JEE Main 2009

LEVELJEE Main

Animated Solution for Mathematics - Matrices and Determinants: Let $A$ be a $2 \times 2$ matrix. \\ Statement-1: $a d j (a d j A) = A$ \\ Statement-2: $∣ a d j A ∣ = ∣ A ∣$

Q: Wait, why do we assume $|A| \neq 0$ in the division step? What if $|A| = 0$?

That is a brilliant question! If $|A| = 0$, the matrix $A$ is singular. The formula $\text{adj}(\text{adj} A) = |A|^{n-2} A$ actually still holds true for all square matrices. If $|A| = 0$ and $n=2$, the formula gives $\text{adj}(\text{adj} A) = |A|^0 A = 1 \cdot A = A$, which is not quite right if $A$ is singular. Actually, for a singular matrix $A$, $\text{adj}(A)$ is also singular, and $\text{adj}(\text{adj} A)$ becomes the zero matrix. The formula $\text{adj}(\text{adj} A) = |A|^{n-2} A$ is the general case, and for $n=2$, it simplifies to $\text{adj}(\text{adj} A) = A$ only when $A$ is non-singular. In the context of JEE problems, we often assume the non-singular case unless specified otherwise, but your intuition about singular matrices is spot on!

Select Answer:

Visualized Solution

Analyzing Statement-2

Let $A$ be a $2 \times 2$ matrix, so $n = 2$ .
The determinant of the adjoint of an $n \times n$ matrix is $∣ adj A ∣ = ∣ A ∣^{n - 1}$ .
For $n = 2$ , $∣ adj A ∣ = ∣ A ∣^{2 - 1} = ∣ A ∣$ .
Thus, Statement-2 is True.

The Base Matrix Property

To evaluate Statement-1, we need to find $adj (adj A)$ .
We start with the fundamental matrix property: $B \cdot adj (B) = ∣ B ∣ I$ .

Substituting the Adjoint

Let's substitute $B = adj (A)$ into our fundamental property.
This gives: $adj (A) \cdot adj (adj A) = ∣ adj A ∣ I$ .

Using Statement-2 as a Bridge

Notice the term $∣ adj A ∣$ on the right side.
From Statement-2, we know $∣ adj A ∣ = ∣ A ∣^{n - 1}$ .
Substituting this in: $adj (A) \cdot adj (adj A) = ∣ A ∣^{n - 1} I$ .

Multiplying by $A$

Pre-multiply both sides by matrix $A$ :
$A \cdot adj (A) \cdot adj (adj A) = A \cdot ∣ A ∣^{n - 1} I$
Since $A \cdot adj (A) = ∣ A ∣ I$ , the left side becomes $∣ A ∣ I \cdot adj (adj A)$ .
This simplifies to: $∣ A ∣ \cdot adj (adj A) = ∣ A ∣^{n - 1} A$ .

The General Double Adjoint Formula

Divide both sides by $∣ A ∣$ (assuming $∣ A ∣ \neq = 0$ ):
$adj (adj A) = \frac{∣ A ∣ ^{n - 1}}{∣ A ∣} A$
$adj (adj A) = ∣ A ∣^{n - 2} A$
This is the general property for the double adjoint.

Final Conclusion for $n = 2$

Substitute $n = 2$ into the general formula:
$adj (adj A) = ∣ A ∣^{2 - 2} A = ∣ A ∣^{0} A$
Since $∣ A ∣^{0} = 1$ , we get $adj (adj A) = A$ .
Statement-1 is True, and Statement-2 is the correct explanation because we used it in the derivation.

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The Sigma Insight: Adjoint and Inverse of a Matrix

Solution Diagram

Analyzing the Setup

We are examining the properties of the adjoint of a matrix, specifically focusing on a

2 \times 2

matrix where the dimension

n = 2

The fundamental property for the determinant of an adjoint matrix is given by:

∣ adj A ∣ = ∣ A ∣^{n - 1}

When we substitute

n = 2

into this expression, we obtain

∣ adj A ∣ = ∣ A ∣^{2 - 1} = ∣ A ∣

. This confirms that Statement-2 is correct.

The Master Equation

To evaluate

adj (adj A)

, we utilize the fundamental identity for any square matrix

B

B \cdot adj (B) = ∣ B ∣ I

To derive the desired result, we substitute

B = adj (A)

into this identity. This yields:

adj (A) \cdot adj (adj A) = ∣ adj A ∣ I

Deriving the General Formula

Using the result from Statement-2, we replace

∣ adj A ∣

with

∣ A ∣^{n - 1}

. The equation now becomes:

adj (A) \cdot adj (adj A) = ∣ A ∣^{n - 1} I

To isolate

adj (adj A)

, we pre-multiply both sides by the original matrix

A

A \cdot adj (A) \cdot adj (adj A) = A \cdot ∣ A ∣^{n - 1} I

Since

A \cdot adj (A) = ∣ A ∣ I

, the left side simplifies to

∣ A ∣ \cdot adj (adj A)

. Consequently, we have:

∣ A ∣ \cdot adj (adj A) = ∣ A ∣^{n - 1} A

Final Calculation

Assuming $|A| eq 0$, we divide both sides by

∣ A ∣

to obtain the general formula:

adj (adj A) = ∣ A ∣^{n - 2} A

For our specific case where

n = 2

, the formula simplifies to:

adj (adj A) = ∣ A ∣^{2 - 2} A = ∣ A ∣^{0} A = A

Thus, Statement-1 is true, and Statement-2 provides the correct explanation for this result.

