# Matrices

• Dec 14th 2007, 11:09 PM
lakillian
Matrices
I am going crazy. it is almost 3:00 in the morning and I have been trying to figure out a problem and I just can't get it.

If you could show me how I would so appreciate it. Iam taking an online math class and the teacher has 7 other classes and never seems to answer the questions on time. Thank you so much!

A =
• 2 0 1
• 2 -1 3
• 4 1 2

B =
• -2 1 0
• 4 3 -2
• 1 2 -1

I need to find (A + B)squared

2 + -2 0 + 1 1 + 0 is 0 1 1

and then squared is the same answer

The book has the answer as

Matrix

• 11 5 2
• 17 13 9
• 23 14 9
I know this is difficult to read but I hope you can understand it. i didn't know how to type any other way :(
• Dec 14th 2007, 11:51 PM
red_dog
$\displaystyle \displaystyle A+B=\begin{pmatrix} 2 & 0 & 1\\ 2 & -1 & 3\\ 4 & 1 & 2\end{pmatrix}+ \begin{pmatrix} -2 & 1 & 0\\ 4 & 3 & -2\\ 1 & 2 & -1\end{pmatrix}= \begin{pmatrix} 0 & 1 & 1\\ 6 & 2 & 1\\ 5 & 3 & 1\end{pmatrix}$
Then, $\displaystyle (A+B)^2=\begin{pmatrix} 11 & 5 & 2\\ 17 & 13 & 9\\ 23 & 14 & 9\end{pmatrix}$
• Dec 15th 2007, 12:00 AM
DivideBy0
$\displaystyle \left(\begin{bmatrix} {2}\;\; {0}\;\;{1}\\ {2}\;\;{-1}\;\;{3} \\ {4}\;\;{1}\;\;{2} \end{bmatrix} +\begin{bmatrix} {-2}\;\; {1}\;\;{0}\\ {4}\;\;{3}\;\;{-2}\\ {1}\;\;{2}\;\;{-1}\\ \end{bmatrix}\right)^2$

$\displaystyle =\left(\begin{bmatrix} {2-2\ }\;\; {\ 0+1\ }\;\;{\ 1+0}\\ {2+4\ }\;\;{\ -1+3\ }\;\;{\ 3-2}\\ {4+1\ }\;\;{\ 1+2\ }\;\;{\ 2-1}\\ \end{bmatrix}\right)^2$

$\displaystyle =\left(\begin{bmatrix} {0}\;\;{1}\;\;{1}\\ {6}\;\;{2}\;\;{1}\\ {5}\;\;{3}\;\;{1}\\ \end{bmatrix}\right)^2$

$\displaystyle =\begin{bmatrix} {0}\;\;{1}\;\;{1}\\ {6}\;\;{2}\;\;{1}\\ {5}\;\;{3}\;\;{1}\\ \end{bmatrix} \cdot \begin{bmatrix} {0}\;\;{1}\;\;{1}\\ {6}\;\;{2}\;\;{1}\\ {5}\;\;{3}\;\;{1}\\ \end{bmatrix}$

$\displaystyle =\begin{bmatrix} {0\cdot 0 + 1 \cdot 6 +1\cdot 5\ }\;\;{\ 0\cdot 1+1\cdot 2 + 1 \cdot 3\ }\;\;{\ 0\cdot 1+1\cdot 1 + 1\cdot 1}\\ {6\cdot 0 + 2 \cdot 6 + 1 \cdot 5\ }\;\;{\ 6\cdot 1+2\cdot 2 + 1 \cdot 3\ }\;\;{\ 6\cdot 1 +2 \cdot 1 + 1 \cdot 1}\\ {5\cdot 0+ 3 \cdot 6 + 1 \cdot 5\ }\;\;{\ 5 \cdot 1 + 3 \cdot 2 + 1 \cdot 3\ }\;\;{\ 5 \cdot 1 + 3 \cdot 1 + 1 \cdot 1}\\ \end{bmatrix}$

$\displaystyle =\begin{bmatrix} {11}\;\;{5}\;\;{2}\\ {17}\;\;{13}\;\;{9}\\ {23}\;\;{14}\;\;{9}\\ \end{bmatrix}$
• Dec 15th 2007, 07:05 AM
lakillian
Thank you so very much Divideby0
I knew there was an easy explaination. I just couldn't for the life of me figure it out. Next time I am just going to come here and post because I really do not like staying up till all hours of the night.

Thank you again really!

Lori