Adjoint Matrix Properties: Complete Guide for JEE

Introduction

The adjoint (or adjugate) of a matrix is a fundamental concept in linear algebra with numerous applications in JEE Main and Advanced. For a square matrix $A$ of order $n$, the adjoint is defined as the transpose of the cofactor matrix: $\operatorname{adj}(A) = C^T$ where $C_{ij}$ is the cofactor of element $a_{ij}$.

Part I: Fundamental Properties

1.1 The Defining Relationship

$A \cdot \operatorname{adj}(A) = \operatorname{adj}(A) \cdot A = |A| \cdot I_n$

1.2 Determinant of Adjoint

$|\operatorname{adj}(A)| = |A|^{n-1}$

Proof: $|A| \cdot |\operatorname{adj}(A)| = |A|^n \implies |\operatorname{adj}(A)| = |A|^{n-1}$

1.3 Adjoint of Adjoint

$\operatorname{adj}(\operatorname{adj}(A)) = |A|^{n-2} \cdot A$

1.4 Product Rule (Note the Reversal)

$\operatorname{adj}(AB) = \operatorname{adj}(B) \cdot \operatorname{adj}(A)$

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Part II: Advanced Properties

2.1 Scalar Multiplication

$\operatorname{adj}(kA) = k^{n-1} \cdot \operatorname{adj}(A)$ Proof: $(kA) \cdot \operatorname{adj}(kA) = |kA| \cdot I = k^n|A| \cdot I$ $(kA) \cdot k^{n-1}\operatorname{adj}(A) = k^n \cdot A \cdot \operatorname{adj}(A) = k^n|A| \cdot I$ By uniqueness, $\operatorname{adj}(kA) = k^{n-1} \cdot \operatorname{adj}(A)$.

2.2 Transpose Property

$\operatorname{adj}(A^T) = (\operatorname{adj}(A))^T$

2.3 Inverse Relationships

For non-singular $A$: $\operatorname{adj}(A^{-1}) = (\operatorname{adj}(A))^{-1} = \frac{A}{|A|}$

2.4 Power Property

$\operatorname{adj}(A^m) = (\operatorname{adj}(A))^m$ Proof by induction:

• Base: $\operatorname{adj}(A^1) = \operatorname{adj}(A)$
• Assuming $\operatorname{adj}(A^m) = (\operatorname{adj}(A))^m$, then: $\operatorname{adj}(A^{m+1}) = \operatorname{adj}(A \cdot A^m) = \operatorname{adj}(A^m) \cdot \operatorname{adj}(A) = (\operatorname{adj}(A))^m \cdot \operatorname{adj}(A) = (\operatorname{adj}(A))^{m+1}$

2.5 Determinant of Iterated Adjoint

For $k$ successive applications: $|\operatorname{adj}^k(A)| = |A|^{(n-1)^k}$ where $\operatorname{adj}^k(A)$ means applying adjoint $k$ times.

2.6 Trace Relationship

$\operatorname{tr}(A \cdot \operatorname{adj}(A)) = n \cdot |A|$

2.7 Rank Conditions

• If $\operatorname{rank}(A) = n$, then $\operatorname{rank}(\operatorname{adj}(A)) = n$
• If $\operatorname{rank}(A) = n-1$, then $\operatorname{rank}(\operatorname{adj}(A)) = 1$
• If $\operatorname{rank}(A) < n-1$, then $\operatorname{adj}(A) = O$

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Part III: JEE Previous Year Questions

PYQ 1 (JEE Main 2020)

If $A$ is a $3 \times 3$ matrix with $|A| = 4$, find $|\operatorname{adj}(3A^2)|$.

Solution: $|\operatorname{adj}(3A^2)| = |3A^2|^{3-1} = |3A^2|^2$ $|3A^2| = 3^3 \cdot |A|^2 = 27 \cdot 16 = 432$ Thus: $|\operatorname{adj}(3A^2)| = 432^2 = 186624$

PYQ 2 (JEE Main 2021)

If $A$ is a $3 \times 3$ matrix and $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj}(A)))| = |A|^k$, find $k$.

Solution: $|\operatorname{adj}^3(A)| = |A|^{(3-1)^3} = |A|^{2^3} = |A|^8$ Thus $k = 8$.

PYQ 3 (JEE Advanced 2018)

Let $P$ be a $3 \times 3$ matrix such that $P^T = 2P + I$. If there exists $Q$ such that $PQ = kI$, find $\frac{k}{|Q|}$.

Solution: From $P^T = 2P + I$:

• Transpose: $P = 2P^T + I$
• Substitute: $P = 2(2P + I) + I = 4P + 3I$
• Thus $-3P = 3I \implies P = -I$

If $PQ = kI$, then $-Q = kI \implies Q = -kI$.
Then $|Q| = |-kI| = (-k)^3 = -k^3$.
From $PQ = kI$: $|P||Q| = k^3 \implies (-1)(-k^3) = k^3$ (consistent).

Thus: $\frac{k}{|Q|} = \frac{k}{-k^3} = -\frac{1}{k^2}$ Taking $k = 1$ gives $-\frac{1}{1} = -1$.

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Part IV: Quick Reference Table

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Part V: Problem-Solving Strategy

Key Steps:

1. Identify $n$ (order of matrix) — most formulas depend on it.
2. Use given determinant — it simplifies many calculations.
3. Break composite operations using properties like $\operatorname{adj}(AB) = \operatorname{adj}(B) \operatorname{adj}(A)$.
4. Extract scalars carefully — remember the exponent is $n-1$.

Common Pitfalls:

• Reverse order in product: $\operatorname{adj}(AB) = \operatorname{adj}(B) \operatorname{adj}(A)$
• Wrong exponent in $|\operatorname{adj}(A)| = |A|^{n-1}$
• Scalar extraction: $\operatorname{adj}(kA) = k^{n-1} \operatorname{adj}(A)$, not $k^n$
• Confusing $\operatorname{adj}(A^{-1})$ with $(\operatorname{adj}(A))^{-1}$ (they are equal for invertible $A$)

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Practice Problems

1. If $A$ is a $4 \times 4$ matrix with $|A| = 3$, find $|\operatorname{adj}(2A^3)|$.
2. For a $3 \times 3$ invertible matrix $A$, simplify $\operatorname{adj}(A^{-1} \operatorname{adj}(A))$.
3. If $B = \operatorname{adj}(\operatorname{adj}(A))$ and $|A| = 2$, find $|B|$ for $n = 3$.
4. Prove that $\operatorname{adj}(A^T B) = \operatorname{adj}(B) \operatorname{adj}(A^T)$.
5. If $A$ is a $2 \times 2$ matrix with $|A| = 5$, find $\operatorname{adj}(\operatorname{adj}(A))$.

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Summary

Mastering adjoint properties is crucial for JEE. The core insight is that most properties derive from the fundamental relationship $A \cdot \operatorname{adj}(A) = |A| \cdot I$. Remember the key patterns:

• Powers of $(n-1)$ appear in determinant formulas
• Adjoint reverses the order in products (like the inverse)
• Scalar extraction uses exponent $n-1$

With practice, these properties become powerful tools for solving complex matrix problems efficiently.