<p>For what values of <em>x</em> the matrix
\[\begin{bmatrix} 3+x & 5 & 2 \\ 1 & 7+x & 6 \\ 2 & 5 & 3+x \end{bmatrix}\] has the rank 2?</p>
Step-by-Step Solution
Key Concept: A matrix has rank 2 when its determinant equals zero (making it singular) but at least one 2×2 minor is non-zero. Find det(A) = 0, then verify 2×2 minors aren't all zero for the solutions.
<p><strong>Step 1:</strong> For rank to be less than 3, the determinant must be zero. Expand det(A):</p><p>det(A) = (3+x)[(7+x)(3+x) - 30] - 5[1(3+x) - 12] + 2[5 - 2(7+x)]</p><p>= (3+x)[21 + 7x + 3x + x² - 30] - 5[3 + x - 12] + 2[5 - 14 - 2x]</p><p>= (3+x)[x² + 10x - 9] - 5[x - 9] + 2[-9 - 2x]</p><p>= x³ + 10x² - 9x + 3x² + 30x - 27 - 5x + 45 - 18 - 4x</p><p>= x³ + 13x² + 12x = x(x² + 13x + 12) = x(x + 1)(x + 12)</p><p><strong>Step 2:</strong> Setting det(A) = 0 gives: x = 0, -1, -12</p><p><strong>Step 3:</strong> Verify rank = 2 (not rank < 2) by checking at least one 2×2 minor is non-zero:</p><p>• For x = 0: Minor = |3 5; 1 7| = 21 - 5 = 16 ≠ 0 ✓</p><p>• For x = -1: Minor = |2 5; 1 6| = 12 - 5 = 7 ≠ 0 ✓</p><p>• For x = -12: Minor = |-9 5; 1 -5| = 45 - 5 = 40 ≠ 0 ✓</p><p>∴ <strong>Answer: x = 0, -1, -12</strong></p>
Correct Answer: x = 0, -1, -12