Thread: Matrices

I need help with this problem:

Find 3A+B where A =
{3 5}
{-3 2}
and B=
{1 1}
{3 3}

Possible solutions are:
{10 16}
{12 9 }
---------
{10 15}
{-6 9 }
---------
{10 16}
{-6 9 }
---------
{10 16}
{-9 9 }

Originally Posted by gretchen

I need help with this problem:

Find 3A+B where A =
{3 5}
{-3 2}
and B=
{1 1}
{3 3}

Possible solutions are:
{10 16}
{12 9 }
---------
{10 15}
{-6 9 }
---------
{10 16}
{-6 9 }
---------
{10 16}
{-9 9 }

Code:

 
Find 3A+B where A = 
{ 3 5} 
{-3 2} 
and B=
{ 1 1}
{ 3 3}
 
3 [ 3 5 ] + [ 1 1 ] = [ 3.3+1 3.5+1 ] = [ 10 16 ]
  [-3 2 ]   [ 3 3 ]   [-3.3+3 3.2+3 ]   [-6   9 ]

RonL

Hello, gretchen!

Didn't they teach you anything about Matrix Arithmetic?

Find $\displaystyle 3A+B$ where $\displaystyle A \:= \:\begin{bmatrix}3 & 5 \\ \text{-}3 & 2\end{bmatrix}$ and $\displaystyle B = \begin{bmatrix}1 & 1 \\ 3 & 3 \end{bmatrix}$

Possible solutions are: .$\displaystyle \begin{bmatrix}10 & 16 \\ 12 & 9\end{bmatrix}\quad \begin{bmatrix}10 & 15 \\ \text{-}6 & 9\end{bmatrix}\quad\begin{bmatrix}10 & 16 \\\text{-}6 & 9\end{bmatrix}\quad\begin{bmatrix}10 & 16 \\ \text{-}9 & 9\end{bmatrix}$

$\displaystyle 3A + B \;=\;3\begin{bmatrix}3 & 5\\ \text{-}3&2\end{bmatrix} + \begin{bmatrix}1 & 1 \\ 3 & 3\end{bmatrix} \;=\;\begin{bmatrix}3(3) & 3(5) \\ 3(\text{-}3) & 3(2)\end{bmatrix} + \begin{bmatrix}1 & 1 \\ 3 & 3\end{bmatrix}$

. . . . . . $\displaystyle = \;\begin{bmatrix}9 & 15 \\ -9 & 6\end{bmatrix} + \begin{bmatrix}1 & 1 \\ 3 & 3\end{bmatrix} \;= \;\begin{bmatrix}9 + 1 & 15 + 1 \\ \text{-}9 + 3 & 6 + 3\end{bmatrix}$

. . . . . . $\displaystyle = \;\begin{bmatrix}10 & 16 \\ \text{-}6 & 9\end{bmatrix}$ . . . third answer choice