Consider the system of equations `x+2y+3z=6,`
`4x+5y+6z=lambda`,
`7x+8y+9z=24`.
Then, the value of `lambda` for which the system has infinite solutions is

Options

A. 15
B. 20
C. 16
D. 30

Answer

To find the value of \(\lambda\) for which the given system of equations has infinitely many solutions, we can use the concept of determinants from linear algebra. The system of equations is:

1. \(x + 2y + 3z = 6\)
2. \(4x + 5y + 6z = \lambda\)
3. \(7x + 8y + 9z = 24\)

### Step 1: Write the system in matrix form

We can express the system in the form \(AX = B\), where:

\[ A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 6 \\ \lambda \\ 24 \end{bmatrix} \]

### Step 2: Calculate the determinant of matrix \(A\)

To find the determinant of \(A\), we can use the formula for the determinant of a \(3 \times 3\) matrix:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix \(A\):

\[ \text{det}(A) = 1(5 \cdot 9 - 6 \cdot 8) - 2(4 \cdot 9 - 6 \cdot 7) + 3(4 \cdot 8 - 5 \cdot 7) \]

Calculating each term:

- \(5 \cdot 9 - 6 \cdot 8 = 45 - 48 = -3\)
- \(4 \cdot 9 - 6 \cdot 7 = 36 - 42 = -6\)
- \(4 \cdot 8 - 5 \cdot 7 = 32 - 35 = -3\)

Thus, substituting back:

\[ \text{det}(A) = 1(-3) - 2(-6) + 3(-3) = -3 + 12 - 9 = 0 \]

### Step 3: Set up the condition for infinitely many solutions

For the system to have infinitely many solutions, the determinant of the coefficient matrix \(A\) must be zero, which we found to be true. Now we need to ensure that the modified determinants (using the constants from \(B\)) are also zero.

### Step 4: Calculate the determinant \(\Delta_1\)

We replace the first column of \(A\) with \(B\):

\[ \Delta_1 = \begin{vmatrix} 6 & 2 & 3 \\ \lambda & 5 & 6 \\ 24 & 8 & 9 \end{vmatrix} \]

Calculating \(\Delta_1\):

\[ \Delta_1 = 6(5 \cdot 9 - 6 \cdot 8) - 2(\lambda \cdot 9 - 6 \cdot 24) + 3(\lambda \cdot 8 - 5 \cdot 24) \]

Substituting the values we calculated earlier:

\[ = 6(-3) - 2(\lambda \cdot 9 - 144) + 3(\lambda \cdot 8 - 120) \]

\[ = -18 - 2(9\lambda - 144) + 3(8\lambda - 120) \]

Expanding:

\[ = -18 - 18\lambda + 288 + 24\lambda - 360 \]

Combining like terms:

\[ = (24\lambda - 18\lambda) + (-18 + 288 - 360) = 6\lambda - 90 \]

### Step 5: Set \(\Delta_1 = 0\)

For infinitely many solutions, we set \(\Delta_1 = 0\):

\[ 6\lambda - 90 = 0 \]

Solving for \(\lambda\):

\[ 6\lambda = 90 \implies \lambda = 15 \]

### Final Answer

Thus, the value of \(\lambda\) for which the system has infinitely many solutions is:

\[ \lambda = 15 \]