Given `2x + 4y + z = 1, lambdax + 2y + z = 2, x+ y - lambdaz = 3,` then one of the value of a such that the given system of equations has no solution, is

Answer

To find the value of \( \lambda \) such that the given system of equations has no solution, we need to analyze the determinant of the coefficients of the system. The system of equations is:

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

We can express this system in matrix form as follows:

\[ \begin{bmatrix} 2 & 4 & 1 \\ \lambda & 2 & 1 \\ 1 & 1 & -\lambda \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \]

### Step 1: Set up the determinant

To find when the system has no solution, we need to calculate the determinant of the coefficient matrix and set it equal to zero:

\[ D = \begin{vmatrix} 2 & 4 & 1 \\ \lambda & 2 & 1 \\ 1 & 1 & -\lambda \end{vmatrix} \]

### Step 2: Calculate the determinant

Using the determinant formula for a 3x3 matrix, we can expand it as follows:

\[ D = 2 \begin{vmatrix} 2 & 1 \\ 1 & -\lambda \end{vmatrix} - 4 \begin{vmatrix} \lambda & 1 \\ 1 & -\lambda \end{vmatrix} + 1 \begin{vmatrix} \lambda & 2 \\ 1 & 1 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \( \begin{vmatrix} 2 & 1 \\ 1 & -\lambda \end{vmatrix} = 2(-\lambda) - 1(1) = -2\lambda - 1 \)

2. \( \begin{vmatrix} \lambda & 1 \\ 1 & -\lambda \end{vmatrix} = \lambda(-\lambda) - 1(1) = -\lambda^2 - 1 \)

3. \( \begin{vmatrix} \lambda & 2 \\ 1 & 1 \end{vmatrix} = \lambda(1) - 2(1) = \lambda - 2 \)

Substituting these back into the determinant expression:

\[ D = 2(-2\lambda - 1) - 4(-\lambda^2 - 1) + 1(\lambda - 2) \]

### Step 3: Simplify the determinant

Now we simplify:

\[ D = -4\lambda - 2 + 4\lambda^2 + 4 + \lambda - 2 \]

Combining like terms:

\[ D = 4\lambda^2 + (-4\lambda + \lambda) + (-2 + 4 - 2) \]
\[ D = 4\lambda^2 - 3\lambda + 0 \]

### Step 4: Set the determinant to zero

To find the values of \( \lambda \) for which the system has no solution, we set the determinant equal to zero:

\[ 4\lambda^2 - 3\lambda = 0 \]

Factoring out \( \lambda \):

\[ \lambda(4\lambda - 3) = 0 \]

### Step 5: Solve for \( \lambda \)

Setting each factor to zero gives us:

1. \( \lambda = 0 \)
2. \( 4\lambda - 3 = 0 \) which gives \( \lambda = \frac{3}{4} \)

Thus, the values of \( \lambda \) for which the system has no solution are \( \lambda = 0 \) and \( \lambda = \frac{3}{4} \).

### Final Answer:

One of the values of \( \lambda \) such that the given system of equations has no solution is \( \lambda = 0 \) or \( \lambda = \frac{3}{4} \).

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