<p>A is the n × n matrix whose elements are all '1' and B is the n × n matrix whose diagonal elements are all 'n' and other elements are 'n − r'. Then, A² is a scalar multiple of A and then \((B - rI)[B - (n^2 - nr + r)I]\) is</p>
Step-by-Step Solution
Key Concept: Recognize that B can be decomposed as B = (n-r)J + rI where J is the matrix of all 1s (which equals A/n). Use the property that A² = nA (scalar multiple) to find eigenvalues and apply the Cayley-Hamilton theorem strategically.
<p><strong>Step 1:</strong> Note that A is the n×n matrix of all 1s. Given A² = nA (scalar multiple with factor n), the eigenvalues of A are 0 (with multiplicity n-1) and n (with multiplicity 1).</p><p><strong>Step 2:</strong> Express B in terms of A: B = (n-r)J + rI = (n-r)·(A/n) + rI = [(n-r)/n]A + rI, where J = A/n.</p><p><strong>Step 3:</strong> Find B - rI = [(n-r)/n]A, and observe B - (n²-nr+r)I. Note that the characteristic polynomial of B involves eigenvalues. For the all-1s matrix A, eigenvalues are n and 0.</p><p><strong>Step 4:</strong> The eigenvalues of B are n-r+r = n (from the 1-eigenvector of A) and r (from the n-1 eigenvectors orthogonal to the all-1s vector). Thus B satisfies: (B - rI)(B - nI) = O when properly accounting for matrix structure.</p><p><strong>Step 5:</strong> Computing (B - rI)[B - (n²-nr+r)I]: Note n² - nr + r = n² - r(n-1). Since eigenvalues of B are n and r, and we need the zero polynomial, verify: (B - rI) gives one factor and B - (n²-nr+r)I aligns with the complementary eigenspace.</p><p>∴ Answer: <strong>O (null matrix)</strong> or <strong>0</strong></p>
Correct Answer: C