Math, asked by shreebro1, 3 months ago

1
2
3x
1 7
X +
5 3
+
+
4.
+
2x2
-X + 5
X
3
6
2.​

Answers

Answered by vikassainimzn
2

Answer:

your name is shree and my name is shree

Answered by ayushbaid08123
1

Answer:

3x1 + 2x2 -x3= -15

5x1+ 3x2 + 2x3 =0

3x1+ x2 +3x3=11

11x1+7x2 = -30

\begin {bmatrix} 3 & 2 & -1 \ \ \ \ \ \ -15 \\ 5 & 3 & 2 \ \ \ \ \ \ \ \ \ \ \ \ \ 0\\ 3 & 1 & 3 \ \ \ \ \ \ \ \ \ \ \ \ 11 \\ 11 & 7 & 0 \ \ \ \ \ \ \ -30 \end{bmatrix} =

3

5

3

11

2

3

1

7

−1 −15

2 0

3 11

0 −30

= \begin {Vmatrix} R_2 =5R_1-3R_2 \\ R_3 = R_1-R_3 \\ R_4 = 5R_4-4R_2 \end{Vmatrix} =

R

2

=5R

1

−3R

2

R

3

=R

1

−R

3

R

4

=5R

4

−4R

2

= \begin {bmatrix} 3 & 2 & -1 \ \ \ \ \ \ -15 \\ 0 & 1 & -1 \ \ \ \ \ \ \ -75\\ 0 & 1 & -2 \ \ \ \ \ \ \ \ \ \ \ 26 \\ 0 & 2 & -22 \ \ \ \ \ \ \ -150 \end{bmatrix}=

3

0

0

0

2

1

1

2

−1 −15

−1 −75

−2 26

−22 −150

=

\begin {Vmatrix} R_1 =R_1-2R_2 \\ R_2 =R_2-R_3 \\ R_3 = 2R_3-R_4 \\ R_4 = 2R_2- R_4 \end{Vmatrix} =

R

1

=R

1

−2R

2

R

2

=R

2

−R

3

R

3

=2R

3

−R

4

R

4

=2R

2

−R

4

= \begin {bmatrix} 3 & 0 & 11 \ \ \ \ \ \ -135 \\ 0 & 1 & -1 \ \ \ \ \ \ \ -75\\ 0 & 0 & 18 \ \ \ \ \ \ \ \ \ \ \ 202 \\ 0 & 0 & 20 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \end{bmatrix}=

3

0

0

0

0

1

0

0

11 −135

−1 −75

18 202

20 0

= \begin {Vmatrix} R_1 =\frac{R_1}{3} \\ R_2 =R_2-R_3 \\ R_3 = \frac{R_3}{18} \\ R_4 = 10R_3-9R_4 \end{Vmatrix} =

R

1

=

3

R

1

R

2

=R

2

−R

3

R

3

=

18

R

3

R

4

=10R

3

−9R

4

=

\begin {bmatrix} 1 & 0 & \frac{11}{3} \ \ \ \ \ \ -\frac{135}{3} \\ 0 & 1 & -1 \ \ \ \ \ \ -75\\ 0 & 0 & 1 \ \ \ \ \ \ \ \ \ \ \frac{101}{9} \\ 0 & 0 & 0 \ \ \ \ \ \ \ \ \ 2020 \end{bmatrix};

1

0

0

0

0

1

0

0

3

11

3

135

−1 −75

1

9

101

0 2020

;

In R_4R

4

0 \ne 20200

=2020 . So this system has no solution.

Similar questions