ADDITION OF MATRICES

\[\color {purple} {Example\ 1\ :}\ \color{red}{If}\ A =\begin{bmatrix} 1 & 2 & 7 \\ 0 & 4 & 5 \\ 3 & 1 & 6 \\ \end{bmatrix}\ ,\ B =\begin{bmatrix} 1 & 3 & 1 \\ 2 & 4 & 0 \\ 1 & 7 & 5 \\ \end{bmatrix}\ \hspace{5cm}\]

\[\color {red}{Find\ A + B}\ \hspace{10cm}\]

\[\color {blue}{Solution:}\ A + B=\begin{bmatrix} 1 + 1 & 2 + 3 & 7 + 1 \\ 0 + 2 & 4 + 4 & 5 + 0 \\ 3 + 1 & 1 + 7 & 6 + 5 \\ \end{bmatrix}\ \hspace{6cm}\]

\[A + B=\begin{bmatrix} 2 & 5 & 8 \\ 2 & 8 & 5 \\ 4 & 8 & 11 \\ \end{bmatrix}\ \hspace{7cm}\]

\[\color {purple} {Example\ 2:}\ \color {red}{If}\ A =\begin{bmatrix} 2 & -3 & 8 \\ 21 & 6 & -6 \\ 4 & -33 & 19 \\ \end{bmatrix}\ ,\ B =\begin{bmatrix} 1 & -29 & -8 \\ 2 & 0 & 3 \\ 17 & 15 & 4 \\ \end{bmatrix}\ \hspace{7cm}\]\[\color {red}{Prove\ that\ (A\ +\ B)^T\ =\ A^T\ +\ B^T}\ \hspace{5cm}\]

\[\color {blue}{Solution:}\ Given\ A =\begin{bmatrix} 2 & -3 & 8 \\ 21 & 6 & -6 \\ 4 & -33 & 19 \\ \end{bmatrix}\ ,\ B =\begin{bmatrix} 1 & -29 & -8 \\ 2 & 0 & 3 \\ 17 & 15 & 4 \\ \end{bmatrix}\ \hspace{5cm}\]

\[A + B=\begin{bmatrix} 2 + 1 & -3 – 29 & 8 – 8 \\ 21 + 2 & 6 + 0 & -6 + 3 \\ 4 + 17 & -33 + 15 & 19 + 4 \\ \end{bmatrix}\ \hspace{6cm}\]

\[A + B=\begin{bmatrix} 3 & -32 & 0 \\ 23 & 6 & -3\\ 21 & -18 & 23\\ \end{bmatrix}\ \hspace{6cm}\]

\[(A + B)^T\ =\ \begin{bmatrix} 3 & 23 & 21 \\ -32 & 6 & – 18 \\ 0 & -3 & 23\\ \end{bmatrix}\ \hspace{2cm}\ ——– (1)\]

\[A^T =\begin{bmatrix} 2 & 21 & 4 \\ -3 & 6 & -33 \\ 8 & -6 & 19 \\ \end{bmatrix}\ \hspace{10cm}\]

\[B^T =\begin{bmatrix} 1 & 2 & 17 \\ -29 & 0 & 15 \\ – 8 & 3 & 4 \\ \end{bmatrix}\ \hspace{10cm}\]

\[A^T\ +\ B^T\ =\begin{bmatrix} 2 + 1 & 21\ +\ 2 & 4\ +\ 17\\ -3\ -\ 29 & 6 + 0 & -33 + 15 \\ 8 – 8 & -6 + 3 & 19 + 4 \\ \end{bmatrix}\ \hspace{6cm}\]

\[A^T\ +\ B^T\ =\begin{bmatrix} 3 & 23 & 21\\ -32 & 6 & – 18 \\ 0 & -3 & 23 \\ \end{bmatrix}\ \hspace{2cm}\ ———- (2)\]

\[From\ (1)\ and\ (2), It\ is\ concluded\ that (A\ +\ B)^T\ =\ A^T\ +\ B^T\]