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

Correct Answer is : [000000.50−0.50]{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0.5 \\ 0 & -0.5 & 0{bmatrix}00000−0.500.50

{bmatrix}0 & 0.5 & 0 \\ -0.5 & 0 & 0 \\ 0 & 0 & 0{bmatrix}

{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0.5 \\ 0 & -0.5 & 0{bmatrix}

{bmatrix}0 & -0.5 & 0.5 \\ 0.5 & 0 & 0 \\ -0.5 & 0 & 0{bmatrix}

{bmatrix}0 & 0.5 & 0\\ -0.5 & 0 & 0.5\\ 0 & -0.5 & 0{bmatrix}

Correct Answer is : [000000.50−0.50]{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0.5 \\ 0 & -0.5 & 0{bmatrix}00000−0.500.50

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TS EAMCET 2017 Code A Solved Paper

Section: Mathematics

Question : 8 of 160

Marks: +1, -0

A=\begin{bmatrix}1 & 0 & 1 \\ 0 & 2 & 0 \\ 1 & -1 & 4\end{bmatrix},

A=B+C,B=B^T

C=-C^T

C=

Solution:

Consider the given matrix A.

A=\begin{bmatrix}1 & 0 & 1 \\ 0 & 2 & 0 \\ 1 & -1 & 4\end{bmatrix}

The transpose of matrix A is,

A^{T}=\begin{bmatrix}1 & 0 & 1 \\ 0 & 2 & -1 \\ 1 & 0 & 4\end{bmatrix}

The value of

(A+A^{T})

is calculated as

A+A^{T}=\begin{bmatrix}2 & 0 & 2 \\ 0 & 4 & -1 \\ 2 & -1 & 8\end{bmatrix}

The value of

(A-A^{T})

is calculated as

A-A^{T}=\begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0\end{bmatrix}

The matrix A can be calculated as,

A=\frac{1}{2}(A+A^{T})+\frac{1}{2}(A-A^{T})

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

The matrix

C

is,

C=\begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0.5 \\ 0 & -0.5 & 0\end{bmatrix}=-C^{T}

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