JEE Advanced 2011

LEVELJEE Main

Animated Solution for Mathematics - Matrices and Determinants: Let $M$ and $N$ be two $3 \times 3$ non-singular skew-symmetric matrices such that $M N = N M$ . If $P^{T}$ denotes the transpose of $P$ , then $M^{2} N^{2} (M^{T} N)^{- 1} (M N^{- 1})^{T}$ is equal to

Select Answer:

Visualized Solution

Understanding the Given Properties

We are given two $3 \times 3$ non-singular skew-symmetric matrices $M$ and $N$ .
Skew-symmetric property: $M^{T} = - M$ and $N^{T} = - N$ .
Commutativity is given as: $M N = N M$ .
Non-singular means their determinants are non-zero, so their inverses $M^{- 1}$ and $N^{- 1}$ exist.

Expanding the Inverse Term $(M^{T} N)^{- 1}$

We use the reversal law of matrix multiplication for inverses: $(A B)^{- 1} = B^{- 1} A^{- 1}$ .
Applying this to our first complex term: $(M^{T} N)^{- 1} = N^{- 1} (M^{T})^{- 1}$ .

Applying Skew-Symmetry to $(M^{T})^{- 1}$

Since $M$ is skew-symmetric, we have $M^{T} = - M$ .
Substituting this: $(M^{T})^{- 1} = (- M)^{- 1}$ .
Using the property $(- A)^{- 1} = - (A^{- 1})$ , we get: $- N^{- 1} M^{- 1}$ .

Expanding the Transpose Term $(M N^{- 1})^{T}$

We use the reversal law of transposes: $(A B)^{T} = B^{T} A^{T}$ .
Applying this to our second complex term: $(M N^{- 1})^{T} = (N^{- 1})^{T} M^{T}$ .

Applying Skew-Symmetry to $(M N^{- 1})^{T}$

We know that transpose and inverse operations commute: $(N^{- 1})^{T} = (N^{T})^{- 1}$ .
Since $N$ is skew-symmetric, $N^{T} = - N$ , so $(N^{T})^{- 1} = (- N)^{- 1} = - N^{- 1}$ .
Also, $M^{T} = - M$ .
Substituting these: $(- N^{- 1}) (- M) = N^{- 1} M$ .

Substituting Simplified Terms back into $E$

Original Expression: $E = M^{2} N^{2} (M^{T} N)^{- 1} (M N^{- 1})^{T}$
Substitute our simplified terms:
$E = M^{2} N^{2} (- N^{- 1} M^{- 1}) (N^{- 1} M)$
Pulling the negative sign to the front: $E = - M^{2} N^{2} N^{- 1} M^{- 1} N^{- 1} M$

Applying Commutativity to Rearrange Terms

We are given that $M$ and $N$ commute: $M N = N M ⟹ M^{- 1} N^{- 1} = N^{- 1} M^{- 1}$ .
Let's rewrite $N^{2}$ as $N \cdot N$ .
Expression becomes: $E = - M^{2} N (N N^{- 1}) (M^{- 1} N^{- 1}) M$
Substitute $M^{- 1} N^{- 1}$ with $N^{- 1} M^{- 1}$ :
$E = - M^{2} N (N N^{- 1}) (N^{- 1} M^{- 1}) M$

Cancelling Terms to Find the Final Result

We know that $A A^{- 1} = I$ (Identity Matrix).
Therefore, $N N^{- 1} = I$ and $M^{- 1} M = I$ .
Substituting these: $E = - M^{2} N (I) (N^{- 1} I)$
Which simplifies to: $E = - M^{2} (N N^{- 1}) = - M^{2} I = - M^{2}$ .
Thus, the correct option is Option 3: $- M^{2}$ .

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The Sigma Insight: Algebraic Operations on Matrices

Solution Diagram

Analyzing the Setup

We are given two

3 \times 3

non-singular skew-symmetric matrices,

M

and

N

, that commute, meaning

M N = N M

. Our goal is to simplify the expression:

E = M^{2} N^{2} (M^{T} N)^{- 1} (M N^{- 1})^{T}

Skew-symmetry implies the properties

M^{T} = - M

and

N^{T} = - N

. Because

M

and

N

commute, their inverses

M^{- 1}

and

N^{- 1}

also commute with each other and with the original matrices.

