Understanding the Adjoint and Inverse of a Matrix

This problem explores the fundamental properties relating an invertible matrix, its adjoint, and its inverse. For an $n \times n$ invertible matrix $A$, these relationships are crucial in linear algebra and frequently tested in JEE. An invertible matrix means its determinant, $|A|$, is non-zero. In this specific question, $A$ is a $3 \times 3$ matrix, so we will use $n=3$ in our formulas.

Key Concepts and Formulas for an $n \times n$ invertible matrix $A$:

1. Definition of Inverse: The product of a matrix and its adjoint is equal to the determinant of the matrix times the identity matrix. $A \cdot \text{adj}(A) = \text{adj}(A) \cdot A = |A| I$ where $I$ is the identity matrix of order $n$.

2. Inverse in terms of Adjoint: This is a direct consequence of the definition above. $A^{-1} = \frac{1}{|A|} \text{adj}(A)$

3. Determinant of Adjoint: The determinant of the adjoint of $A$ is related to the determinant of $A$. $|\text{adj}(A)| = |A|^{n-1}$

4. Adjoint of Adjoint: The adjoint of the adjoint of $A$ can be expressed in terms of $A$ and its determinant. $\text{adj}(\text{adj}(A)) = |A|^{n-2} A$

5. Inverse of Adjoint: The inverse of the adjoint of $A$ can also be expressed simply. We know $A^{-1} = \frac{1}{|A|} \text{adj}(A)$. From this, we can write $A = |A| A^{-1}$. Now, consider $(\text{adj}(A))^{-1}$. Using the formula for inverse in terms of adjoint (Formula 2) but applied to $\text{adj}(A)$ instead of $A$: $(\text{adj}(A))^{-1} = \frac{1}{|\text{adj}(A)|} \text{adj}(\text{adj}(A))$ Substitute $|\text{adj}(A)| = |A|^{n-1}$ (Formula 3) and $\text{adj}(\text{adj}(A)) = |A|^{n-2} A$ (Formula 4): $(\text{adj}(A))^{-1} = \frac{1}{|A|^{n-1}} (|A|^{n-2} A)$ $(\text{adj}(A))^{-1} = |A|^{n-2-(n-1)} A = |A|^{-1} A = \frac{1}{|A|} A$ So, $(\text{adj}(A))^{-1} = \frac{1}{|A|} A$.

For this problem, $A$ is a $3 \times 3$ matrix, so we substitute $n=3$ into these formulas. Let's evaluate each option to find which one is not always true.

---

Analysis of Option (A): $\text{adj}(A) = |A| \cdot A^{-1}$

Step-by-step derivation: We begin with the fundamental formula for the inverse of an invertible matrix (Key Concept 2): $A^{-1} = \frac{1}{|A|} \text{adj}(A)$ To isolate $\text{adj}(A)$, we multiply both sides of this equation by the scalar $|A|$: $|A| \cdot A^{-1} = |A| \cdot \left( \frac{1}{|A|} \text{adj}(A) \right)$ $|A| \cdot A^{-1} = \text{adj}(A)$ This precisely matches Option (A).

Conclusion for (A): Option (A) is a direct rearrangement of the definition of the inverse using the adjoint. Therefore, it is always true for any invertible matrix $A$, regardless of its dimension.

---

Analysis of Option (B): $\text{adj}(\text{adj}(A)) = |A| \cdot A$

Step-by-step derivation: We use the general formula for the adjoint of an adjoint matrix (Key Concept 4): $\text{adj}(\text{adj}(A)) = |A|^{n-2} A$ Since $A$ is a $3 \times 3$ matrix, we substitute $n=3$ into the formula: $\text{adj}(\text{adj}(A)) = |A|^{3-2} A$ $\text{adj}(\text{adj}(A)) = |A|^1 A$ $\text{adj}(\text{adj}(A)) = |A| A$ This matches Option (B).

Conclusion for (B): Option (B) is derived directly from a standard property of adjoints by substituting $n=3$. Therefore, it is always true for any $3 \times 3$ invertible matrix $A$.

