# Math Help - Matrix

1. ## Matrix

1)There are 4 switches numbered 1 through 4. Is it possible to set the switches such that:
i) Switch 3 is off.
ii) An odd number of switches 1, 2 and 3 are OFF.
iii) An odd number of switches 2, 3 and 4 are ON.
iv) An even number of switches 1 and 3 are ON.
Switch 1: ?
Switch 2: ?
Switch 3: ?
Switch 4:?

2)Find a non-trivial solution to the following homogeneous system of equations:
(4+i)x −1y = 0
(−18+4i)x +(4−2i)y = 0

3)A student finds \$1.05 in dimes, nickels
and pennies. If there are 17 coins in all, how many coins of each
type can he have?
This is what i got..
0.10 x + 0.05 y + 0.01 z = 1.05
x + y + z = 17 ,

The first question, I have no idea how to do it....
For second, I tired to use matrix, but the answer doesn't make sense to me. the answer have to be in a+bi form...
The third , I don't know what to do next, because there is infinite solution... but how can I find all the solution?
ty!

2. Generally, you shouldn't post more than two problems in a thread, as per Rule 8.

For # 1, I would just write out a truth table, and check to see which rows, if any, satisfy all the conditions.

For # 2, can you post your work so far?

For # 3, I would recommend working in units of cents, not dollars. Then impose the condition that x, y, and z must all be integers. That might narrow down the solution space a bit for you.

3. I got #3 now
#1 ,
switch 1 2 3 4
ii)------1 1 0..=0
iii)-----...1 0 0=1
iv)-----1....0...=1
switch 1) on
2)on
3) off
4)off
is that right?
for 2,
i use matrix
(4+i)x −1y = 0
(−18+4i)x +(4−2i)y = 0

4+i 1 0
-9+2i 2-i 0

I got
4+i 1 0
17 i 0
and then i tried to plug it into the x+y+z form, and solve for both equation, i got -4iy+18y=0, which doesn't make sense to me....

4. #1: Your proposed switch configuration does not satisfy condition iv).

#2: You did not correctly produce the augmented matrix from the simultaneous equations. That is,

4+i 1 0
-9+2i 2-i 0

is incorrect. Look more closely.

5. -9+2i 2-i 0
yea, I divided the row by 2...I thought i could do that....?

iv) An even number of switches 1 and 3 are ON.
1+0=1

why isn't that satisfy?

6. # 1 You have on switch 1 on and switch 3 off. Therefore, one switch of switches 1 and 3 are on. One is an odd number. Therefore, condition iv is not satisfied.

# 2 You can divide by 2. In fact, your second row of the augmented matrix contains no errors. The error is in the first row of the matrix. That is,

(4+i)x −1y = 0 does not go into the augmented matrix as

4+i 1 0. You've made an error there. Do you see it?

7. I am confused...
I don't think i understand the question
the even number, is it mean 1 plus 3 is on, or 1 and 3 is on , on? or they are saying that how many switches are there?, there is two switches , so the number is even?

Yea , I see the mistake... I will redo it ,ty!

8. They're requiring the following:

Get yourself tunnel vision, and focus in on switches 1 and 3. That's all you're looking at. The requirement iv) is that an even number of those switches are on. There are only two ways to do this: Switches 1 and 3 are both off (zero of those switches are on, and zero is an even number), or Switches 1 and 3 are both on (two of those switches are on, and two is an even number). Does that clear things up for you?

How is # 3 going?

9. I think it's make sense to me... ,
since switch 3 is off, then switch 1 must be off, two switch is off, two is even
Yea i got it

#3
I multiplied 0.10 x + 0.05 y + 0.01 z = 1.05 by 100, then use the matrix, and backward substitution,
then I got something like x=?+?z and y=?+?z, I notice that z can only be 0,5 or 10, 15...otherwise x and y will be a fraction, and z can only be 0 or 5, then i just solve for x and y
it is much easier when i multiplied by 100, ty for for telling me

10. So you've solved all the problems, then? What are your answers?

11. No... i still didn't get #2
this is what i got..
-9+2i 2-i 0
4+i -1 0

row 1 minus two time row 2
-17 4-i 0
4+i -1 0

1 (4+i)/(-17) 0
1 (-1)/(4+i) 0

Then I tried to minus row 1from row 2
then i got

1 (4+i)/(-17) 0
0 0 0

so y=s , x=s(4-i)/17
but i think this is wrong ,cause the answer have to in a+bi form

#3
x=4,y 13,z 0
x 8,y 4,z 5
#1
1off
2off
3off
4on

With respect to # 3, I would say you can rule out the first solution, because of the wording of the problem. It says there are dimes, nickels, AND pennies.

For # 2, I don't think you understand complex arithmetic correctly. Let me do one or two computations for you so you get the idea.

$\left[\begin{matrix}4+i &-1\\-9+2i &2-i\end{matrix}\Big|
\begin{matrix}0\\0\end{matrix}\right]$

First step: need a 1 in the 1,1 position. Divide first row by $4+i$ to obtain

$\left[\begin{matrix}1 &\frac{-1}{4+i}\\-9+2i &2-i\end{matrix}\Big|
\begin{matrix}0\\0\end{matrix}\right]\to\left[\begin{matrix}1 &\frac{-1}{4+i}\cdot\frac{4-i}{4-i}\\-9+2i &2-i\end{matrix}\Big|
\begin{matrix}0\\0\end{matrix}\right]\to\left[\begin{matrix}1 &\frac{-4+i}{17}\\-9+2i &2-i\end{matrix}\Big|
\begin{matrix}0\\0\end{matrix}\right].$

So here you can see that division by a complex number will require multiplication by a fraction that is equal to 1: a special fraction. In general,

$\dfrac{a+bi}{c+di}=\dfrac{a+bi}{c+di}\cdot\dfrac{c-di}{c-di}=\dfrac{ac-adi+bci+bd}{c^{2}+d^{2}}=\dfrac{ac+bd+(bc-ad)i}{c^{2}+d^{2}}.$

Don't memorize this formula. Instead, remember the procedure: multiply top and bottom by the complex conjugate, and then foil it out.

Moving on, the next step will be to multiply row 1 by $9-2i$, add it to row 2, and store the result in row 2. What do you get when you do that?

13. from
multiply row 1 by , add it to row 2
1 (-4+i)/17 0
0 (2-i)+(-4+i)(9-2i)/17 0

1 (-4+i)/17 0
0 (2-i)+(-2+i) 0

1 (-4+i)/17 0
0 0 0

I still got 0 in the second row??

14. Excellent. This is what you should expect, if you were at all able to get a nontrivial solution of a homogeneous linear equation. Now what?

15. Do you mean y=s , x=s(4-i)/17 is the answer?
but there is two unique answer y = -46/5-(6 i)/5 and x = -38/17+(22 i)/85
and y = 67/10-(67 i)/10 and x = 201/170-(67 i)/34, from wolfram

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