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Mathematics 14 Online
OpenStudy (ellamoyseyuk):

Complete the multiplication: 2EA? E=[1,2] A=[3,0,2,-1] the answer choices:. a) [7 −2] B. [14 −4] C. [14 4]

OpenStudy (ellamoyseyuk):

@jim_thompson5910 can u help me

OpenStudy (ellamoyseyuk):

i think the answer is b is that correct?

jimthompson5910 (jim_thompson5910):

one moment

jimthompson5910 (jim_thompson5910):

The given matrices are these? \[\Large E = \begin{bmatrix}1\\2\end{bmatrix}\] \[\Large A = \begin{bmatrix}3 & 0\\2 & -1\end{bmatrix}\] Is that correct?

jimthompson5910 (jim_thompson5910):

or is matrix E supposed to look like this? \[\Large E = \begin{bmatrix}1 & 2\end{bmatrix}\]

OpenStudy (ellamoyseyuk):

e is supposed to be like the second answer you put

jimthompson5910 (jim_thompson5910):

Ok so we have \[\Large E = \begin{bmatrix}1 & 2\end{bmatrix}\] \[\Large A = \begin{bmatrix}3 & 0\\2 & -1\end{bmatrix}\]

jimthompson5910 (jim_thompson5910):

What results when you multiply E*A ?

jimthompson5910 (jim_thompson5910):

you should get a 1x2 matrix

jimthompson5910 (jim_thompson5910):

are you familiar with matrix multiplication?

OpenStudy (ellamoyseyuk):

7-2 = 2(7-2) = 14-4

OpenStudy (ellamoyseyuk):

yes i am

jimthompson5910 (jim_thompson5910):

You should find that \[\Large E*A = \begin{bmatrix}1 & 2\end{bmatrix}*\begin{bmatrix}3 & 0\\2 & -1\end{bmatrix}\] \[\Large E*A = \begin{bmatrix}1*3+2*2 & 1*0+2*(-1)\end{bmatrix}\] \[\Large E*A = \begin{bmatrix}7 & -2\end{bmatrix}\] Double both sides to get \[\Large E*A = \begin{bmatrix}7 & -2\end{bmatrix}\] \[\Large 2*E*A = 2*\begin{bmatrix}7 & -2\end{bmatrix}\] \[\Large 2*E*A = \begin{bmatrix}2*7 & 2*(-2)\end{bmatrix}\] \[\Large 2*E*A = \begin{bmatrix}14 & -4\end{bmatrix}\]

jimthompson5910 (jim_thompson5910):

I think that's what you wrote above. Or part of it anyway

OpenStudy (ellamoyseyuk):

yeah thanks

jimthompson5910 (jim_thompson5910):

sure thing

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