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Let A = [a_ij] be a 3x3 matrix with |A| = 3. Find the value of |3 adj(adj(3A))|.

1. 3¹⁵
2. 3¹⁹
3. 3²²
4. 3²⁵

Correct answer: 3²⁵

Solution

n=3. |3A| = 3³ * |A| = 27*3 = 81 = 3⁴. adj(3A) is a 3x3 matrix with |adj(3A)| = |3A|^(n-1) = (3⁴)² = 3⁸. adj(adj(3A)) is a 3x3 matrix with |adj(adj(3A))| = |adj(3A)|^(n-1) = (3⁸)² = 3¹⁶. 3*adj(adj(3A)) means scalar 3 times the matrix, so |3*adj(adj(3A))| = 3³ * |adj(adj(3A))| = 3³ * 3¹⁶ = 3¹⁹. Hmm, that gives 3¹⁹. But wait: let me reconsider the expression |3 adj(adj(3A))| — is the 3 a scalar multiplier of the whole matrix? |k*M| = kⁿ * |M| with k=3, n=3: |3*adj(adj(3A))| = 3³ * 3¹⁶ = 3¹⁹. But the answer option 3²⁵ — let me recheck. |adj(adj(M))| for n x n: |adj(M)| = |M|^(n-1). |adj(adj(M))| = |adj(M)|^(n-1) = (|M|^(n-1))^(n-1) = |M|^((n-1)²). For n=3: |adj(adj(3A))| = |3A|^((3-1)²) = |3A|⁴ = (3⁴)⁴ = 3¹⁶. Then |3 adj(adj(3A))| = 3³ * 3¹⁶ = 3¹⁹. So answer is 3¹⁹.

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