Understanding Conditions for Infinitely Many Solutions

We are given a system of three linear equations with three variables x, y, and z, involving parameters λ and μ:

1. $2x + 3y + 5z = 9$
2. $7x + 3y - 2z = 8$
3. $2x + 3y + \lambda z = \mu$

We need to find the conditions on λ and μ for this system to have infinitely many solutions.

Matrix Representation of the System

A system of linear equations can be written in matrix form $Ax = B$, where A is the coefficient matrix, x is the column vector of variables, and B is the column vector of constants.

Condition for Infinitely Many Solutions

A system of linear equations $Ax = B$ with n variables has infinitely many solutions if and only if the following two conditions are met:

• The determinant of the coefficient matrix A is zero, i.e., $\det(A) = 0$.
• The rank of the coefficient matrix A is equal to the rank of the augmented matrix $[A|B]$, and this rank is less than the number of variables (n).

In our case, n=3. So, we need $\operatorname{rank}(A) = \operatorname{rank}([A|B]) < 3$.

Calculating the Determinant of Matrix A

Let's calculate the determinant of matrix A:

$\det(A) = \begin{vmatrix} 2 & 3 & 5 \\ 7 & 3 & -2 \\ 2 & 3 & \lambda \end{vmatrix}$

Using cofactor expansion along the first row:

$\det(A) = 2 \begin{vmatrix} 3 & -2 \\ 3 & \lambda \end{vmatrix} - 3 \begin{vmatrix} 7 & -2 \\ 2 & \lambda \end{vmatrix} + 5 \begin{vmatrix} 7 & 3 \\ 2 & 3 \end{vmatrix}$

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

$\det(A) = 2(3\lambda + 6) - 3(7\lambda + 4) + 5(21 - 6)$

$\det(A) = 6\lambda + 12 - 21\lambda - 12 + 5(15)$

$\det(A) = -15\lambda + 75$

Finding the Condition on λ

For infinitely many solutions, $\det(A)$ must be equal to 0.

$-15\lambda + 75 = 0$

$-15\lambda = -75$

$\lambda = \frac{-75}{-15} = 5$

So, the first condition for infinitely many solutions is $\lambda = 5$. If $\lambda \neq 5$, $\det(A) \neq 0$, which means the system has a unique solution.

Finding the Condition on μ when λ = 5

Now, let's consider the system when $\lambda = 5$. The system becomes:

1. $2x + 3y + 5z = 9$
2. $7x + 3y - 2z = 8$
3. $2x + 3y + 5z = \mu$

Observe the first and third equations:

Equation 1: $2x + 3y + 5z = 9$

Equation 3: $2x + 3y + 5z = \mu$

The left-hand sides of these two equations are identical. For the system to be consistent (have any solution, including infinitely many), the right-hand sides must also be equal.

Therefore, for the first and third equations to be consistent, we must have $\mu = 9$.

If $\mu \neq 9$ while $\lambda = 5$, the first and third equations become $2x + 3y + 5z = 9$ and $2x + 3y + 5z = \mu$ (where $\mu \neq 9$). This is a contradiction (e.g., $9 = \mu$ is false), meaning the system is inconsistent and has no solution.

If $\mu = 9$ and $\lambda = 5$, the third equation becomes identical to the first equation. The system effectively reduces to:

1. $2x + 3y + 5z = 9$
2. $7x + 3y - 2z = 8$

This is a system of two linear equations in three variables. As long as these two equations are linearly independent, the system will have infinitely many solutions. We can check for linear independence by looking at the coefficients or using row operations on the augmented matrix.

When $\lambda=5$ and $\mu=9$, the augmented matrix is:

$\begin{pmatrix} 2 & 3 & 5 & 9 \\ 7 & 3 & -2 & 8 \\ 2 & 3 & 5 & 9 \end{pmatrix}$

Notice that R3 is identical to R1. Performing R3 → R3 - R1 results in a row of zeros:

$\begin{pmatrix} 2 & 3 & 5 & 9 \\ 7 & 3 & -2 & 8 \\ 0 & 0 & 0 & 0 \end{pmatrix}$

The rank of A (first three columns) is 2 (assuming the first two rows are linearly independent, which they are). The rank of $[A|B]$ is also 2. Since the rank (2) is less than the number of variables (3), the system has infinitely many solutions.

Conclusion on Conditions

The conditions for the system to have infinitely many solutions are $\lambda = 5$ and $\mu = 9$.

Matching with Options

Let's compare our derived conditions with the given options:

• Option 1: λ = 5 and μ ≠ 9 - No solution (inconsistent).
• Option 2: λ = 5 and μ = 9 - Infinitely many solutions (matches our finding).
• Option 3: λ = 9 and μ = 5 - Unique solution ($\det(A) \neq 0$).
• Option 4: λ = 9 and μ ≠ 5 - Unique solution ($\det(A) \neq 0$).

The condition that results in infinitely many solutions is indeed λ = 5 and μ = 9.