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If "AA"^T=I where A^T is transpose of ma...

If `"AA"^T=I` where `A^T` is transpose of matrix A then, `1/2A[(A+A^T)^2+(A-A^T)^2]=?`

A

`A^2`

B

`A^3+1`

C

`A^2+1`

D

`A^3+A^T`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem \( \frac{1}{2} A \left[ (A + A^T)^2 + (A - A^T)^2 \right] \) given that \( AA^T = I \), we will follow these steps: ### Step 1: Expand the expressions We start by expanding \( (A + A^T)^2 \) and \( (A - A^T)^2 \). 1. **Expanding \( (A + A^T)^2 \)**: \[ (A + A^T)^2 = A^2 + AA^T + A^TA + (A^T)^2 \] 2. **Expanding \( (A - A^T)^2 \)**: \[ (A - A^T)^2 = A^2 - AA^T - A^TA + (A^T)^2 \] ### Step 2: Combine the expanded expressions Now, we can combine the two expansions: \[ (A + A^T)^2 + (A - A^T)^2 = \left( A^2 + AA^T + A^TA + (A^T)^2 \right) + \left( A^2 - AA^T - A^TA + (A^T)^2 \right) \] ### Step 3: Simplify the combined expression When we combine the two expansions, we notice that the \( AA^T \) and \( A^TA \) terms cancel each other out: \[ = 2A^2 + 2(A^T)^2 \] ### Step 4: Substitute back into the original expression Now, we substitute this back into the original expression: \[ \frac{1}{2} A \left[ 2A^2 + 2(A^T)^2 \right] = A \left[ A^2 + (A^T)^2 \right] \] ### Step 5: Use the given condition \( AA^T = I \) Since \( AA^T = I \), we can replace \( A^TA \) with \( I \): \[ = A \left[ A^2 + I \right] \] ### Step 6: Final expression Thus, the final expression simplifies to: \[ = A^3 + A \] ### Conclusion The final answer is: \[ A^3 + A \]
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  • Let A*B=A^(T)B^(-1) where A^(T) represents transpose of matrix A and B^(-1) represents inverse of square matrix B . This operation is defined when the number of rows of A is equal to the number of rows of B . Matrix A is said to be orthogonal if A^(-1)=A^(T) If A*B is defined then choose the incorrect statement

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    C
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    D
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