solve the system of equations using row operations of matrices

Question

solve the system of equations using row operations of matrices show a step by step process.  matrices must be in row echelon form

 

x-y=3

6x-5z=36

6y+2z=18

Answer 1

You have the following augmented matrix:

 

1   -1   0   3

6    0  -5   36

0    6   2   18

 

1) Multiply first row by 6 and subtract from the second row:

 

1   -1   0   3

0    6  -5   18

0    6   2   18

 

2)  Subtract second row from the third row:

 

1   -1   0    3

0    6   -5   18

0    0   7    0

 

Now we have 7z=0 from the last equation. Therefore, z=0

 

From the second equation we obtain:

6y-5*0=18 or 6y=18, so y=3;

 

From the first equation, x-y=3, so x=y+3 or x=6;

 

Thus, the solution is (6, 3, 0)

 

Answer : (6, 3, 0)

Answer 2

  x -  y + 0z  =  3
6x -  0y - 2z = 36 
0x + 6y + 2z = 18

(1) 1   -1    0       3 
(2) 6    0   -2     36 
(3) 0    6    2     18

Let's (3) add to (2) and divide by 6 

(1) 1   -1    0       3
(2) 1    1    0       9
(3) 0    6    2      18

Now, let's (2) add  to (1) and divide by 2

(1) 1   0   0     6
(2) 1   1   0     9 
(3) 0   6   2    18 

Let's (3) ÷ 2 

(1) 1   0   0     6
(2) 1   1   0     9
(3) 0   3   1     9  

Let's (1) subtract from (2)

(1) 1   0   0     6
(2) 0   1   0     3
(3) 0   3   1     9

Finally, let's (2) multiply by 3 and subtract from (3)

(1) 1   0   0     6
(2) 0   1   0     3
(3) 0   0   1     0

Thereby

x = 6, y = 3 and z = 0

The answer is (6, 3, 0)

We have to verify the answer

6 - 3 = 3   ===> 3 = 3
6*6 - 5*0 = 36   ===> 36 = 36
6*3 + 2*0 = 18    ===> 18 = 18