Step-by-step Solution

1. Understand the given information:

• M is a 3x3 invertible matrix with real entries.
• I is the 3x3 identity matrix.
• The core equation is M⁻¹ = adj(adj M). (Let's call this Equation 1)

2. Recall key properties of matrices: For an n x n invertible matrix A:

• adj(adj A) = (det A)ⁿ⁻² A
• det(A⁻¹) = 1 / det(A)
• det(kA) = kⁿ det(A) for a scalar k.

Since M is a 3x3 matrix (n=3), the property for the adjugate of the adjugate becomes: adj(adj M) = (det M)³⁻² M = (det M) M

3. Simplify the given equation: Substitute the simplified expression for adj(adj M) into Equation 1: M⁻¹ = (det M) M (Let's call this Equation 2)

This is the central relationship derived from the problem statement. We will use it to test the given options.

4. Evaluate Option B: det M = 1 Take the determinant of both sides of Equation 2: det(M⁻¹) = det((det M) M)

Let k = det M. The equation becomes: det(M⁻¹) = det(kM)

Using the properties from step 2:

• The left side is det(M⁻¹) = 1 / det M = 1/k.
• The right side is det(kM) = k³ det(M) = k³ * k = k⁴ (since M is 3x3).

Equating the two sides: 1/k = k⁴ 1 = k⁵

So, (det M)⁵ = 1. Since M has real entries, det M must be a real number. The only real solution to x⁵ = 1 is x = 1. Therefore, det M = 1. Conclusion: Statement B is ALWAYS TRUE.

5. Evaluate Option C: M² = I Now that we have established det M = 1, we can substitute this back into Equation 2: M⁻¹ = (1) M M⁻¹ = M

To eliminate the inverse, we can pre-multiply (or post-multiply) both sides by M: M * M⁻¹ = M * M I = M²

Conclusion: Statement C is ALWAYS TRUE.

6. Evaluate Option D: (adj M)² = I Recall the formula for the inverse of a matrix: M⁻¹ = (1 / det M) adj(M)

Since we know det M = 1, this simplifies to: M⁻¹ = adj(M)

From step 5, we also know that M⁻¹ = M. Therefore, we can conclude that adj(M) = M.

Now let's consider the expression (adj M)²: (adj M)² = M²

From step 5, we found that M² = I. So, (adj M)² = I.

Conclusion: Statement D is ALWAYS TRUE.

7. Evaluate Option A: M = I We have the condition M² = I. This does not necessarily mean M = I. For example, a rotation matrix or a reflection matrix can satisfy this. Let's find a counterexample. Consider the matrix: M = [[-1, 0, 0], [0, -1, 0], [0, 0, 1]]

Let's check if this M satisfies the initial conditions.

• It is a 3x3 matrix with real entries.
• det M = (-1)(-1)(1) = 1, so it's invertible.
• M² = [[(-1)² + 0 + 0, 0, 0], [0, (-1)² + 0, 0], [0, 0, 1²]] = [[1, 0, 0], [0, 1, 0], [0, 0, 1]] = I.

Since det M = 1 and M² = I, all the derived conditions (det M = 1, M⁻¹ = M, adj(M) = M) are satisfied. This means this M is a valid solution. However, M ≠ I. Therefore, the statement M = I is not always true.

Conclusion: Statement A is NOT ALWAYS TRUE.

Summary

The statements that are always true are B, C, and D.