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KIM [24]
2 years ago
14

Use matrices to solve the system of equations if possible. Use Gaussian elimination with back substitution or gauss Jordan elimi

nation. -x+y-z=-20,2x-y+z=29, 3x+2y+z=29
Mathematics
1 answer:
CaHeK987 [17]2 years ago
5 0

In matrix form, the system is given by

\begin{bmatrix} -1 & 1 & -1 \\ 2 & -1 & 1 \\ 3 & 2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -20 \\ 29 \\ 29 \end{bmatrix}

I'll use G-J elimination. Consider the augmented matrix

\left[ \begin{array}{ccc|c} -1 & 1 & -1 & -20 \\ 2 & -1 & 1 & 29 \\ 3 & 2 & 1 & 29 \end{array} \right]

• Multiply through row 1 by -1.

\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 2 & -1 & 1 & 29 \\ 3 & 2 & 1 & 29 \end{array} \right]

• Eliminate the entries in the first column of the second and third rows. Combine -2 (row 1) with row 2, and -3 (row 1) with row 3.

\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 5 & -2 & -31 \end{array} \right]

• Eliminate the entry in the second column of the third row. Combine -5 (row 2) with row 3.

\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 0 & 3 & 24 \end{array} \right]

• Multiply row 3 by 1/3.

\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 0 & 1 & 8 \end{array} \right]

• Eliminate the entry in the third column of the second row. Combine row 2 with row 3.

\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 8 \end{array} \right]

• Eliminate the entries in the second and third columns of the first row. Combine row 1 with row 2 and -1 (row 3).

\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 9 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 8 \end{array} \right]

Then the solution to the system is

\boxed{x=9, y=-3, z=8}

If you want to use G elimination and substitution, you'd stop at the step with the augmented matrix

\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 0 & 1 & 8 \end{array} \right]

The third row tells us that z=8. Then in the second row,

y-z = -11 \implies y=-11 + 8 = -3

and in the first row,

x-y+z=20 \implies x=20 + (-3) - 8 = 9

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In a certain school there are 180 pupils in the year7,110 pupils study french,88 study German , and 65 indonesian.40 pupils stud
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Answer:

(a) All three languages  - 9

(b) Indonesian only  - 10

(c) none of the languages - 12

(d) At least one language  - 168

(e) Either one or two of the three languages - 159​

Step-by-step explanation:

The question require the knowledge of Set theory and its Formulas.

Total pupils = 180

French (Let F) = 110  

German (Let G) = 88

Indonesian (Let I) =  65

French and German (F intersection G) =  40

German and Indonesian (G intersection I) =  38

French and Indonesian (F intersection I)  = 26

German only  = 19

(a) All three languages

We are given only German speaking people are 19

Only German = n(G) - n( F intersection G) - n(G intersection I) + n(F intersection G intersection I)

 19 = 88 - 40-38 + n(G intersection I) + n(F intersection G intersection I)

n(G intersection I) + n(F intersection G intersection I) = 9

n(G intersection I) + n(F intersection G intersection I) represents the number of pupils speaking who study all the three languages.

(b)Indonesian only

   n( I) - n(G intersection I) + n(F intersection I) + n(G intersection I) + n(F intersection G intersection I)

65 - 38 - 26 + 9 = 10

So 10 pupils speak Indonesian only

(c)none of the languages

It will be equal to Total - pupils speaking any of the three languages

Any of the three languages = (F union G union I)

  = n(F) +n(G) + n(I) -n(F intersection G) - n(G intersection I) -n(F intersection I)

    +  n(F intersection G intersection I)

n(G intersection I)

  = 110 + 88 + 65 -40 -38-26 + 9

  = 263 - 95

   = 168

So 168 pupils speak any of the three languages

None speakers = 180 - 168 = 12 pupils

So 12 pupils do not speak any of the three languages.

(d)at least one language

At least one language has the meaning that the person can either speak one two or all three languages, so it will be same as we proceeded above

Any of the three languages = (F union G union I)

  = n(F) +n(G) + n(I) -n(F intersection G) - n(G intersection I) -n(F intersection I)

    +  n(F intersection G intersection I)

n(G intersection I)

  = 110 + 88 + 65 -40 -38-26 + 9

  = 263 - 95

   = 168

(e) either one or two of the three languages​

Pupils can speak one or two language but not all the three so will subtract all the three language speaker from the total.

  Total speaker = 168

 All three languages speaker  = 9

 Either one or two of the three languages​ speaker = 168 - 9 = 159

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3 years ago
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