<p>Let \(A\) be an \(m \times m\) matrix with all elements equal to 1 such that \(A^n = 16^{17}\, A\), \(m, n \in N\). Find the sum of possible values of \(n\).</p>
Step-by-Step Solution
Key Concept: For an m×m matrix of all 1's, A has eigenvalue m with multiplicity 1 and eigenvalue 0 with multiplicity (m-1). Thus A^n = m^(n-1)·A, which must equal 16^17·A, giving m^(n-1) = 16^17 = 2^68.
<p><strong>Step 1:</strong> For an m×m matrix with all elements equal to 1, denote it as A = J (all-ones matrix).</p><p><strong>Step 2:</strong> Find A². Each element of A² equals the dot product of a row with a column: A² = m·A (since each row sums to m ones).</p><p><strong>Step 3:</strong> By induction, A^n = m^(n-1)·A for n ≥ 1.</p><p><strong>Step 4:</strong> Given condition: A^n = 16^17·A, so m^(n-1)·A = 16^17·A.</p><p><strong>Step 5:</strong> Since A ≠ 0, we have m^(n-1) = 16^17 = (2^4)^17 = 2^68.</p><p><strong>Step 6:</strong> Therefore m must be a power of 2. Let m = 2^k where k ≥ 1.</p><p><strong>Step 7:</strong> Then 2^(k(n-1)) = 2^68, so k(n-1) = 68.</p><p><strong>Step 8:</strong> Find divisors of 68: 68 = 4 × 17, so divisors are {1, 2, 4, 17, 34, 68}.</p><p><strong>Step 9:</strong> For each divisor d of 68, we have k = d and n-1 = 68/d, giving n = 1 + 68/d.</p><p><strong>Step 10:</strong> Possible values: n ∈ {69, 35, 18, 5, 3, 2}.</p><p><strong>Step 11:</strong> Sum = 69 + 35 + 18 + 5 + 3 + 2 = 132.</p><p>∴ Answer: <strong>132</strong></p>
Correct Answer: 132