Answer :
To solve for [tex]\( A - B \)[/tex] given that [tex]\( \left[\begin{array}{ccc}1 & 3 & 0 \\ 1 & 0 & -2 \\ -4 & -4 & 4\end{array}\right]=A+B \)[/tex] where [tex]\( A \)[/tex] is a symmetric matrix and [tex]\( B \)[/tex] is a skew-symmetric matrix, we follow these steps:
1. Understand the properties of symmetric and skew-symmetric matrices:
- A symmetric matrix [tex]\( A \)[/tex] satisfies [tex]\( A^T = A \)[/tex].
- A skew-symmetric matrix [tex]\( B \)[/tex] satisfies [tex]\( B^T = -B \)[/tex] and the diagonal elements of [tex]\( B \)[/tex] are zero.
2. Formulate the given matrix [tex]\( A + B \)[/tex]:
[tex]\[ M = \left[\begin{array}{ccc} 1 & 3 & 0 \\ 1 & 0 & -2 \\ -4 & -4 & 4 \end{array}\right] \][/tex]
3. Derive the symmetric matrix [tex]\( A \)[/tex]:
Since [tex]\( A \)[/tex] is symmetric, we can use the property of symmetric matrices [tex]\( A = \frac{1}{2}(M + M^T) \)[/tex] to find [tex]\( A \)[/tex]. Calculating the transpose of [tex]\( M \)[/tex] and then averaging:
[tex]\[ M^T = \left[\begin{array}{ccc} 1 & 1 & -4 \\ 3 & 0 & -4 \\ 0 & -2 & 4 \end{array}\right] \][/tex]
[tex]\[ A = \frac{1}{2} \left( \left[\begin{array}{ccc} 1 & 3 & 0 \\ 1 & 0 & -2 \\ -4 & -4 & 4 \end{array}\right] + \left[\begin{array}{ccc} 1 & 1 & -4 \\ 3 & 0 & -4 \\ 0 & -2 & 4 \end{array}\right] \right) = \frac{1}{2} \left[\begin{array}{ccc} 2 & 4 & -4 \\ 4 & 0 & -6 \\ -4 & -6 & 8 \end{array}\right] = \left[\begin{array}{ccc} 1 & 2 & -2 \\ 2 & 0 & -3 \\ -2 & -3 & 4 \end{array}\right] \][/tex]
4. Derive the skew-symmetric matrix [tex]\( B \)[/tex]:
Since [tex]\( B \)[/tex] is skew-symmetric, we can use the property of skew-symmetric matrices [tex]\( B = \frac{1}{2}(M - M^T) \)[/tex] to find [tex]\( B \)[/tex]:
[tex]\[ B = \frac{1}{2} \left( \left[\begin{array}{ccc} 1 & 3 & 0 \\ 1 & 0 & -2 \\ -4 & -4 & 4 \end{array}\right] - \left[\begin{array}{ccc} 1 & 1 & -4 \\ 3 & 0 & -4 \\ 0 & -2 & 4 \end{array}\right] \right) = \frac{1}{2} \left[\begin{array}{ccc} 0 & 2 & 4 \\ -2 & 0 & 2 \\ -4 & -2 & 0 \end{array}\right] = \left[\begin{array}{ccc} 0 & 1 & 2 \\ -1 & 0 & 1 \\ -2 & -1 & 0 \end{array}\right] \][/tex]
5. Compute [tex]\( A - B \)[/tex]:
Finally, we subtract [tex]\( B \)[/tex] from [tex]\( A \)[/tex]:
[tex]\[ A - B = \left[\begin{array}{ccc} 1 & 2 & -2 \\ 2 & 0 & -3 \\ -2 & -3 & 4 \end{array}\right] - \left[\begin{array}{ccc} 0 & 1 & 2 \\ -1 & 0 & 1 \\ -2 & -1 & 0 \end{array}\right] = \left[\begin{array}{ccc} 1 & 1 & -4 \\ 3 & 0 & -4 \\ 0 & -2 & 4 \end{array}\right] \][/tex]
Thus, the matrix [tex]\( A - B \)[/tex] is:
[tex]\[ \left[\begin{array}{ccc} 1 & 1 & -4 \\ 3 & 0 & -4 \\ 0 & -2 & 4 \end{array}\right] \][/tex]
1. Understand the properties of symmetric and skew-symmetric matrices:
- A symmetric matrix [tex]\( A \)[/tex] satisfies [tex]\( A^T = A \)[/tex].
