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A is a square matrix of order 4 and B = adj(A), where |B| = 27. Find the value of |A^(-1) * adj(3AB)|.
- 3²⁰
- 3²¹
- 3²²
- 3²³
Correct answer: 3²¹
Solution
|B| = |adj A| = |A|³ = 27 => |A| = 3. B = adj(A), so AB = A*adj(A) = |A|*I = 3I (for 4x4 matrix). So 3AB = 3*3I = 9I. adj(3AB) = adj(9I) = 9³ * I (since adj of scalar multiple of I). |adj(9I)| = (9⁴)³... Let us compute directly. |3AB| = |9I| = 9⁴. adj(3AB): for matrix M, adj(M) has det = |M|^(n-1). |adj(3AB)| = |3AB|³ = (9⁴)³ = 9¹² = 3²⁴. A^(-1) = adj(A)/|A| = B/3. |A^(-1)| = 1/|A| = 1/3. |A^(-1) * adj(3AB)| = |A^(-1)| * |adj(3AB)| = (1/3) * 3²⁴ = 3²³.
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