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.
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)
\[(1234)
I've never done that before - I'm quite pleased with myself!
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!!
A=(1234)B=(26−10)
AB+B=(A+I)B
Identity Matrix I=(1001)
\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)
AB+B=(212118)
.