+0  
 
0
976
12
avatar+105 

A = 1 2   B = 2 6

      3 4        -1 0

Solve the following:

 AB + B

So heres my issue, i already solved AB. which = 0, 6, 2, 18. but now i dont know how to add the + B. 

 Jun 16, 2014

Best Answer 

 #5
avatar+118587 
+10

$$\[ \left( \begin{array}{cc}
1 & 2 \\
3 & 4
\end{array} \right)
%
\left( \begin{array}{cc}
2 & 6 \\
-1 & 0
\end{array} \right)=
%
\left( \begin{array}{cc}
0 & 6 \\
2 & 18
\end{array} \right)
\]
\[ \left( \begin{array}{cc}
0 & 6 \\
2 & 18
\end{array} \right)+
%
\left( \begin{array}{cc}
2 & 6 \\
-1 & 0
\end{array} \right)=
%
\left( \begin{array}{cc}
0 & 12 \\
1 & 18
\end{array} \right)
\]$$

 

I've never done that before - I'm quite pleased with myself!

 Jun 16, 2014
 #1
avatar+118587 
0

You just add the ones that are in the same position.

 Jun 16, 2014
 #2
avatar+105 
0

so would that be..

AB= 0 2     +   B= 2 6

      6 18              -1 0

 Jun 16, 2014
 #3
avatar+118587 
0

Hang on I'm trying to work out how to do matrix input.    Ummm

 Jun 16, 2014
 #4
avatar+105 
0

i do it like

AB11= 1X2, 2X-1 = 0

AB12= 1X6, 2X0  = 6

AB21= 3X2, 4X-1 = 2

AB22=-3X6 4X0 = 18

So im just wondering if i put it   0(AB11)   2(AB21)

                                              6(AB12)   18(AB22)

 Jun 16, 2014
 #5
avatar+118587 
+10
Best Answer

$$\[ \left( \begin{array}{cc}
1 & 2 \\
3 & 4
\end{array} \right)
%
\left( \begin{array}{cc}
2 & 6 \\
-1 & 0
\end{array} \right)=
%
\left( \begin{array}{cc}
0 & 6 \\
2 & 18
\end{array} \right)
\]
\[ \left( \begin{array}{cc}
0 & 6 \\
2 & 18
\end{array} \right)+
%
\left( \begin{array}{cc}
2 & 6 \\
-1 & 0
\end{array} \right)=
%
\left( \begin{array}{cc}
0 & 12 \\
1 & 18
\end{array} \right)
\]$$

 

I've never done that before - I'm quite pleased with myself!

Melody Jun 16, 2014
 #6
avatar+118587 
0

Thanks for the thumbs up 

 Jun 16, 2014
 #7
avatar+105 
+5

Thanks for the help! Appreciate it Melody! 

 Jun 16, 2014
 #8
avatar+128079 
+5

Good job, Melody......you're exactly correct with the nultiplication........

Note that it is easy to remember how to find a paricular element in matrix multiplication. To find element aij in the resultant matrix, we just multiply the "ith" row of the first matrix times the "jth" column of the second!!

 

 Jun 16, 2014
 #9
avatar+26364 
+6

$$\\A=\left(
\begin{array}{cc}1 & 2 \\3 & 4
\end{array}
\right)\qquad
B=\left(
\begin{array}{cc}2 & 6 \\-1 & 0
\end{array}
\right)$$

$$\boxed{AB+B=(A+\textcolor[rgb]{0,0,1}{I})B}$$

$$\mbox{Identity Matrix }\textcolor[rgb]{0,0,1}{I=\left(
\begin{array}{cc}1 & 0 \\0 & 1
\end{array}
\right)}\\$$

$$\mbox{A+\textcolor[rgb]{0,0,1}{I}} =\left(
\begin{array}{cc}1 & 2 \\3 & 4
\end{array}
\right)
+
\textcolor[rgb]{0,0,1}{\left(
\begin{array}{cc}1 & 0 \\0 & 1
\end{array}
\right)}
}
=
\left(
\begin{array}{cc}1+\textcolor[rgb]{0,0,1}{1} & 2+\textcolor[rgb]{0,0,1}{0} \\3+\textcolor[rgb]{0,0,1}{0} & 4+\textcolor[rgb]{0,0,1}{1}
\end{array}
\right)
=
\left(
\begin{array}{cc}2 & 2 \\3 & 5
\end{array}
\right)$$

$$\mbox{(A+\textcolor[rgb]{0,0,1}{I})B}=\left(
\begin{array}{cc}2 & 2 \\3 & 5
\end{array}
\right)
}
\left(
\begin{array}{cc}2 & 6 \\-1 & 0
\end{array}
\right)
=
\left(
\begin{array}{cc}\textcolor[rgb]{1,0,0}{2} & 12 \\1 & 18
\end{array}
\right)$$

$$\boxed{AB+B=
\left(
\begin{array}{cc}\textcolor[rgb]{1,0,0}{2} & 12 \\1 & 18
\end{array}
\right)
}$$

.
 Jun 16, 2014
 #10
avatar+128079 
0

I'm always impressed with your ability to use LaTex, heureka !!!   ......I wish I could learn that, but I'm probably too old !!!!

 

 Jun 16, 2014
 #11
avatar+26364 
+6

Hi Melody,

0 + 2 = 2    n o t  0

 Jun 16, 2014
 #12
avatar+118587 
0

Thank you Heureka.

I got that. Back to kindergarten for me!

 Jun 16, 2014

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