The equations `lambdax-y=2,2x-3y=-lambda and 3x-2y=-1` are consistent for

Options

A. `lambda=-4`
B. `lambda=-1,4`
C. `lambda=-1`
D. `lambda=1,-4`

Answer

To determine the values of \(\lambda\) for which the given equations are consistent, we need to find the determinant of the coefficients of the equations and set it equal to zero.

The equations are:
1. \(\lambda x - y = 2\)
2. \(2x - 3y = -\lambda\)
3. \(3x - 2y = -1\)

### Step 1: Write the equations in standard form
We can rearrange the equations as follows:
1. \(\lambda x - y - 2 = 0\)
2. \(2x - 3y + \lambda = 0\)
3. \(3x - 2y + 1 = 0\)

### Step 2: Form the coefficient matrix
The coefficient matrix \(A\) for the system of equations is:
\[ A = \begin{pmatrix} \lambda & -1 & -2 \\ 2 & -3 & \lambda \\ 3 & -2 & 1 \end{pmatrix} \]

### Step 3: Calculate the determinant of the matrix
The determinant of the matrix \(A\) can be calculated using the formula for the determinant of a 3x3 matrix:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where the matrix is:
\[ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \]

For our matrix:
\[ \text{det}(A) = \lambda \cdot (-3 \cdot 1 - \lambda \cdot (-2)) - (-1) \cdot (2 \cdot 1 - \lambda \cdot 3) + (-2) \cdot (2 \cdot (-2) - (-3) \cdot 3) \]

Calculating each term:
1. \(\lambda \cdot (-3 + 2\lambda) = -3\lambda + 2\lambda^2\)
2. \(-(-1)(2 + 3\lambda) = 2 + 3\lambda\)
3. \(-2( -4 + 9) = -2 \cdot 5 = -10\)

Putting it all together:
\[ \text{det}(A) = -3\lambda + 2\lambda^2 + 2 + 3\lambda - 10 \]
\[ = 2\lambda^2 - 8 \]

### Step 4: Set the determinant equal to zero
To find the values of \(\lambda\) for which the equations are consistent, we set the determinant equal to zero:
\[ 2\lambda^2 - 8 = 0 \]

### Step 5: Solve for \(\lambda\)
Dividing the equation by 2:
\[ \lambda^2 - 4 = 0 \]
Factoring:
\[ (\lambda - 2)(\lambda + 2) = 0 \]

Thus, we find:
\[ \lambda = 2 \quad \text{or} \quad \lambda = -2 \]

### Conclusion
The equations are consistent for \(\lambda = 2\) and \(\lambda = -2\).