Step 1: Concept Definition:
Matrix rank is the dimension of the column space (or row space). A square matrix has a rank less than its dimension if and only if it is singular (determinant is zero).
Step 2: Methodology:
For a 3x3 matrix to have rank 2, its determinant must be 0, and at least one 2x2 minor must be non-zero.
1. Verify rank is at least 2 by finding a non-zero 2x2 minor.
2. Equate the 3x3 matrix determinant to zero and solve for \( \lambda \).
Step 3: Detailed Procedure:
Consider the matrix \( A \):
\[A = \begin{pmatrix}
1 & 2 & 3 \\
4 & 5 & 6 \\
0 & 7 & \lambda
\end{pmatrix}\]
1. Rank Verification (\(\geq 2\)):
Calculate the top-left 2x2 minor:
\[\det \begin{pmatrix} 1 & 2 \\ 4 & 5 \end{pmatrix}
= (1)(5) - (2)(4) = 5 - 8 = -3\]
This non-zero minor confirms the matrix rank is at least 2.
2. Determinant Calculation for Rank = 2:
\[\det(A) =
1 \begin{vmatrix} 5 & 6 \\ 7 & \lambda \end{vmatrix}
- 2 \begin{vmatrix} 4 & 6 \\ 0 & \lambda \end{vmatrix}
+ 3 \begin{vmatrix} 4 & 5 \\ 0 & 7 \end{vmatrix}\]
\[= 1(5\lambda - 42) - 2(4\lambda - 0) + 3(28 - 0)\]
\[= (5\lambda - 42) - 8\lambda + 84\]
\[= -3\lambda + 42\]
Set determinant to zero:
\[-3\lambda + 42 = 0 \quad \implies \quad \lambda = 14\]
Step 4: Conclusion:
The value of \( \lambda \) yielding a rank of 2 for the matrix is:
\[\boxed{14}\]