Question

Let \( A \) be a symmetric matrix and \( B \) be a skew symmetric matrix. If

\[ A + B = \begin{pmatrix} 1 & 3 \\ -2 & 5 \end{pmatrix}, \] then \( A - B \) is equal to:

• A. \[ \begin{pmatrix} 1 & 3 \\ -2 & 5 \end{pmatrix} \]
• B. \[ \begin{pmatrix} 1 & -2 \\ 3 & -5 \end{pmatrix} \]
• C. \[ \begin{pmatrix} 1 & -2 \\ -3 & -5 \end{pmatrix} \]
• D. \[ \begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix} \]
• E. \[ \begin{pmatrix} -1 & 3 \\ 2 & -5 \end{pmatrix} \]

Hint:

To find \( A - B \) when \( A + B \) is given, simply subtract the matrices using the known properties of symmetric and skew-symmetric matrices. Use the relationship \( A + B + A - B = 2A \) to help.

Answer 1

The Correct Option is D

Solution:

Given that \( A \) is a symmetric matrix and \( B \) is a skew-symmetric matrix, we know the following properties:

- A symmetric matrix satisfies \( A^T = A \).
- A skew-symmetric matrix satisfies \( B^T = -B \).

We are given that:

\[ A + B = \begin{pmatrix} 1 & 3 \\ -2 & 5 \end{pmatrix} \] We need to find \( A - B \).

Let's define:

\[ A = \begin{pmatrix} a & b \\ b & d \end{pmatrix}, \quad B = \begin{pmatrix} 0 & e \\ -e & 0 \end{pmatrix} \] Since \( A \) is symmetric, the off-diagonal elements are equal, and for \( B \), the diagonal elements are zero, and the off-diagonal elements are negatives of each other.

Now, from the equation \( A + B = \begin{pmatrix} 1 & 3 \\ -2 & 5 \end{pmatrix} \), we can write:

\[ \begin{pmatrix} a & b \\ b & d \end{pmatrix} + \begin{pmatrix} 0 & e \\ -e & 0 \end{pmatrix} = \begin{pmatrix} 1 & 3 \\ -2 & 5 \end{pmatrix} \] This gives us the following system of equations:

1. \( a = 1 \)
2. \( b + e = 3 \)
3. \( b - e = -2 \)
4. \( d = 5 \)

From equations 2 and 3, we can solve for \( b \) and \( e \):

\[ b + e = 3 \quad \text{and} \quad b - e = -2 \] Adding these two equations gives:

\[ 2b = 1 \quad \Rightarrow \quad b = \frac{1}{2} \] Substitute \( b = \frac{1}{2} \) into \( b + e = 3 \):

\[ \frac{1}{2} + e = 3 \quad \Rightarrow \quad e = \frac{5}{2} \] Now, we can find \( A - B \):

\[ A - B = \begin{pmatrix} a & b \\ b & d \end{pmatrix} - \begin{pmatrix} 0 & e \\ -e & 0 \end{pmatrix} = \begin{pmatrix} 1 & \frac{1}{2} \\ \frac{1}{2} & 5 \end{pmatrix} - \begin{pmatrix} 0 & \frac{5}{2} \\ -\frac{5}{2} & 0 \end{pmatrix} \] This gives:

\[ A - B = \begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix} \] Thus, the correct answer is option (D), \( \begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix} \).