Q48 / 160 · TS EAMCET 2017 · Mathematics
If $A=\left[\begin{array}{ccc}1 & 0 & 1 \\ 0 & 2 & 0 \\ 1 & -1 & 4\end{array}\right], \quad A=B+C, B=B^T$ and $C=-C^T$, then $C=$
- A$\left[\begin{array}{ccc}0 & 0.5 & 0 \\ -0.5 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$
- Correct answerB$\left[\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & 0.5 \\ 0 & -0.5 & 0\end{array}\right]$
- C$\left[\begin{array}{ccc}0 & -0.5 & 0.5 \\ 0.5 & 0 & 0 \\ -0.5 & 0 & 0\end{array}\right]$
- D$\left[\begin{array}{ccc}0 & 0.5 & 0 \\ -0.5 & 0 & 0.5 \\ 0 & -0.5 & 0\end{array}\right]$
Solution
We have,
$$
\begin{aligned}
A & =\left[\begin{array}{ccc}
1 & 0 & 1 \\
0 & 2 & 0 \\
1 & -1 & 4
\end{array}\right] \\
A^T & =\left[\begin{array}{ccc}
1 & 0 & 1 \\
0 & 2 & -1 \\
1 & 0 & 4
\end{array}\right] \\
A+A^T & =\left[\begin{array}{ccc}
2 & 0 & 2 \\
0 & 4 & -1 \\
2 & -1 & 8
\end{array}\right] A-A^T=\left[\begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & 1 \\
0 & -1 & 0
\end{array}\right] \\
A & =\frac{1}{2}\left(A+A^T\right)+\frac{1}{2}\left(A-A^T\right) \\
A & =\left[\begin{array}{ccc}
1 & 0 & 1 \\
0 & 2 & -0.5 \\
1 & -0.5 & 4
\end{array}\right]+\left[\begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & 0.5 \\
0 & -0.5 & 0
\end{array}\right] \\
A & =B+C
\end{aligned}
$$
$$
C=\left[\begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & 0.5 \\
0 & -0.5 & 0
\end{array}\right]=-C^T
$$
Hence, the correct answer is option: 'B'.