JEE Main 2012

LEVELJEE Main

Animated Solution for Mathematics - Matrices and Determinants: Let $P$ and $Q$ be $3 \times 3$ matrices $P \neq = Q$ . If $P^{3} = Q^{3}$ and $P^{2} Q = Q^{2} P$ then determinant of $(P^{2} + Q^{2})$ is equal to:

Select Answer:

Visualized Solution

Analyze the Given Matrix Equations

Given matrices: $P$ and $Q$ of order $3 \times 3$ .
Condition 1: $P \neq = Q$ .
Equation 1: $P^{3} = Q^{3}$ .
Equation 2: $P^{2} Q = Q^{2} P$ .

Subtracting the Equations

Subtract Equation 2 from Equation 1:
$P^{3} - P^{2} Q = Q^{3} - Q^{2} P$

Factoring the Matrix Expressions

Factor $P^{2}$ on the left side.
Factor $- Q^{2}$ on the right side.
$P^{2} (P - Q) = - Q^{2} (P - Q)$

Grouping into a Single Equation

Rearrange all terms to one side:
$P^{2} (P - Q) + Q^{2} (P - Q) = O$
Factor out $(P - Q)$ from the right:
$(P^{2} + Q^{2}) (P - Q) = O$

Applying the Determinant Property

Take the determinant on both sides:
$∣ (P^{2} + Q^{2}) (P - Q) ∣ = ∣ O ∣$
Using the property $∣ A B ∣ = ∣ A ∣∣ B ∣$ :
$∣ P^{2} + Q^{2} ∣ \cdot ∣ P - Q ∣ = 0$

The Final Logical Conclusion

Assume $∣ P^{2} + Q^{2} ∣ \neq = 0$ . Then $(P^{2} + Q^{2})$ is invertible.
Multiplying $(P^{2} + Q^{2}) (P - Q) = O$ by $(P^{2} + Q^{2})^{- 1}$ gives:
$(P^{2} + Q^{2})^{- 1} (P^{2} + Q^{2}) (P - Q) = (P^{2} + Q^{2})^{- 1} O ⟹ P - Q = O ⟹ P = Q$ .
This contradicts the given condition $P \neq = Q$ .
Therefore, $∣ P^{2} + Q^{2} ∣$ must be $0$ .

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The Sigma Insight: Properties of Determinants

Analyzing the Setup

We are given two

3 \times 3

matrices,

P

and

Q

, satisfying the following conditions: 1. $P eq Q$ 2.

P^{3} = Q^{3}

P^{2} Q = Q^{2} P

Our objective is to determine the value of the determinant

∣ P^{2} + Q^{2} ∣

. Because matrix multiplication is non-commutative, we must proceed with algebraic caution.

The Algebraic Dance

We begin by manipulating the given equations. Subtracting the second equation from the first, we obtain:

P^{3} - P^{2} Q = Q^{3} - Q^{2} P

Factoring the terms on both sides, we get:

P^{2} (P - Q) = - Q^{2} (P - Q)

Rearranging the terms to one side yields:

P^{2} (P - Q) + Q^{2} (P - Q) = O

Factoring out the common term

(P - Q)

from the right, we arrive at the fundamental identity:

(P^{2} + Q^{2}) (P - Q) = O

where

O

represents the

3 \times 3

null matrix.

The Determinant Insight

Taking the determinant of both sides of the equation

(P^{2} + Q^{2}) (P - Q) = O

, we apply the property

∣ A B ∣ = ∣ A ∣∣ B ∣

∣ P^{2} + Q^{2} ∣ \cdot ∣ P - Q ∣ = ∣ O ∣

Since the determinant of the null matrix is

0

, we have:

∣ P^{2} + Q^{2} ∣ \cdot ∣ P - Q ∣ = 0

This implies that at least one of the two determinants must be zero. We must now evaluate the condition $P eq Q$.

Final Deduction

If $|P^2 + Q^2| eq 0$, then the matrix

(P^{2} + Q^{2})

is invertible. Multiplying both sides of the equation

(P^{2} + Q^{2}) (P - Q) = O

by the inverse

(P^{2} + Q^{2})^{- 1}

would result in:

(P - Q) = O \Rightarrow P = Q

However, the problem explicitly states that $P eq Q$. This creates a contradiction, meaning our assumption that $|P^2 + Q^2| eq 0$ must be false.

Therefore, the only logical conclusion is that: $∣ P^{2} + Q^{2} ∣ = 0$

Common Traps & FAQs

Wait, why can't we just say P^3 = Q^3 implies P = Q?

That is a classic trap! In scalar algebra, x^3 = y^3 implies x = y. However, matrix algebra is different. Matrices can have zero divisors, meaning the product of two non-zero matrices can be the null matrix. We cannot assume P = Q just from P^3 = Q^3.

Why do we assume |P^2 + Q^2| != 0 to prove it is 0?

This is a standard proof technique called 'proof by contradiction.' We assume the opposite of what we want to prove. If |P^2 + Q^2| were non-zero, the matrix would be invertible. Multiplying by its inverse would force P = Q, which contradicts the problem statement. Therefore, our assumption must be false, and the determinant must be 0.

Is it possible that |P - Q| = 0 as well?

It is possible, but we don't need to determine that. The equation |P^2 + Q^2| * |P - Q| = 0 only requires that at least one of the determinants is zero. Since we proved that |P^2 + Q^2| cannot be non-zero, it must be zero regardless of the value of |P - Q|.

Animated Solution for Mathematics - Matrices and Determinants: Let $P$ and $Q$ be $3 \times 3$ matrices $P \neq = Q$ . If $P^{3} = Q^{3}$ and $P^{2} Q = Q^{2} P$ then determinant of $(P^{2} + Q^{2})$ is equal to:

Select Answer:

Visualized Solution

Analyze the Given Matrix Equations

Subtracting the Equations

Factoring the Matrix Expressions

Grouping into a Single Equation

Applying the Determinant Property

The Final Logical Conclusion

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Algebraic Dance

The Determinant Insight

Final Deduction

Similar Questions

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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 P and Q be 3×3 matrices P=Q. If P3=Q3 and P2Q=Q2P then determinant of (P2+Q2) is equal to:

Select Answer:

Visualized Solution

Analyze the Given Matrix Equations

Subtracting the Equations

Factoring the Matrix Expressions

Grouping into a Single Equation

Applying the Determinant Property

The Final Logical Conclusion

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Algebraic Dance

The Determinant Insight

Final Deduction

Similar Questions

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