Question

If the rank of matrix \[ \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 0 & 7 & \lambda \end{pmatrix} \] is 2, then the value of \(\lambda\) is:

• A. 10
• B. 12
• C. 14
• D. 16

Hint:

For a 3x3 matrix, the condition "rank = 2" is almost always equivalent to "determinant = 0". It's a quick and reliable way to solve such problems in an exam setting.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The rank of a matrix is the dimension of the vector space spanned by its columns (or rows). For a square matrix (like this 3x3), the rank is less than its dimension if and only if the matrix is singular, which means its determinant is zero.
Step 2: Key Formula or Approach:
A 3x3 matrix has rank 2 if its determinant is 0, provided that at least one of its 2x2 minors is non-zero. 1. Check that the rank is at least 2 by finding a non-zero 2x2 minor. 2. Set the determinant of the 3x3 matrix equal to zero and solve for \( \lambda \).
Step 3: Detailed Explanation:
Let the matrix be \( A \): \[ A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 0 & 7 & \lambda \end{pmatrix} \] 1. Check for Rank \(\geq 2\):
Consider the top-left 2x2 minor: \[ \det \begin{pmatrix} 1 & 2 \\ 4 & 5 \end{pmatrix} = (1)(5) - (2)(4) = 5 - 8 = -3 \] Since this minor is non-zero, the rank of the matrix is at least 2.

2. Set Determinant to Zero for Rank = 2:
\[ \det(A) = 1 \begin{vmatrix} 5 & 6 \\ 7 & \lambda \end{vmatrix} - 2 \begin{vmatrix} 4 & 6 \\ 0 & \lambda \end{vmatrix} + 3 \begin{vmatrix} 4 & 5 \\ 0 & 7 \end{vmatrix} \] \[ = 1(5\lambda - 42) - 2(4\lambda - 0) + 3(28 - 0) \] \[ = (5\lambda - 42) - 8\lambda + 84 \] \[ = -3\lambda + 42 \] Setting determinant = 0: \[ -3\lambda + 42 = 0 \quad \implies \quad \lambda = 14 \] Step 4: Final Answer:
The value of \( \lambda \) for which the rank of the matrix is 2 is: \[ \boxed{14} \]