# Compute the following a) begin{bmatrix}-5 & -4&3&-10&-3&6 6&-10&5&9&4&-1 end{bmatrix}+begin{bmatrix}-7 & 3&10&0&8&8 8&0&4&-3&-8&0 end{bmatrix}b) -5begin{bmatrix}8 & -10&7 0 & -9&710&-5&-101&5&-10 end{bmatrix}c)begin{bmatrix}3 & 0&-8 6 & -4&-26&0&-8-9&-7&-7 end{bmatrix}^T

Compute the following
a) $$\begin{bmatrix}-5 & -4&3&-10&-3&6 \\6&-10&5&9&4&-1 \end{bmatrix}+\begin{bmatrix}-7 & 3&10&0&8&8 \\8&0&4&-3&-8&0 \end{bmatrix}$$
b) $$-5\begin{bmatrix}8 & -10&7 \\0 & -9&7\\10&-5&-10\\1&5&-10 \end{bmatrix}$$
c)$$\begin{bmatrix}3 & 0&-8 \\6 & -4&-2\\6&0&-8\\-9&-7&-7 \end{bmatrix}^T$$

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Gennenzip
Part (a)
Given:
$$\begin{bmatrix}-5 & -4&3&-10&-3&6 \\6&-10&5&9&4&-1 \end{bmatrix}+\begin{bmatrix}-7 & 3&10&0&8&8 \\8&0&4&-3&-8&0 \end{bmatrix}$$
By using matrix addition
$$\begin{bmatrix}-5 & -4&3&-10&-3&6 \\6&-10&5&9&4&-1 \end{bmatrix}+\begin{bmatrix}-7 & 3&10&0&8&8 \\8&0&4&-3&-8&0 \end{bmatrix}$$
$$=\begin{bmatrix}-5+(-7) & -4+3&3+10&-10+0&-3+8&6+8 \\6+8&-10+0&5+4&9+(-3)&4+(-8)&-1+0 \end{bmatrix}$$
$$\begin{bmatrix}-12 & -1&13&-10&5&14 \\14&-10&9&6&-4&-1 \end{bmatrix}$$
Part (b)
Given:
$$-5\begin{bmatrix}8 & -10&7 \\0 & -9&7\\10&-5&-10\\1&5&-10 \end{bmatrix}$$
Using scalar multiplication of matrices.
$$-5\begin{bmatrix}8 & -10&7 \\0 & -9&7\\10&-5&-10\\1&5&-10 \end{bmatrix}=\begin{bmatrix}-5(8) & -5(-10)&-5(7) \\-5(0) & -5(-9)&-5(7)\\-5(10)&-5(-5)&-5(-10)\\-5(1)&-5(5)&-5(-10) \end{bmatrix}$$
$$\begin{bmatrix}-40 & 50&-35 \\0 & 45&-35\\-50&25&50\\-5&-25&50 \end{bmatrix}$$
Part (c)
Given:
$$\begin{bmatrix}3 & 0&-8 \\6 & -4&-2\\6&0&-8\\-9&-7&-7 \end{bmatrix}^T$$
Transpose of a matrix is obtained by flipping matrix along diagonal,
$$\begin{bmatrix}3 & 0&-8 \\6 & -4&-2\\6&0&-8\\-9&-7&-7 \end{bmatrix}^T=\begin{bmatrix}3 & 6&6&-9 \\0 & -4&0&-7\\-8&-2&-8&-7 \end{bmatrix}$$