If the system of equationsx-3y+5z=3 x-2y+4z=4 2x-7y+λz=5has infinite…

The correct answer is Option (4) → 11

The given system of equations is:

$x - 3y + 5z = 3$

$x - 2y + 4z = 4$

$2x - 7y + \lambda z = 5$

This can be written in matrix form as:

$A \cdot X = B$

Where

$A = \begin{bmatrix} 1 & -3 & 5 \\ 1 & -2 & 4 \\ 2 & -7 & \lambda \end{bmatrix}$

$X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 3 \\ 4 \\ 5 \end{bmatrix}$

For the system to have infinite solutions, the coefficient matrix $A$ must be singular, i.e.,

$|A| = 0$

Compute the determinant of $A$:

$|A| = \begin{vmatrix} 1 & -3 & 5 \\ 1 & -2 & 4 \\ 2 & -7 & \lambda \end{vmatrix}$

Expand along the first row:

$= 1 \cdot \begin{vmatrix} -2 & 4 \\ -7 & \lambda \end{vmatrix} - (-3) \cdot \begin{vmatrix} 1 & 4 \\ 2 & \lambda \end{vmatrix} + 5 \cdot \begin{vmatrix} 1 & -2 \\ 2 & -7 \end{vmatrix}$

$= 1 \cdot (-2 \cdot \lambda - (-7 \cdot 4)) + 3 \cdot (1 \cdot \lambda - 2 \cdot 4) + 5 \cdot (1 \cdot -7 - 2 \cdot -2)$

$= 1 \cdot (-2\lambda + 28) + 3 \cdot (\lambda - 8) + 5 \cdot (-7 + 4)$

$= -2\lambda + 28 + 3\lambda - 24 - 15$

$= \lambda - 11$

Set determinant to zero for infinite solutions:

$\lambda - 11 = 0$

$\lambda = 11$