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Elis [28]
3 years ago
7

The augmented matrix of a system of equations has been transformed to an equivalent matrix in​ row-echelon form. Using​ x, y

, and z as​ variables, write the system of equations corresponding to the following matrix. If the system is​ consistent, solve it.
left bracket Start 3 By 3 Matrix 1st Row 1st Column 1 2nd Column 2 3rd Column 2 2nd Row 1st Column 0 2nd Column 1 3rd Column negative 3 3rd Row 1st Column 0 2nd Column 0 3rd Column 1 EndMatrix Start 3 By 1 Table 1st Row 1st Column 4 2nd Row 1st Column 9 3rd Row 1st Column negative 2 EndTable right bracket

1 2 2
0 1 −3
0 0 1
4
9
−2
Which of the following system of equations corresponds to the given​ matrix?

A.

x plus 2 y plus 2 zx+2y+2z

equals=44

y minus 3 zy−3z

equals=99

z

equals=minus−22

B.

2 x plus 2 y2x+2y

equals=44

negative 3 y−3y

equals=99

z

equals=minus−22

C.

2 x plus 2 y plus 4 z2x+2y+4z

equals=1

negative 3 y plus 9 z−3y+9z

equals=1

minus−z

equals=1

D.

x

equals=44

2 x plus y2x+y

equals=99

2 x minus 3 y plus z2x−3y+z

equals=minus−22

Select the correct choice below and fill in any answer boxes in your choice.

A.

The solution set is

StartSet nothing EndSet{}.

​(Type an ordered​ triple.)

B.

The solution is the empty set.
Mathematics
1 answer:
jenyasd209 [6]3 years ago
4 0

Answer: The system of equations is:

x + 2y + 2 = 4

y - 3z = 9

z = - 2

The solution is: x = -22; y = 15; z = -2;

Step-by-step explanation: ONe way of solving a system of equations is using the Gauss-Jordan Elimination.

The method consists in transforming the system into an augmented matrix, which is writing the system in form of a matrix and then into a <u>Row</u> <u>Echelon</u> <u>Form,</u> which satisfies the following conditions:

  • There is a row of all zeros at the bottom of the matrix;
  • The first non-zero element of any row is 1, which called leading role;
  • The leading row of the first row is to the right of the leading role of the previous row;

For this question, the matrix is a Row Echelon Form and is written as:

\left[\begin{array}{ccc}1&2&2\\0&1&3\\0&0&1\end{array}\right]\left[\begin{array}{ccc}4\\9\\-2\end{array}\right]

or in system form:

x + 2y + 2z = 4

       y + 3z = 9

               z = -2

Now, to determine the variables:

z = -2

y + 3(-2) = 9

y = 15

x + 30 - 4 = 4

x = - 22

The solution is (-22,15,-2).

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