Question

Solve the matrix problem below:

Answer 1

I don't understand the question, is your x a matrix or a number? if it's a matrix, it makes more sense to me, if it is a number, how to subtract a number by a matrix?

Answer 2

That's just it, it HAS to be a matrix for it to make sense, @Loser66 ^.^

Answer 3

thanks for your medal, help her please. I am super dummy in teaching

Answer 4

It is an upper case X, it is a matrix...

Answer 5

oh, okay...... @RichGrl20
In this setup, dealing with 2x2 matrices is very similar to just dealing with numbers, you just deal with 4 numbers at a time :D

So much like with numbers, you can add and subtract matrices on both sides of the equation.

Allow me to rewrite your problem in a more OpenStudy-friendly manner :)

\[\Large 3\color{blue}X -2\left[\begin{matrix} 1.5 & -2.4 \\ 3.6 & -0.5\end{matrix}\right]=\left[\begin{matrix} 1.2 & -3.6 \\ 0.3 & 0.4\end{matrix}\right]\]

Answer 6

So, much like with normal equations, we can bring this bit...
\[\Large 3X\color{red} {-2\left[\begin{matrix} 1.5 & -2.4 \\ 3.6 & -0.5\end{matrix}\right]}=\left[\begin{matrix} 1.2 & -3.6 \\ 0.3 & 0.4\end{matrix}\right]\]

to the right side, like so...
\[\Large 3\color{blue}X =\left[\begin{matrix} 1.2 & -3.6 \\ 0.3 & 0.4\end{matrix}\right]+2\left[\begin{matrix} 1.5 & -2.4 \\ 3.6 & -0.5\end{matrix}\right]\]

Catch me so far?

Answer 7

@terenzreignz : I understand thusfar. What do I do from there?

Answer 8

Well, an operation involving matrices, one that is called "scalar multiplication".
\[\Large 3X =\left[\begin{matrix} 1.2 & -3.6 \\ 0.3 & 0.4\end{matrix}\right]+\boxed{\color{red}2}\left[\begin{matrix} 1.5 & -2.4 \\ 3.6 & -0.5\end{matrix}\right]\]

You see this 2?
You have to sort of "distribute" it to the entries of the matrix it is multiplied with...
in other words, multiply 2 to all the numbers in the matrix next to it...
So we get...
\[\Large 3X =\left[\begin{matrix} 1.2 & -3.6 \\ 0.3 & 0.4\end{matrix}\right]+\left[\begin{matrix} 3 & -4.8 \\ 7.2 & -1\end{matrix}\right]\]

Catch me so far? :)

Answer 9

@terenzreignz So far, yes. Please continue.

Answer 10

When adding two matrices, you simply add their individual components. Can you try that now? It's simple, it just results in a matrix of the same size, where, for instance, the top-left entry is the sum of the top-left entries of the two matrices you added. So can you please add these matrices? Just tell me the entries, and where....
\[\Large 3X =\color{blue}{\left[\begin{matrix} 1.2 & -3.6 \\ 0.3 & 0.4\end{matrix}\right]+\left[\begin{matrix} 3 & -4.8 \\ 7.2 & -1\end{matrix}\right]}\]

Answer 11

@terenzreignz So... I would add, for example, 1.2 and 3 to get the top left number in the matrix?

Answer 12

Yup. Precisely. :)

Answer 13

@terenzreignz Alright... So would the matrix look like this ?
| 4.2 -8.4|
|7.5 -0.6|

Answer 14

Very good :)
\[\Large 3X =\left[\begin{matrix} 4.2 & -8.4 \\ 7.5 & -0.6\end{matrix}\right]\]

All that left... well, sidenote, if you have
3x = 15
what you do is divide both sides by 3, right? Or multiply both sides by 1/3, it doesn't make a difference

Well, in this scenario, it's no different, we divide both sides by 3, and we finally get
\[\Large X =\frac13\left[\begin{matrix} 4.2 & -8.4 \\ 7.5 & -0.6\end{matrix}\right]\]

And from here, just do scalar multiplication again, (much like I did with the 2, earlier, if you scroll up)

go ahead now :)

PS
I'm here, there's no need to keep tagging me every time :)
And call me Terence ^.^

Answer 15

Okay, thank you, Terence. ^u^

So... I think I'll end up with
|1.4 -2.8|
X= |2.5 -0.2|

Is this correct?

Answer 16

Excellent :)

Answer 17

Great! Thank you~

Answer 18

No problem :)

Answer 19

And welcome to the wide, wonderful world of matrices! Where math takes on new dimensions, like 3 by 3 and \(\mathbb{R}^n\)

Answer 20

Wonderful world of matrices?
I prefer the "Trials of Matrix Analysis"
Just you wait when they get to Matrix multiplication... heads will roll XD

But no pressure, @RichGrl20 ^.^

Answer 21

I'm still stuck wondering when letters became a part of Algebra. Lol~

Answer 22

Don't worry, I have confidence in you XD

Answer 23

Bah. Multiplication is not that bad. Now Ghram-Schmidt on a \(\ge 5\times 5\) is a pain in the anatomy!

The letters? Oh, about 3000 years ago that happened.

Answer 24

Well, I have OpenStudy to help me out. ^.^