The Reversal Laws

We must simplify the complex terms

(M^{T} N)^{- 1}

and

(M N^{- 1})^{T}

using matrix algebra laws. The reversal law for inverses states that

(A B)^{- 1} = B^{- 1} A^{- 1}

Applying this to the first term:

(M^{T} N)^{- 1} = N^{- 1} (M^{T})^{- 1}

Since

M^{T} = - M

, this becomes

N^{- 1} (- M)^{- 1}

. Factoring out the scalar

- 1

, we obtain:

(M^{T} N)^{- 1} = - N^{- 1} M^{- 1}

Next, we address the transpose term using the law

(A B)^{T} = B^{T} A^{T}

(M N^{- 1})^{T} = (N^{- 1})^{T} M^{T}

We know that

(N^{- 1})^{T} = (N^{T})^{- 1}

. Given

N^{T} = - N

, this becomes

(- N)^{- 1} M^{T}

. Substituting

M^{T} = - M

, we get:

(M N^{- 1})^{T} = (- N)^{- 1} (- M) = N^{- 1} M

The Grand Simplification

Now, we substitute these simplified components back into the original expression

E

E = M^{2} N^{2} (- N^{- 1} M^{- 1}) (N^{- 1} M)

Pulling the negative sign to the front, we have:

E = - M^{2} N^{2} N^{- 1} M^{- 1} N^{- 1} M

Since

M

and

N

commute, we can rearrange the terms to group

N

with

N^{- 1}

. We rewrite

N^{2}

N \cdot N

E = - M^{2} N (N N^{- 1}) (M^{- 1} N^{- 1}) M

Since

N N^{- 1} = I

(the identity matrix), the expression simplifies to:

E = - M^{2} N (I) (N^{- 1} M^{- 1}) M

E = - M^{2} (N N^{- 1}) (M^{- 1} M)

Applying the identity property again, where

N N^{- 1} = I

and

M^{- 1} M = I

, we arrive at the final result:

$E = - M^{2}$

Common Traps & FAQs

Wait, why does

(M^{T})^{- 1}

become

- M^{- 1}

? Shouldn't it be

1/ (- M)

That is a great question! Remember that in matrix algebra, we do not use division. Instead, we use the inverse. Since

M^{T} = - M

, we have

(M^{T})^{- 1} = (- M)^{- 1}

. Because

- 1

is a scalar, we can factor it out of the inverse operation:

(- M)^{- 1} = (- 1)^{- 1} M^{- 1} = - M^{- 1}

. It is just like factoring a constant out of a matrix equation!

Can we just assume

M

and

N

are diagonal matrices since they commute?

While it is true that commuting matrices often share properties, assuming they are diagonal is a dangerous shortcut! Not all commuting matrices are diagonal. It is much safer and more rigorous to use the algebraic properties of skew-symmetry and commutativity directly, as we did in the solution.

Why does

(N^{- 1})^{T}

equal

(N^{T})^{- 1}

This is a fundamental property of matrices! The transpose of an inverse is always equal to the inverse of the transpose. You can prove this by starting with

N N^{- 1} = I

. If you take the transpose of both sides, you get

(N N^{- 1})^{T} = I^{T} = I

. Using the reversal law,

(N^{- 1})^{T} N^{T} = I

. Multiplying both sides by

(N^{T})^{- 1}

on the right gives

(N^{- 1})^{T} = (N^{T})^{- 1}

. It is a very useful identity to keep in your toolkit!

Animated Solution for Mathematics - Matrices and Determinants: Let $M$ and $N$ be two $3 \times 3$ non-singular skew-symmetric matrices such that $M N = N M$ . If $P^{T}$ denotes the transpose of $P$ , then $M^{2} N^{2} (M^{T} N)^{- 1} (M N^{- 1})^{T}$ is equal to

Select Answer:

Visualized Solution

Understanding the Given Properties

Expanding the Inverse Term $(M^{T} N)^{- 1}$

Applying Skew-Symmetry to $(M^{T})^{- 1}$

Expanding the Transpose Term $(M N^{- 1})^{T}$

Applying Skew-Symmetry to $(M N^{- 1})^{T}$

Substituting Simplified Terms back into $E$

Applying Commutativity to Rearrange Terms

Cancelling Terms to Find the Final Result

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The Reversal Laws

The Grand Simplification

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Select Answer:

Visualized Solution

Understanding the Given Properties

Expanding the Inverse Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (MTN)−1

Applying Skew-Symmetry to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (MT)−1

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Applying Commutativity to Rearrange Terms

Cancelling Terms to Find the Final Result

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The Reversal Laws

The Grand Simplification

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