Question

Let \(S\) be a symmetric matrix obtained from \[ A= \begin{bmatrix} 1 & 2 & -3\\ 2 & -2 & 1\\ 3 & 1 & -1 \end{bmatrix} \] and \(T\) be a skew-symmetric matrix obtained from \[ B= \begin{bmatrix} 4 & 2 & 0\\ 1 & -1 & 3\\ 0 & 2 & -3 \end{bmatrix} \] If trace of \(S=-4\) and the non-zero elements of \(T\) are \(-1,1\), then \(S+T=\)

• A. \[ \begin{bmatrix} 2 & 5 & 0\\ 3 & -4 & 3\\ 0 & 1 & -2 \end{bmatrix} \]
• B. \[ \begin{bmatrix} 2 & 5 & 0\\ 3 & 3 & 3\\ 0 & 1 & 2 \end{bmatrix} \]
• C. \[ \begin{bmatrix} 2 & 5 & 0\\ 3 & -3 & 3\\ 0 & 1 & -2 \end{bmatrix} \]
• D. \[ \begin{bmatrix} 2 & -5 & 0\\ 3 & -3 & 3\\ 0 & 1 & -2 \end{bmatrix} \]

Hint:

For any square matrix \(A\): \[ \text{Symmetric part}=\frac{A+A^T}{2} \] \[ \text{Skew-symmetric part}=\frac{A-A^T}{2} \] These formulas are extremely important in matrix decomposition problems.

Answer 1

The Correct Option is A

Concept: For any square matrix \(A\), \[ A=\frac{A+A^T}{2}+\frac{A-A^T}{2} \] where:

• \(\dfrac{A+A^T}{2}\) is symmetric

• \(\dfrac{A-A^T}{2}\) is skew-symmetric
We use these standard decompositions to obtain matrices \(S\) and \(T\).

Step 1: Find the symmetric matrix \(S\).
Given: \[ A= \begin{bmatrix} 1 & 2 & -3\\ 2 & -2 & 1\\ 3 & 1 & -1 \end{bmatrix} \] Transpose of \(A\): \[ A^T= \begin{bmatrix} 1 & 2 & 3\\ 2 & -2 & 1\\ -3 & 1 & -1 \end{bmatrix} \] Now, \[ S=\frac{A+A^T}{2} \] \[ S= \frac{1}{2} \begin{bmatrix} 2 & 4 & 0\\ 4 & -4 & 2\\ 0 & 2 & -2 \end{bmatrix} \] \[ S= \begin{bmatrix} 1 & 2 & 0\\ 2 & -2 & 1\\ 0 & 1 & -1 \end{bmatrix} \] Trace: \[ 1+(-2)+(-1)=-2 \]

Step 2: Find the skew-symmetric matrix \(T\).
Given: \[ B= \begin{bmatrix} 4 & 2 & 0\\ 1 & -1 & 3\\ 0 & 2 & -3 \end{bmatrix} \] Transpose: \[ B^T= \begin{bmatrix} 4 & 1 & 0\\ 2 & -1 & 2\\ 0 & 3 & -3 \end{bmatrix} \] Now, \[ T=\frac{B-B^T}{2} \] \[ T= \frac{1}{2} \begin{bmatrix} 0 & 1 & 0\\ -1 & 0 & 1\\ 0 & -1 & 0 \end{bmatrix} \] Thus the non-zero entries are \(-1\) and \(1\).

Step 3: Compute \(S+T\).
Adding the matrices: \[ S+T= \begin{bmatrix} 2 & 5 & 0\\ 3 & -4 & 3\\ 0 & 1 & -2 \end{bmatrix} \]

Step 4: Choose the correct option.
Hence the correct answer is: \[ \boxed{ \begin{bmatrix} 2 & 5 & 0\\ 3 & -4 & 3\\ 0 & 1 & -2 \end{bmatrix} } \]