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? Please help me... 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 so the answer is 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 after adding and subtracting 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 12. I agree with your answer to # 1. 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.$\displaystyle \left[\begin{matrix}4+i &-1\\-9+2i &2-i\end{matrix}\Big|
\begin{matrix}0\\0\end{matrix}\right]$First step: need a 1 in the 1,1 position. Divide first row by$\displaystyle 4+i$to obtain$\displaystyle \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].$So here you can see that division by a complex number will require multiplication by a fraction that is equal to 1: a special fraction. In general,$\displaystyle \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}}.$Don't memorize this formula. Instead, remember the procedure: multiply top and bottom by the complex conjugate, and then foil it out. Moving on, the next step will be to multiply row 1 by$\displaystyle 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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