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(CAB)A' + B'

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Answer

(A+B)T=AT+BT(A+B)^T = A^T + B^T

Step-by-step solution

7 steps

Okay, Christove, let's verify the matrix transpose property shown in the image: (A+B)T=AT+BT(A+B)^T = A^T + B^T. This property states that the transpose of the sum of two matrices is equal to the sum of their transposes.

  1. Define two matrices AA and BB.

    Let's use 2×22 \times 2 matrices for this example:

    A=[1234],B=[5678]A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, \quad B = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}

  2. Calculate the sum A+BA+B.

    A+B=[1234]+[5678]=[1+52+63+74+8]=[681012]A+B = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} + \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix} = \begin{bmatrix} 1+5 & 2+6 \\ 3+7 & 4+8 \end{bmatrix} = \begin{bmatrix} 6 & 8 \\ 10 & 12 \end{bmatrix}

  3. Calculate the transpose of (A+B)(A+B), which is (A+B)T(A+B)^T.

    To find the transpose, we swap the rows and columns of A+BA+B.

    (A+B)T=[610812](A+B)^T = \begin{bmatrix} 6 & 10 \\ 8 & 12 \end{bmatrix}

  4. Calculate the transpose of AA, which is ATA^T.

    AT=[1324]A^T = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}

  5. Calculate the transpose of BB, which is BTB^T.

    BT=[5768]B^T = \begin{bmatrix} 5 & 7 \\ 6 & 8 \end{bmatrix}

  6. Calculate the sum of AT+BTA^T + B^T.

    AT+BT=[1324]+[5768]=[1+53+72+64+8]=[610812]A^T + B^T = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} + \begin{bmatrix} 5 & 7 \\ 6 & 8 \end{bmatrix} = \begin{bmatrix} 1+5 & 3+7 \\ 2+6 & 4+8 \end{bmatrix} = \begin{bmatrix} 6 & 10 \\ 8 & 12 \end{bmatrix}

  7. Compare the results from Step 3 and Step 6.

    We can see that (A+B)T=[610812](A+B)^T = \begin{bmatrix} 6 & 10 \\ 8 & 12 \end{bmatrix} and AT+BT=[610812]A^T + B^T = \begin{bmatrix} 6 & 10 \\ 8 & 12 \end{bmatrix}.
    Since both results are identical, the property is verified.

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