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olchik [2.2K]
3 years ago
13

Please Help Me With Matrices! Fully explain to me what to do, for I have no clue what I am doing. I can substitute, and eliminat

e, but Lord help me for I can not do matrices.
Solve the system using matrices (row operations)
−6x−y−5z=−10
− 5x+6y+4z=−7
2x−3y−2z=3
Also, how many solutions are there to this system?
Mathematics
1 answer:
prohojiy [21]3 years ago
8 0
Can you find an explanation of "row operations" with examples in any of your learning materials, online or in print?

Once you get the hang of row ops, it's not terribly hard.  This does, however, take a lot of arithmetic.

<span>−6x−y−5z=−10
− 5x+6y+4z=−7 
2x−3y−2z=3 

can be represented by the matrix  

-6  -1  -5  -10
-5    6  4   -7
 2   -3  -2    3

Our goal is to transform this 3 x 4 matrix so that it ends up looking like:

1  0  0  a
0  1  0  b
0  0  1  c

and the solution you want is the vector (a, b, c) (three numeric values).

</span>I have more or less arbitrarily chosen to start with the third row:   
2   -3  -2    3.  We want this row to begin with a 1, so we multiply each of the original four digits by (1/2), obtaining 1   -3/2   -2/2   3/2, or 1  -3/2   -1   3/2.

We can present the original matrix in any order without changing its value.  Thus, the original 

-6  -1  -5  -10
-5    6  4   -7
 2   -3  -2    3

becomes 

-6  -1    -5  -10
-5    6     4   -7
 1  -3/2  -1   3/2

We want that "1" to appear in the upper, left hand corner of the matrix.  We are free to interchange rows, so we interchange the first and 3rd rows, obtaining 

1  -3/2  -1   3/2
-5    6     4   -7
-6  -1    -5  -10

Next, we manipulate the first row (which begins with 1) so as to get the first element of the 2nd and 3rd rows to be 0.

To achieve this for the 2nd row, we multiply the 1st row by 5, obtaining

5   -15/2   -5   15/2

and then we add this to the existing 2nd row.  The result will be an "0"
in the first column:

0   (6-15/2)   ( 4-5)  (-7+15/2), or   0   -3/2   -1   1/2.

Substitute this new 2nd row for the original 2nd row.  We'll now have:

  1  -3/2  -1   3/2
  0  -3/2   -1   1/2
-6  -1    -5  -10

Now we have to "fix" the 3rd row, so that it starts with a zero (0):
To accomplish this, mult. the first row by 6 and add the resulting new row to the existing 3rd row.  Result should be  0  -10  -11  -1, and the revised matrix will be 

 1  -3/2  -1   3/2
  0  -3/2   -1   1/2
 0  -10    -11  -1

Next steps involve transforming the 2nd column so that it looks lilke

0
1
0.

To do this, mult. the entire 2nd row by -2/3,  Here's the expected result:

0    1     2/3    -1/3

Replace the existing 2nd row with this revised 2nd row:

 1  -3/2  -1   3/2
  0  -3/2   -1   1/2
 0  -10    -11  -1  becomes

 1  -3/2  -1   3/2
 0    1    2/3   -1/3
 0  -10    -11  -1

In the end we want this matrix to look like 

1  0  0  a
0  1  0  b
0  0  1  c

and the solution you want is the vector (a, b, c) (three numeric values).

Use this new 2nd row to further fix the 2nd column, so that it looks like

0 
1
0.


I ask that you go thru this discussion and work out each set of calculations yourself, to verify what I have done so far.  Reply with any questions that arise.  We'll find a way to finish this solution.

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Peter works at Silly's and makes $33.19 an hour. He also collects a 5.3% commission on final weekly sales. Last month, he worked
Romashka-Z-Leto [24]

Answer:

$10,318.62

Step-by-step explanation:

We have been given that works at Silly's and makes $33.19 an hour. He also collects a 5.3% commission on final weekly sales. If he works over 40 hours during any week, he earns double his hourly wage.

Peter's hourly charges during 1st and 2nd week would be $33.19 and hourly charges during 3rd and 4th week would be 2 times $33.19.

First of all, we will find total hourly income as:

\text{Total hourly income}=\$33.19(30+40)+2\times \$33.19(49+55)

\text{Total hourly income}=\$33.19(70)+\$66.38(104)

\text{Total hourly income}=\$2323.3+\$6903.52

\text{Total hourly income}=\$9226.82

Now, we will find income from commission on sales as:

\text{Income from commission on sales}=\frac{5.3}{100}(4500+5500+3800+6800)

\text{Income from commission on sales}=0.053(20600)

\text{Income from commission on sales}=1091.8

Peter's total earnings is the last month: \$9226.82+\$1091.8=\$10,318.62

Therefore, Peter's total earnings in last month was $10,318.62.

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Step-by-step explanation:

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An account earns simple interest. (a) Find the interest earned. (b) Find the
Debora [2.8K]

Answer:

#a. $80

#b. $1680

Step-by-step explanation:

We are given;

  • Amount invested (principal) is $1600
  • Rate of interest is 5%
  • Time = 1 year

We are required to determine the amount of simple interest earned and the amount or balance in the account after 1 year.

#a. Interest earned

To calculate simple interest we use the formula;

I = (PRT) ÷ 100

Where, P is the principal, R is the rate, T is the time and I is the simple interest.

Therefore;

I = (1600 × 5 × 1) ÷ 100

 = $80

Therefore, simple interest earned is $80

#b. Balance of the account (Amount accrued)

We are going to use the formula;

A = P + I , where A is the amount accrued, P is the principal and I is the simple interest earned.

Therefore;

Account balance = $1600 + $80

                            = $1680

Thus, the account balance after 1 year will be $1680

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