Let M be a `3xx3` invertible matrix with real entries and let I denote the `3xx` matrix. If `M^(-1)` adj(adjM). Then which of the following statements, is/are ALWAYS TRUE ?

Options

A. `M=I`
B. `"det"M=1`
C. `M^(2)=I`
D. `(adj M)^(2)=I`

Answer

To solve the problem, we need to analyze the given equation \( M^{-1} = \text{adj}(\text{adj}(M)) \) and derive implications from it.

### Step-by-Step Solution:

**Step 1: Understanding the adjoint of a matrix**
The adjoint of a matrix \( M \), denoted as \( \text{adj}(M) \), is defined such that:
\[ M \cdot \text{adj}(M) = \det(M) \cdot I \]
where \( I \) is the identity matrix.

**Step 2: Applying the property of adjoint**
From the property of the adjoint, we can write:
\[ \text{adj}(M) = \det(M) \cdot M^{-1} \]
Thus, substituting this into our equation, we have:
\[ M^{-1} = \text{adj}(\text{adj}(M)) \]

**Step 3: Finding the adjoint of the adjoint**
Using the property of the adjoint again, we know:
\[ \text{adj}(\text{adj}(M)) = \det(M)^{n-1} \cdot M \]
For a \( 3 \times 3 \) matrix, \( n = 3 \), so:
\[ \text{adj}(\text{adj}(M)) = \det(M)^2 \cdot M \]

**Step 4: Equating both sides**
Now, substituting back into our equation:
\[ M^{-1} = \det(M)^2 \cdot M \]

**Step 5: Multiplying both sides by \( M \)**
Multiplying both sides by \( M \) gives:
\[ I = \det(M)^2 \cdot M^2 \]

**Step 6: Rearranging the equation**
Rearranging this, we find:
\[ M^2 = \frac{1}{\det(M)^2} I \]
This implies that \( M^2 \) is a scalar multiple of the identity matrix.

**Step 7: Analyzing the implications**
Since \( M^2 = \frac{1}{\det(M)^2} I \), we can conclude that \( M \) must be an invertible matrix whose square is a scalar multiple of the identity matrix. This means that \( M \) could be either a rotation or a reflection matrix, depending on the determinant.

### Conclusion:
From the above steps, we can derive that:
- \( M^2 = I \) if \( \det(M) = \pm 1 \).
- The eigenvalues of \( M \) must be \( \pm 1 \).