) = -7$
• $C_{33} = \begin{vmatrix} 1 & 4 \\ 7 & 9 \end{vmatrix} = (1)(9) - (4)(7) = 9 - 28 = -19$

Now, form the cofactor matrix $C$: $C = \begin{bmatrix} 18 & -14 & -38 \\ -9 & 7 & 19 \\ 9 & -7 & -19 \end{bmatrix}$ The adjoint matrix $\text{adj } A$ is the transpose of the cofactor matrix $C^T$: $\text{adj } A = C^T = \begin{bmatrix} 18 & -9 & 9 \\ -14 & 7 & -7 \\ -38 & 19 & -19 \end{bmatrix}$ Now, we apply the condition $(\text{adj } A)B = 0$: $\begin{bmatrix} 18 & -9 & 9 \\ -14 & 7 & -7 \\ -38 & 19 & -19 \end{bmatrix} \begin{bmatrix} \lambda \\ -3 \\ -1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$ Performing the matrix multiplication, each row of the resulting matrix must be zero. We can use any one row to find $\lambda$. Let's use the first row: $18\lambda + (-9)(-3) + 9(-1) = 0$ $18\lambda + 27 - 9 = 0$ $18\lambda + 18 = 0$ $18\lambda = -18$ $\lambda = -1$ To ensure consistency and verify our calculation, let's check with the second row as well: $-14\lambda + 7(-3) + (-7)(-1) = 0$ $-14\lambda - 21 + 7 = 0$ $-14\lambda - 14 = 0$ $-14\lambda = 14$ $\lambda = -1$ Both rows yield the same value for $\lambda$, confirming our result.

Step 4: Calculate the Required Expression

We have found the values $\mu = 0$ and $\lambda = -1$. The problem asks for the value of $(2\mu + 3\lambda)$. Substitute the values: $2\mu + 3\lambda = 2(0) + 3(-1)$ $= 0 - 3$ $= -3$

The final answer is $\boxed{-3}$.

Key Takeaways and Tips:

• Systematic Approach: Always start by identifying the type of solution required (unique, no solution, infinitely many) and the corresponding mathematical conditions.
• Determinant Calculation: Be extremely careful with signs and arithmetic when calculating determinants. A single error can propagate through the entire solution.
• Adjoint Matrix: Remember that the adjoint matrix is the transpose of the cofactor matrix. A common mistake is to forget the transpose.
• Consistency Check: When solving for a variable (like $\lambda$ here) using the $(\text{adj } A)B = 0$ condition, if you have multiple equations, use at least two to cross-check your result. If they give different values, there's a calculation error.
• Cramer's Rule Alternative: For infinitely many solutions, an equivalent condition to $(\text{adj } A)B = 0$ is that all determinants $|A_x|, |A_y|, |A_z|$ (where columns of $A$ are replaced by $B$) must also be zero, in addition to $|A|=0$. This can sometimes be a quicker way to find $\lambda$ after $\mu$ is determined.