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Let M and N be two even order non sing...

Let M and N be two even order non singular skew symmetric matrices than MN=NM. If `P^(T)` denotes the transpose of P, then `M^(2)N^(2)(M^(T)N)^(-1)(MN^(-1))^(T)` is equal to

A

`M^(2)`

B

`-N^(2)`

C

`-M^(2)`

D

`MN`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to simplify the expression \( M^2 N^2 (M^T N)^{-1} (M N^{-1})^T \). ### Step-by-step Solution: 1. **Start with the given expression:** \[ M^2 N^2 (M^T N)^{-1} (M N^{-1})^T \] 2. **Use the property of transpose:** The transpose of a product of matrices is the product of their transposes in reverse order. Thus, \[ (M N^{-1})^T = N^{-T} M^T \] Since \( N \) is skew-symmetric, we have \( N^T = -N \), which implies \( N^{-T} = -N^{-1} \). Therefore, \[ (M N^{-1})^T = -N^{-1} M^T \] 3. **Substituting back into the expression:** The expression now becomes: \[ M^2 N^2 (M^T N)^{-1} (-N^{-1} M^T) \] 4. **Factor out the negative sign:** This gives us: \[ -M^2 N^2 (M^T N)^{-1} N^{-1} M^T \] 5. **Simplify \( (M^T N)^{-1} \):** Using the property of inverses, we can express: \[ (M^T N)^{-1} = N^{-1} (M^T)^{-1} \] Since \( M \) is skew-symmetric, \( (M^T)^{-1} = -M^{-1} \). Thus, \[ (M^T N)^{-1} = N^{-1} (-M^{-1}) = -N^{-1} M^{-1} \] 6. **Substituting this back:** Now substituting this into our expression: \[ -M^2 N^2 (-N^{-1} M^{-1}) N^{-1} M^T \] This simplifies to: \[ M^2 N^2 N^{-1} M^{-1} M^T \] 7. **Combine terms:** Since \( N^2 N^{-1} = N \): \[ M^2 N M^{-1} M^T \] 8. **Using the skew-symmetric property:** Recall that \( M^T = -M \): \[ M^2 N (-M^{-1}) = -M^2 N M^{-1} \] 9. **Final simplification:** We have: \[ -M^2 N M^{-1} \] 10. **Since \( M \) and \( N \) commute:** We can replace \( M^2 N M^{-1} \) with \( M N \) (because \( MN = NM \)): \[ -M N \] ### Final Answer: Thus, the final result is: \[ -MN \]
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