Question

solve this q
PLZZ do not scam urgent​

Answer 1

Answer:

Step-by-step explanation:

A =    $\left[\begin{array}{ccc}13& -11\\5&8\end{array}\right]$      B =  $\left[\begin{array}{ccc}1&3\\4&2\end{array}\right]$    C = $\left[\begin{array}{ccc}-2&1\\6&3\end{array}\right]$

1) A - B + 2C =  

=       $\left[\begin{array}{ccc}13& -11\\5&8\end{array}\right]$      -    $\left[\begin{array}{ccc}1&3\\4&2\end{array}\right]$    + 2 $\left[\begin{array}{ccc}-2&1\\6&3\end{array}\right]$

=       $\left[\begin{array}{ccc}13-1&-11 -3\\5-4&8-2\end{array}\right]$     +  $\left[\begin{array}{ccc}2* - 2 & 2 * 1\\2 * 6 &2 * 3\end{array}\right]$

=        $\left[\begin{array}{ccc}12&-14\\1&6\end{array}\right]$    + $\left[\begin{array}{ccc}-4&2\\12&6\end{array}\right]$

=        $\left[\begin{array}{ccc}12 - 4&-14 +2\\1 + 12&6 + 6\end{array}\right]$

=        $\left[\begin{array}{ccc}8&-12\\13&12\end{array}\right]$

2) PT (A+B) + C = A + (B+C)

A + B =  $\left[\begin{array}{ccc}13& -11\\5&8\end{array}\right]$   +   $\left[\begin{array}{ccc}1&3\\4&2\end{array}\right]$  = $\left[\begin{array}{ccc}13+1&-11+3\\5+4&8+2\end{array}\right]$  =  $\left[\begin{array}{ccc}14&-8\\9&10\end{array}\right]$

(A + B) + C =   $\left[\begin{array}{ccc}14&-8\\9&10\end{array}\right]$  + $\left[\begin{array}{ccc}-2&1\\6&3\end{array}\right]$ =      $\left[\begin{array}{ccc}14-2&-8+1\\9+6&10+3\end{array}\right]$  = $\left[\begin{array}{ccc}12&-7\\15&13\end{array}\right]$

B+C =   $\left[\begin{array}{ccc}1&3\\4&2\end{array}\right]$   +  $\left[\begin{array}{ccc}-2&1\\6&3\end{array}\right]$   = $\left[\begin{array}{ccc}1-2&3+1\\4+6&2+3\end{array}\right]$  = $\left[\begin{array}{ccc}-1&4\\10&5\end{array}\right]$

A + (B+C) =  $\left[\begin{array}{ccc}13& -11\\5&8\end{array}\right]$  +  $\left[\begin{array}{ccc}-1&4\\10&5\end{array}\right]$ = $\left[\begin{array}{ccc}13-1&-11+4\\5+10&8+5\end{array}\right]$ = $\left[\begin{array}{ccc}12&-7\\15&13\end{array}\right] \left$

=> (A+B) + C = A + (B+C)

Hence Proved.