Question

If $A=\begin{bmatrix}3&1\\4&2\end{bmatrix}$ and $B_p$ is another matrix defined as.

$B_p = A^P (\underbrace{adj(adj(adj.....(adj}_{p-times} A))))$ where $p \in N$, then identify the correct statement(s):

Answer 1

For any natural number p: • If p is even, then Bₚ = A^(p+1). • If p is odd, then Bₚ = 2·A^(p–1).

Answer 2

Solution:

For any invertible 2×2 matrix A we have:   adj A = det A · A⁻¹.

Thus,   adj A = det A · A⁻¹,   adj(adj A) = det(adj A) · (adj A)⁻¹. For a 2×2 matrix, det(adj A) = det A (since for an n×n matrix, det(adj A) = (det A)^(n–1) and here n = 2). Also, since   adj A = det A · A⁻¹  ⟹  (adj A)⁻¹ = A/(det A), it follows that   adj(adj A) = det A · (A/(det A)) = A.

Therefore the pattern is:   adj¹ A = det A · A⁻¹,   adj² A = A,   adj³ A = adj(adj² A) = adj A = det A · A⁻¹,   adj⁴ A = A, … and so on.

Let Bₚ be defined as:   Bₚ = Aᵖ · (adj (adj ( … (adj A)))  with p iterations of adj.

Thus: • When p is even, the p-fold adjugate returns A. So,   Bₚ = Aᵖ · A = A^(p+1).

• When p is odd, the p-fold adjugate returns (det A) · A⁻¹. So,   Bₚ = Aᵖ · (det A · A⁻¹) = (det A) · A^(p–1).

Now, for the given matrix   A = [ [3, 1], [4, 2] ], we compute:   det A = 3×2 – 1×4 = 6 – 4 = 2.

Therefore, the answer is:   • For even p: Bₚ = A^(p+1).   • For odd p: Bₚ = 2 · A^(p–1).

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Minimal Explanation: Using for 2×2 matrices: adj A = (det A) A⁻¹, so that adj² A = A and adj³ A = adj A. This gives:  if p is even, Bₚ = Aᵖ·A = A^(p+1);  if p is odd, Bₚ = Aᵖ·(det A·A⁻¹) = (det A)·A^(p–1), with det A = 2 for the given matrix.