1. Tardigrade
  2. Question
  3. Mathematics
  4. If A = [1&0&1 0&2&0 1&-1&4], A = B + C , B = BT and C = - CT , then C =

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Q. If $A = \begin{bmatrix}1&0&1\\ 0&2&0\\ 1&-1&4\end{bmatrix}, A = B + C , B = B^{T} $ and $C = - C^{T} $, then $ C = $

TS EAMCET 2017

Solution:

$A =\begin{bmatrix}1 & 0 & 1 \\ 0 & 2 & 0 \\ 1 & -1 & 4\end{bmatrix} $
$A^{T} =\begin{bmatrix}1 & 0 & 1 \\ 0 & 2 & -1 \\ 1 & 0 & 4\end{bmatrix} $
$A+A^{T} =\begin{bmatrix}2 & 0 & 2 \\ 0 & 4 & -1 \\ 2 & -1 & 8\end{bmatrix} A-A^{T}=\begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0\end{bmatrix} $
$A =\frac{1}{2}\left(A+A^{T}\right)+\frac{1}{2}\left(A-A^{T}\right) $
$A=\begin{bmatrix}1 & 0 & 1 \\ 0 & 2 & -0.5 \\ 1 & -0.5 & 4\end{bmatrix}+\begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0.5 \\ 0 & -0.5 & 0\end{bmatrix}$
$A=B+C$
$C=\begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0.5 \\ 0 & -0.5 & 0\end{bmatrix}=-C^{T}$