Question

Matrix Multiplication Help!!

Answer 1

\[2X + 2 \left[\begin{matrix}2 & -8 \\ -4 & 2\end{matrix}\right] = \left[\begin{matrix}4 & -6 \\ 2 & -8\end{matrix}\right]\]

Answer 2

So you're solving for the matrix X, correct? First, I'd advise factoring a 2 out of the matrix on the right hand side, and then cancelling the twos.

Answer 3

I am assuming yes?

Answer 4

Yep. Factor a 2 out of the right hand side matrix (factor it out of each individual entry, then write it out front like the two matrices on the left hand side.

Answer 5

So the individual 2 times it by both matrix's?

Answer 6

No, we want to get rid of the 2's. Factor a 2 out of the matrix on the right.

Answer 7

This sounds dumb but I am seriously lost... :(

Answer 8

\[\left[\begin{matrix}2 & 4 \\ 6 & 8\end{matrix}\right] = 2\left[\begin{matrix}1 & 2 \\ 3 & 4\end{matrix}\right]\]

Use this as an example. Factor a 2 out of the right hand side matrix.

Answer 9

\[2X + 2\left[\begin{matrix}2 & -8 \\ -4 & 2\end{matrix}\right] = 2\left[\begin{matrix}2 & -6 \\ 2 & -8\end{matrix}\right]\]

Answer 10

Not quite. When you factor out the 2, you have to divide all the terms in the matrix by 2.

Answer 11

ok so the one on the left would be \[2\left[\begin{matrix}2 & -3 \\ 1 & -4\end{matrix}\right]\]

Answer 12

Very good!

Now, since each matrix has a 2 in front of it, we can just cancel them.

Answer 13

So, now we have:
\[X + \left[\begin{matrix}2 & -8 \\ -4 & 2\end{matrix}\right] = \left[\begin{matrix}2 & -3 \\ 1 & -4\end{matrix}\right]\]

When we add matrices, we just add the entries that are in the same location. That is:

\[\left[\begin{matrix}a & b \\ c & d\end{matrix}\right] + \left[\begin{matrix}e & f \\ g & h\end{matrix}\right] = \left[\begin{matrix}a+e & b+f \\ c+g & d+h\end{matrix}\right]\]

Right?

Answer 14

And when we subtract, we do the same. We subtract entries that are in the same position. So subtract the matrix that is being added to X from both sides.

Answer 15

So \[X = \left[\begin{matrix}0 & -11 \\ -5 & -2\end{matrix}\right]\]

Answer 16

Careful with your subtraction involving negative numbers.

Answer 17

\[X = \left[\begin{matrix}2 & -3 \\ 1 & -4\end{matrix}\right] - \left[\begin{matrix}2 & -8 \\ -4 & 2\end{matrix}\right]\]

Answer 18

ok I see what I did wrong there
\[X = \left[\begin{matrix} 0 & 5 \\ -5 & 6\end{matrix}\right]\]

Answer 19

Almost. Watch your sign.

Answer 20

Where?

Answer 21

Bottom row

Answer 22

1 - (-4) = ?

-4 - 2 = ?

Answer 23

oh its +5 and -6

Answer 24

There ya go :)

Answer 25

Thank you!! could you help with one or two more? since your here? Please!

Answer 26

Sure

Answer 27

okay \[\left[\begin{matrix}-5 & -2 \\ -8 & -5\end{matrix}\right]\left[\begin{matrix}-5 & 7 \\ -9 & -5\end{matrix}\right]\]
Would you just multiply this?

Answer 28

Ok, now this is matrix multiplication.

Just follow this recipe:

\[\left[\begin{matrix}a & b \\ c & d\end{matrix}\right]\left[\begin{matrix}e & f \\ g & h\end{matrix}\right] = \left[\begin{matrix}a*e+b*g & a*f+b*h \\ c*e+d*g & c*f+d*h\end{matrix}\right]\]

Answer 29

Easy way to remember it is: first row times first column. Multiply entries in the same spots, add them up. This gives the first entry in the result.

Then, first row times second column, gives the entry next to it.

Second row times first column gives the entry in the second row, first column.

Second row times second column gives the entry in the second row, second column.

Do you see the pattern? If so, you can use it to multiply two matrices of any size.

Answer 30

(After typing that, I realize it may not seem easy to remember, but after you do it a few times, it's simple!)

Answer 31

haha ya I was a little confused why does it jump around like that when multiplying? and would the answer be

\[\left[\begin{matrix}43 & -25 \\ 85 & -31\end{matrix}\right]\]

Answer 32

It's how matrix multiplication is defined, as it's "forced" to obey certain rules to be mathematically viable. This type of multiplication is actually called an "inner product."

At the risk of confusing you, there are other types of multiplication, including something called an "outer product." But we won't cover that today :)

Answer 33

And that's what I get for my answer as well. Well done!

Answer 34

haha okay! thank you!! so what if real quick on another problem if there are two columns in one matrix and three in the second and you have to multiply?

Answer 35

The order that you do it matters, but it can be done.

Answer 36

For matrices of different sizes, you have to check a few things.

First, we'll have to multiply them in a specific order. I like to remember it as (fat)(tall)
That is: The one with more columns (usually a square matrix), then the other one.

Second, the number of columns in the first matrix has to equal the number of rows in the second matrix. If this isn't true, then you can't multiply them in this way.

Answer 37

\[\left[\begin{matrix}1 & 1 & -4 \\ 5& 6& 0\end{matrix}\right]\left(\begin{matrix}9 \\ 1\\ -7\end{matrix}\right)\]

would this be defined or undefined? and how would you know?

Answer 38

Well, we have the "fat" one first, then the skinny one, so that's good.

Next, we see that the number of columns in the first matrix is equal to the number of rows in the second matrix.

So, we can multiply them.

Answer 39

How??

Answer 40

\[\left[\begin{matrix}a & b & c\\ d & e &f\end{matrix}\right]\left(\begin{matrix}g \\ h \\ i\end{matrix}\right) = \left(\begin{matrix}a*g+b*h+c*i \\ d*g+e*h+f*i\end{matrix}\right)\]

Answer 41

So
\[\left(\begin{matrix}38 \\ 51\end{matrix}\right)\]

Answer 42

yep

Answer 43

So how would you determine if it is defined or undefined?

Answer 44

Just check that it satisfies the rules. If it does, it's defined.

Answer 45

I would say it does? then if it is how would you figure if it is
3X3
2X1
2X3?

Answer 46

The result will have the same number of rows as the first matrix, and the same number of columns as the second matrix.

You had a 2x3 and a 3x1. The result will have the same number of rows as the first matrix. In this case, 2. And the same number of columns as the second matrix. In this case, 1.

So the result is a 2x1

Answer 47

Thank you SOOOOO much!!!

Answer 48

My pleasure :)