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HooKarez
2009-05-03, 08:02 PM
Kay. So I wasn't in class on Friday because of Physics day at Six Flags, so my math teacher gives me this graded classwork on Thursday which covers the stuff we did on Friday. And I dunno what we did, so yeah. Two questions I don't really understand how to do.

Solve with Matrices:

Melody has \$45,000 to invest and wishes to receive an annual income of \$4,290 from this money. She has chosen investments that pay 5%, 8% and 12% simple interest. Melody wants to have the amount invested at 12% to be double the amount invested at 8%. How much should she invest at each rate?

A company manufactures three types of wooden chairs: the Kitui, the Goa, and the Santa Fe. To make a Kitui chair requires 1 hour of cutting time, 1.5 hours of assembly time, and 1 hour of finishing time. A Goa chair requires 1.5 hours of cutting time, 2.5 hours of assembly time, and 2 hours of finishing time. A Santa Fe chair requires 1.5 hours of cutting time, 3 hours of assembly time and 3 hours of finishing time. If 41 hours of cutting time, 70 hours of assembly time and 58 hours of finishing time were used in one week, how many of each type of chair were produced?

whybanme
2009-05-03, 09:10 PM
No clue for the first one, it's worded too weirdly, but for the second problem:

Basically set up a system of equations and put them in an augmented matrix. Put that matrix in RREF form unless you can't because there is no solution or there is infinite many solutions. Then write out your answer in terms of X, Y, and Z. (Each variable is a different chair)

EDIT: Okay, maybe I don't know what I'm talking about. I just tried to work it out and I guess I just suck at math. lulz.

KajitiSouls
2009-05-03, 09:46 PM
Melody has \$45,000 to invest and wishes to receive an annual income of \$4,290 from this money. She has chosen investments that pay 5%, 8% and 12% simple interest. Melody wants to have the amount invested at 12% to be double the amount invested at 8%. How much should she invest at each rate?

Humm....

Mathematical equations
x + y + z = 45000
.05x + .08y + .12z = 4290
-2y + z = 0

Matrix representing this problem
1 1 1 45000
.05 .08 .12 4290
0 -2 1 0

Solve

Kalovale
2009-05-04, 12:11 AM
Humm....

Mathematical equations
x + y + z = 45000
.05x + .08y + .12z = 4290
-2y + z = 0

Matrix representing this problem
1 1 1 45000
.05 .08 .12 4290
0 -2 1 0

Solve

How different are Matrix and Simultaneous Equations then? Sorry for the dumb question, I didn't learn Matrix in highschool here.

Russt
2009-05-04, 12:59 AM
Matrices can be used to represent simultaneous equations.

If you have:
ax + by = e
cx + dy = f

Then that's equivalent to the matrix equation:
[a b]*
[c d][e]=
[f][x]
[y]

I think. It's been a while, I may have switched the order around.

Edit: Yeah, other way around.
[a b]*
[c d][x]=
[y][e]
[f]

HooKarez
2009-05-04, 05:54 AM
Humm....

Mathematical equations
x + y + z = 45000
.05x + .08y + .12z = 4290
-2y + z = 0

Matrix representing this problem
1 1 1 45000
.05 .08 .12 4290
0 -2 1 0

Solve

Ah. Starting to get it now.

For the second one then we'd have an equation for each.. type of work? Okay then each variable is for a different chair..

x + 1.5y + 1.5z = 41
1.5x + 2.5x + 3z = 70
x + 2y + 3z = 58

The rref'd matrix I get is:

| 1 0 -3 -10|
| 0 1 3 34|
| 0 0 0 0 |

-shrug-

KajitiSouls
2009-05-04, 11:49 AM
How different are Matrix and Simultaneous Equations then? Sorry for the dumb question, I didn't learn Matrix in highschool here.

They're the same thing as far as I know. There's a lot of terminology and mathematics in matrices though; solving a two-variable two-equation system is just the tip of the iceberg.

Dusk
2009-05-04, 03:17 PM
Ah. Starting to get it now.

For the second one then we'd have an equation for each.. type of work? Okay then each variable is for a different chair..

x + 1.5y + 1.5z = 41
1.5x + 2.5x + 3z = 70
x + 2y + 3z = 58

The rref'd matrix I get is:

| 1 0 -3 -10|
| 0 1 3 34|
| 0 0 0 0 |

-shrug-

In rref form the left side of the matrix should be an identity matrix.

KajitiSouls
2009-05-04, 05:04 PM
In rref form the left side of the matrix should be an identity matrix.

Actually it only needs to be in "Reduced Row Echelon" form. The Identity matrix is part of that group of matrices =P
All nonzero rows are above any rows of all zeroes. The leading coefficient (also called pivot) of a row is always strictly to the right of the leading coefficient of the row above it. Every leading coefficient is 1 and is the only nonzero entry in its column.

In the matrix:

| 1 0 -3 -10|
| 0 1 3 34|
| 0 0 0 0 |

the threes in column three just signify that it's a free variable. You can define the set of equations being represented by the matrix as a 1-parameter family of solutions, or a line in 3-space dimension.

Of course, an Identity matrix means there's only 1 solution, which is preferred =P But if you have the Reduced Row Echelon form of a matrix, and it's not an Identity matrix, then it has infinite solutions instead. You'll never get an Identity matrix in that case.

EDIT: I got the same solution HooKarez found for problem 2. In this case, you can state the solution as:

x = 3z - 10
y = 34 - 3z

Since we're dealing with the real world in this problem, there must be a domain where the set of functions is valid. Usually it's the unique solution or [0,+infinity). Here however, x, y, and z ≥ 0 since I find it extremely hard to have a negative amount of chairs (and this isn't economics where debt exists). Therefore:

3z - 10 ≥ 0
34 - 3z ≥ 0

Or,

10 ≤ 3z ≤ 34

Or,

3 < z < 12

since 3*3 = 9, which is less than 10, and 3*12 = 36, which is greater than 34, and that can't happen.

Dusk
2009-05-04, 05:24 PM
Actually it only needs to be in "Reduced Row Echelon" form.
All nonzero rows are above any rows of all zeroes. The leading coefficient (also called pivot) of a row is always strictly to the right of the leading coefficient of the row above it. Every leading coefficient is 1 and is the only nonzero entry in its column.

In the matrix:

| 1 0 -3 -10|
| 0 1 3 34|
| 0 0 0 0 |

the threes in column three just signify that it's a free variable. You can define the set of equations being represented by the matrix as a 1-parameter family of solutions, or a line in 3-space dimension.

Of course, an Identity matrix means there's only 1 solution, which is preferred =P But if you have the Reduced Row Echelon form of a matrix, and it's not an Identity matrix, then it has infinite solutions instead.
I know what REF is, I've taken Linear Algebra, and even though I completely failed at it, I haven't forgotten how to do that :P

He said RREF, so I was being technical D:

KajitiSouls
2009-05-04, 05:37 PM
I know what REF is, I've taken Linear Algebra, and even though I completely failed at it, I haven't forgotten how to do that :P

He said RREF, so I was being technical D:

RREF =/= Identity matrix, even though Identity matrix = RREF. They're not commutative.