Common Traps & FAQs

Wait, why do we assume

∣ A ∣ \neq = 0

in the division step? What if

∣ A ∣ = 0

That is a brilliant question! If

∣ A ∣ = 0

, the matrix

A

is singular. The formula

adj (adj A) = ∣ A ∣^{n - 2} A

actually still holds true for all square matrices. If

∣ A ∣ = 0

and

n = 2

, the formula gives

adj (adj A) = ∣ A ∣^{0} A = 1 \cdot A = A

, which is not quite right if

A

is singular. Actually, for a singular matrix

A

adj (A)

is also singular, and

adj (adj A)

becomes the zero matrix. The formula

adj (adj A) = ∣ A ∣^{n - 2} A

is the general case, and for

n = 2

, it simplifies to

adj (adj A) = A

only when

A

is non-singular. In the context of JEE problems, we often assume the non-singular case unless specified otherwise, but your intuition about singular matrices is spot on!

Why is Statement-2 considered the correct explanation for Statement-1?

In the world of JEE, a 'correct explanation' means that the logic used to prove Statement-1 directly relies on the truth of Statement-2. As we saw in our derivation, we used the property

∣ adj A ∣ = ∣ A ∣^{n - 1}

(which is Statement-2) to substitute into the equation

adj (A) \cdot adj (adj A) = ∣ adj A ∣ I

. Without Statement-2, we wouldn't have been able to simplify the right-hand side to

∣ A ∣^{n - 1} I

, which was the crucial step to isolate

adj (adj A)

. Thus, Statement-2 is the logical bridge.

Is the formula

adj (adj A) = ∣ A ∣^{n - 2} A

true for all matrices?

Yes, it is! It is a universal property for any square matrix of order

n \times n

. It is one of those elegant, powerful results in linear algebra that makes you appreciate the deep structure of matrices. Whenever you see an adjoint of an adjoint, you can immediately jump to this formula. It turns a potentially complex problem into a simple substitution.

Animated Solution for Mathematics - Matrices and Determinants: Let $A$ be a $2 \times 2$ matrix. \\ Statement-1: $a d j (a d j A) = A$ \\ Statement-2: $∣ a d j A ∣ = ∣ A ∣$

Select Answer:

Visualized Solution

Analyzing Statement-2

The Base Matrix Property

Substituting the Adjoint

Using Statement-2 as a Bridge

Multiplying by $A$

The General Double Adjoint Formula

Final Conclusion for $n = 2$

The Sigma Insight: Adjoint and Inverse of a Matrix

Analyzing the Setup

The Master Equation

Deriving the General Formula

Final Calculation

Similar Questions

Let $A$ be a $3 \times 3$ real matrix. If $det (2 Adj (2 Adj (Adj (2 A)))) = 2^{41}$ , then the value of $det (A^{2})$ equal .

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Animated Solution for Mathematics - Matrices and Determinants: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A be a 2×2 matrix. \\ Statement-1: adj(adjA)=A \\ Statement-2: ∣adjA∣=∣A∣

Select Answer:

Visualized Solution

Analyzing Statement-2

The Base Matrix Property

Substituting the Adjoint

Using Statement-2 as a Bridge

Multiplying by .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

The General Double Adjoint Formula

Final Conclusion for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=2

The Sigma Insight: Adjoint and Inverse of a Matrix

Analyzing the Setup

The Master Equation

Deriving the General Formula

Final Calculation

Similar Questions

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Animated Solution for Mathematics - Matrices and Determinants: Let $A$ be a $2 \times 2$ matrix. \\ Statement-1: $a d j (a d j A) = A$ \\ Statement-2: $∣ a d j A ∣ = ∣ A ∣$

Multiplying by $A$

Final Conclusion for $n = 2$

Let $A$ be a $3 \times 3$ real matrix. If $det (2 Adj (2 Adj (Adj (2 A)))) = 2^{41}$ , then the value of $det (A^{2})$ equal .

Let $M$ be a $3 \times 3$ invertible matrix with real entries and let $I$ denote the $3 \times 3$ identity matrix. If $M^{- 1} = adj (adj M)$ , then which of the following statement is/are ALWAYS TRUE?

Let $A$ be a $3 \times 3$ matrix such that $∣ adj (adj (adj A)) ∣ = 1 2^{4}$ . Then $∣ A^{- 1} adj A ∣$ is equal to

If A, B, and $(a d j (A^{- 1}) + a d j (B^{- 1}))$ are non-singular matrices of same order, then the inverse of $A (a d j (A^{- 1}) + a d j (B^{- 1}))^{- 1} B$ , is equal to

If $A = \frac{1}{5 ! 6 ! 7 !} 5! 6! 7! 6! 7! 8! 7! 8! 9!$ , then $∣ a d j (a d j (2 A)) ∣$ is equal to

Let $B = 11 α 32 α α 34$ , $α > 2$ be the adjoint of a matrix $A$ and $∣ A ∣ = 2$ . Then $[α - 2 α α] B α - 2 α α$ is equal to

Let $A$ be a matrix of order $3 \times 3$ and $∣ A ∣ = 5$ . If $∣2 adj (3 A adj (2 A)) ∣ = 2^{α} \cdot 3^{β} \cdot 5^{γ}, α, β, γ \in N$ then $α + β + γ$ is equal to

Let $A = 210 12 - 1 0 - 1 2$ . If $∣ a d j (a d j (a d j 2 A)) ∣ = (16)^{n}$ , then $n$ is equal to