Question
Download Solution PDFIf the matrix A = \(\left[ {\begin{array}{*{20}{c}} 0&1&{ - 1} \\ 4&{ - 3}&4 \\ 3&{ - 3}&4 \end{array}} \right]\) = B + C, where B is symmetric and C is a skew-symmetric matrix, find the matrix B
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
Given, the matrix A = \(\left[ {\begin{array}{*{20}{c}} 0&1&{ - 1} \\ 4&{ - 3}&4 \\ 3&{ - 3}&4 \end{array}} \right]\)
A = B + C
Here matrix A is expressed as the sum of symmetric and skew-symmetric matrices.
Then, \(B=\frac{1}{2}(A+A^{T}) \ and\ C=\frac{1}{2}(A-A^{T})\)
Where B is symmetric and C is a skew-symmetric matrix.
To Find: The matrix B
A = \(\left[ {\begin{array}{*{20}{c}} 0&1&{ - 1} \\ 4&{ - 3}&4 \\ 3&{ - 3}&4 \end{array}} \right]\)
\(A^{T}=\begin{bmatrix} 0 & 4& 3 \\ 1 & -3 & -3 \\ -1& 4& 4 \\ \end{bmatrix}\)
\(A+ A^{T}=\left[ {\begin{array}{*{20}{c}} 0&1&{ - 1} \\ 4&{ - 3}&4 \\ 3&{ - 3}&4 \end{array}} \right] +\begin{bmatrix} 0 & 4& 3 \\ 1 & -3 & -3 \\ -1& 4& 4 \\ \end{bmatrix}\)
\(A+ A^{T}=\begin{bmatrix} 0 &5 & 2 \\ 5& -6 & 1 \\ 2& 1 & 8 \\ \end{bmatrix}\)
\(B=\frac{1}{2}(A+ A^{T})=\frac{1}{2}\begin{bmatrix} 0 &5 & 2 \\ 5& -6 & 1 \\ 2& 1 & 8 \\ \end{bmatrix}\)
Last updated on Jun 19, 2025
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