If the system of equation

$x - y + z = 4$

$x - 2y - 2z = 9$

$2x + y + \lambda z = 1$

has a unique solution, then

A system has a unique solution when there is exactly one set of values $(x, y, z)$ that satisfies all three equations. For a system to have a unique solution, the determinant of the coefficient matrix must be non-zero.

If the determinant equals zero, the system either has no solution or infinite solutions (not unique).

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From the three equations, the coefficient matrix is:

$A = \begin{pmatrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & \lambda \end{pmatrix}$

Row 1: coefficients from $x - y + z = 4$ give $(1, -1, 1)$

Row 2: coefficients from $x - 2y - 2z = 9$ give $(1, -2, -2)$

Row 3: coefficients from $2x + y + \lambda z = 1$ give $(2, 1, \lambda)$

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Using the first row expansion:

$\det(A) = 1 \times \begin{vmatrix} -2 & -2 \\ 1 & \lambda \end{vmatrix} - (-1) \times \begin{vmatrix} 1 & -2 \\ 2 & \lambda \end{vmatrix} + 1 \times \begin{vmatrix} 1 & -2 \\ 2 & 1 \end{vmatrix}$

Calculate each $2 \times 2$ determinant:

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

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

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

Combining these:

$\det(A) = 1(-2\lambda + 2) + 1(\lambda + 4) + 1(5)$

$= -2\lambda + 2 + \lambda + 4 + 5$

$= -\lambda + 11$

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For a unique solution:

$\det(A) \neq 0$

$-\lambda + 11 \neq 0$

$\lambda \neq 11$

Therefore, the system has a unique solution when $\lambda \neq 11$.