# Matrices Question

• Oct 29th 2006, 01:22 PM
jenneedshelp
Matrices Question
Hi. I have a couple questions about matrices that I am unable to solve.
if:
A= [2 -1 6] and B+ [1 -2 3]
-5 8 -10 6 10 -3
-9 2 1

find, if possible:

A) A squared B) B squared.

A) not possible

B) [-38 -16 12]
93 82 -15
-6 40 -32

but I have no clue how to get that.

Jenn
• Oct 29th 2006, 01:50 PM
topsquark
Quote:

Originally Posted by jenneedshelp
Hi. I have a couple questions about matrices that I am unable to solve.
if:
A= [2 -1 6] and B+ [1 -2 3]
-5 8 -10 6 10 -3
-9 2 1

find, if possible:

A) A squared B) B squared.

A) not possible

B) [-38 -16 12]
93 82 -15
-6 40 -32

but I have no clue how to get that.

Jenn

$\displaystyle A = \left ( \begin{array}{ccc}2 & -1 & 6 \\ -5 & 8 & -10 \end{array} \right )$

Note that A is a 3 x 2 matrix. If we can square this matrix we will be multiplying a 3 x 2 matrix by a 3 x 2 matrix. But the "2" of the first matrix (the number of rows) must be equal to the "3" of the second matrix (the number of columns). Since these aren't equal we can't do A x A. Thus A cannot be squared.

$\displaystyle B = \left ( \begin{array}{ccc}1 & -2 & 3 \\ 6 & 10 & -3 \\ -9 & 2 & 1 \end{array} \right )$

Note that B is a 3 x 3 matrix, so since the number of rows is equal to the number of columns, we can square this.

I presume you know the rules of matrix multiplication, but I'll run through a couple of the entries for you, since you seemed a bit confused about the answer.

I'm going to calculate the 1,1 component of B^2. We go across the first (top) row of the first matrix (B): $\displaystyle \left ( \begin{array}{ccc}1 & -2 & 3 \end{array} \right )$ and go down the first (leftmost) column of the second matrix (B): $\displaystyle \left ( \begin{array}{c}1 \\ 6 \\ -9 \end{array} \right )$. Now we form the sum:
$\displaystyle 1 \cdot 1 + -2 \cdot 6 + 3 \cdot -9 = -38$

This forms the 1,1 component of B^2 and so sits in the top left corner.

Let me do the 3,2 component. This is the bottom row middle column component. We take the 3rd row of B: $\displaystyle \left ( \begin{array}{ccc}-9 & 2 & 1 \end{array} \right )$ and the second column of B: $\displaystyle \left ( \begin{array}{c}-2 \\ 10 \\ 2 \end{array} \right )$. Now we form the sum:
$\displaystyle -9 \cdot -2 + 2 \cdot 10 + 1 \cdot 2 = 40$

That should give a good refresher on how to multiply the matrices. The finished problem ends up being:
$\displaystyle B^2 = \left ( \begin{array}{ccc}1 & -2 & 3 \\ 6 & 10 & -3 \\ -9 & 2 & 1 \end{array} \right ) \left ( \begin{array}{ccc}1 & -2 & 3 \\ 6 & 10 & -3 \\ -9 & 2 & 1 \end{array} \right )$ = $\displaystyle \left ( \begin{array}{ccc}-38 & -16 & 12 \\ 93 & 82 & -15 \\ -6 & 40 & -32 \end{array} \right )$

-Dan
• Oct 29th 2006, 03:17 PM
jenneedshelp
Thank you so much! I can't believe I didn't notice that. It makes complete sense now!

Thanks :)