For any matrix A, `A-A'` is _________ matrix.

Answer

To determine the nature of the matrix \( A - A' \) where \( A' \) is the transpose of matrix \( A \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Transpose**:
The transpose of a matrix \( A \) is obtained by interchanging its rows and columns. If \( A \) is a matrix given by:
\[ A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \]
then its transpose \( A' \) is:
\[ A' = \begin{pmatrix} a & c \\ b & d \end{pmatrix} \]

2. **Calculating \( A - A' \)**:
We can now compute \( A - A' \):
\[ A - A' = \begin{pmatrix} a & b \\ c & d \end{pmatrix} - \begin{pmatrix} a & c \\ b & d \end{pmatrix} \]
Performing the subtraction:
\[ A - A' = \begin{pmatrix} a - a & b - c \\ c - b & d - d \end{pmatrix} = \begin{pmatrix} 0 & b - c \\ c - b & 0 \end{pmatrix} \]

3. **Identifying the Matrix Type**:
The resulting matrix is:
\[ \begin{pmatrix} 0 & b - c \\ c - b & 0 \end{pmatrix} \]
Notice that the diagonal elements are zero, and the off-diagonal elements are negatives of each other (\( b - c \) and \( c - b \)). This structure is characteristic of a skew-symmetric matrix.

4. **Condition for Skew-Symmetric Matrix**:
A matrix \( M \) is skew-symmetric if \( M' = -M \). We can check this condition for our matrix:
\[ (A - A')' = \begin{pmatrix} 0 & b - c \\ c - b & 0 \end{pmatrix}' = \begin{pmatrix} 0 & c - b \\ b - c & 0 \end{pmatrix} = -\begin{pmatrix} 0 & b - c \\ c - b & 0 \end{pmatrix} \]
This confirms that \( A - A' \) is indeed skew-symmetric.

### Conclusion:
Thus, for any matrix \( A \), the expression \( A - A' \) results in a **skew-symmetric matrix**.