---

Analysis of Option (C): $\text{adj}(\text{adj}(A)) = |A|^2 \cdot (\text{adj}(A))^{-1}$

Step-by-step derivation: To verify this option, we will use two established results for a $3 \times 3$ matrix:

1. From our analysis of Option (B), we know: $\text{adj}(\text{adj}(A)) = |A| A \quad \dots (1)$
2. From Key Concept 5, we know the formula for the inverse of an adjoint matrix: $(\text{adj}(A))^{-1} = \frac{1}{|A|} A$ From this, we can express $A$ in terms of $(\text{adj}(A))^{-1}$ by multiplying both sides by $|A|$: $A = |A| (\text{adj}(A))^{-1} \quad \dots (2)$ Now, substitute the expression for $A$ from equation (2) into equation (1): $\text{adj}(\text{adj}(A)) = |A| \cdot \left( |A| (\text{adj}(A))^{-1} \right)$ $\text{adj}(\text{adj}(A)) = |A|^2 (\text{adj}(A))^{-1}$ This matches Option (C).

Conclusion for (C): Option (C) is derived by combining standard properties of adjoints and inverses for $n=3$. Therefore, it is always true for any $3 \times 3$ invertible matrix $A$.

---

Analysis of Option (D): $\text{adj}(\text{adj}(A)) = |A| \cdot (\text{adj}(A))^{-1}$

Step-by-step derivation: From our analysis of Option (C), we established the correct expression for $\text{adj}(\text{adj}(A))$ for a $3 \times 3$ matrix: $\text{adj}(\text{adj}(A)) = |A|^2 (\text{adj}(A))^{-1}$ Now, let's compare this derived true statement with what Option (D) proposes: Option (D) states: $\text{adj}(\text{adj}(A)) = |A| (\text{adj}(A))^{-1}$

For Option (D) to be always true, the following equality must hold: $|A|^2 (\text{adj}(A))^{-1} = |A| (\text{adj}(A))^{-1}$ Since $A$ is an invertible matrix, its determinant $|A|$ is non-zero. Also, its adjoint $\text{adj}(A)$ is invertible (because $|\text{adj}(A)| = |A|^{n-1} = |A|^2 \neq 0$). Therefore, the inverse $(\text{adj}(A))^{-1}$ exists and is an invertible matrix itself. We can multiply both sides of the equation by $\text{adj}(A)$ (from the right) to simplify: $|A|^2 I = |A| I$ Where $I$ is the identity matrix. This implies: $|A|^2 = |A|$ Rearranging the terms, we get: $|A|^2 - |A| = 0$ Factoring out $|A|$: $|A|(|A| - 1) = 0$ Since $A$ is an invertible matrix, its determinant $|A|$ cannot be zero. Therefore, we must have: $|A| - 1 = 0$ $|A| = 1$ This means Option (D) is true only if the determinant of $A$ is $1$. However, the problem states that $A$ is any $3 \times 3$ invertible matrix. An invertible matrix can have any non-zero determinant (e.g., $|A|=2$, $|A|=-5$, etc.). Since $|A|$ is not necessarily $1$, Option (D) is not always true.

Conclusion for (D): Option (D) is only true under the specific condition that $|A|=1$. Since this condition is not always met for an any invertible matrix, Option (D) is not always true.

---

Final Answer and Summary

We have established that Options (A), (B), and (C) are always true for any $3 \times 3$ invertible matrix $A$. Option (D), however, is true only when $|A|=1$, which is not a universal property of all invertible matrices.

Therefore, the statement that is not always true is Option (D).

The final answer is $\boxed{\text{D}}$.

---

JEE Pro Tip: Memorizing the key formulas for adjoints and inverses (especially $\text{adj}(\text{adj}(A)) = |A|^{n-2} A$ and $(\text{adj}(A))^{-1} = \frac{1}{|A|} A$) is crucial for quickly solving such problems. When faced with "not always true" questions, try to find a counterexample or a specific condition that must be met for the statement to hold. If that condition is not universally true, then the statement is "not always true."