Properties of Invertible Matrix Satisfying Double Adjugate Condition

To determine which of the statements are always true, we use the standard properties of adjugate and determinant for an $n \times n$ invertible matrix.

For any $3 \times 3$ invertible matrix $M$, the formula for the double adjugate is given by: $\text{adj}(\text{adj } M) = (\det M)^{3-2} M = (\det M) M$

Given the condition in the problem: $M^{-1} = \text{adj}(\text{adj } M)$

We can substitute the identity derived above into this relation: $M^{-1} = (\det M) M$

Step 1: Determine $\det M$ Taking the determinant on both sides of $M^{-1} = (\det M) M$: $\det(M^{-1}) = \det((\det M) M)$

Since $M$ is a $3 \times 3$ matrix, scaling it by a constant $k = \det M$ scales its determinant by $k^3$: $\frac{1}{\det M} = (\det M)^3 \cdot \det M$ $\frac{1}{\det M} = (\det M)^4$ $(\det M)^5 = 1$

Since $M$ has real entries, $\det M$ must be a real number. Therefore: $\det M = 1$

Hence, Option (B) is ALWAYS TRUE.

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Step 2: Determine $M^2$ Substituting $\det M = 1$ back into $M^{-1} = (\det M) M$: $M^{-1} = (1) M \implies M^{-1} = M$

Multiplying both sides by $M$: $M^2 = I$

Hence, Option (C) is ALWAYS TRUE.

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Step 3: Determine $(\text{adj } M)^2$ For any $3 \times 3$ matrix $M$, we have: $\text{adj } M = (\det M) M^{-1}$

Since $\det M = 1$: $\text{adj } M = M^{-1}$

Squaring both sides: $(\text{adj } M)^2 = (M^{-1})^2 = (M^2)^{-1}$

Using the result $M^2 = I$: $(\text{adj } M)^2 = I^{-1} = I$

Hence, Option (D) is ALWAYS TRUE.

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Step 4: Check Option (A) Consider the diagonal matrix $M = \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix}$.

• $\det M = (1)(-1)(-1) = 1$
• $M^2 = I$, so $M^{-1} = M$
• $\text{adj}(\text{adj } M) = (\det M) M = M = M^{-1}$

Thus, this matrix satisfies the given condition, but $M \neq I$. Therefore, Option (A) is NOT ALWAYS TRUE.

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Conclusion: The correct statements are (B), (C), and (D).