Let the determinant of a square matrix A of order m be m - n, where m and n satisfy 4m + n = 22 and 17m + 4n = 93. If det (n adj (adj (mA))) = 3<sup>a</sup> 5<sup>b</sup> 6<sup>c</sup>. Then a + b + c is equal to:
Step-by-Step Solution
Key Concept: Solve the system of linear equations for m and n, then use the property det(adj(adj(M))) = (det M)^(m-1)^2 and det(kM) = k^m det(M).
Given 4m + n = 22 and 17m + 4n = 93. Multiplying the first by 4 gives 16m + 4n = 88. Subtracting this from the second gives m = 5. Then 4(5) + n = 22, so n = 2. The order of matrix A is m = 5. The determinant of A is |A| = m - n = 5 - 2 = 3. We need to find det(n adj(adj(mA))). Since n = 2 and m = 5, this is det(2 adj(adj(5A))). Using det(kA) = k^m det(A), we have det(2 adj(adj(5A))) = 2^5 det(adj(adj(5A))). Using det(adj(adj(M))) = (det M)^(m-1)^2, we have det(adj(adj(5A))) = (det(5A))^(5-1)^2 = (5^5 * det A)^16 = (5^5 * 3)^16 = 5^80 * 3^16. Thus, the expression is 2^5 * 5^80 * 3^16. This is 3^16 * 5^80 * 2^5. Since 6^c = (2*3)^c = 2^c * 3^c, we have 3^16 * 5^80 * 2^5 = 3^a * 5^b * 2^c * 3^c = 3^(a+c) * 5^b * 2^c. Comparing powers, c = 5, b = 80, a+c = 16 => a = 11. Then a + b + c = 11 + 80 + 5 = 96. Wait, re-evaluating: 3^a * 5^b * 6^c = 3^a * 5^b * 2^c * 3^c = 3^(a+c) * 5^b * 2^c. We have 3^16 * 5^80 * 2^5. So c=5, b=80, a+c=16 => a=11. a+b+c = 11+80+5 = 96.
Correct Answer: 3