- A skew-symmetric matrix [tex]\( B \)[/tex] satisfies [tex]\( B^T = -B \)[/tex] and the diagonal elements of [tex]\( B \)[/tex] are zero.
2. Formulate the given matrix [tex]\( A + B \)[/tex]:
[tex]\[ M = \left[\begin{array}{ccc} 1 & 3 & 0 \\ 1 & 0 & -2 \\ -4 & -4 & 4 \end{array}\right] \][/tex]
3. Derive the symmetric matrix [tex]\( A \)[/tex]:
Since [tex]\( A \)[/tex] is symmetric, we can use the property of symmetric matrices [tex]\( A = \frac{1}{2}(M + M^T) \)[/tex] to find [tex]\( A \)[/tex]. Calculating the transpose of [tex]\( M \)[/tex] and then averaging:
[tex]\[ M^T = \left[\begin{array}{ccc} 1 & 1 & -4 \\ 3 & 0 & -4 \\ 0 & -2 & 4 \end{array}\right] \][/tex]
[tex]\[ A = \frac{1}{2} \left( \left[\begin{array}{ccc} 1 & 3 & 0 \\ 1 & 0 & -2 \\ -4 & -4 & 4 \end{array}\right] + \left[\begin{array}{ccc} 1 & 1 & -4 \\ 3 & 0 & -4 \\ 0 & -2 & 4 \end{array}\right] \right) = \frac{1}{2} \left[\begin{array}{ccc} 2 & 4 & -4 \\ 4 & 0 & -6 \\ -4 & -6 & 8 \end{array}\right] = \left[\begin{array}{ccc} 1 & 2 & -2 \\ 2 & 0 & -3 \\ -2 & -3 & 4 \end{array}\right] \][/tex]
4. Derive the skew-symmetric matrix [tex]\( B \)[/tex]:
Since [tex]\( B \)[/tex] is skew-symmetric, we can use the property of skew-symmetric matrices [tex]\( B = \frac{1}{2}(M - M^T) \)[/tex] to find [tex]\( B \)[/tex]:
[tex]\[ B = \frac{1}{2} \left( \left[\begin{array}{ccc} 1 & 3 & 0 \\ 1 & 0 & -2 \\ -4 & -4 & 4 \end{array}\right] - \left[\begin{array}{ccc} 1 & 1 & -4 \\ 3 & 0 & -4 \\ 0 & -2 & 4 \end{array}\right] \right) = \frac{1}{2} \left[\begin{array}{ccc} 0 & 2 & 4 \\ -2 & 0 & 2 \\ -4 & -2 & 0 \end{array}\right] = \left[\begin{array}{ccc} 0 & 1 & 2 \\ -1 & 0 & 1 \\ -2 & -1 & 0 \end{array}\right] \][/tex]
5. Compute [tex]\( A - B \)[/tex]:
Finally, we subtract [tex]\( B \)[/tex] from [tex]\( A \)[/tex]:
[tex]\[ A - B = \left[\begin{array}{ccc} 1 & 2 & -2 \\ 2 & 0 & -3 \\ -2 & -3 & 4 \end{array}\right] - \left[\begin{array}{ccc} 0 & 1 & 2 \\ -1 & 0 & 1 \\ -2 & -1 & 0 \end{array}\right] = \left[\begin{array}{ccc} 1 & 1 & -4 \\ 3 & 0 & -4 \\ 0 & -2 & 4 \end{array}\right] \][/tex]
Thus, the matrix [tex]\( A - B \)[/tex] is:
[tex]\[ \left[\begin{array}{ccc} 1 & 1 & -4 \\ 3 & 0 & -4 \\ 0 & -2 & 4 \end{array}\right] \][/tex]
