Download Put in Row Echelon Form

Solve the system 2x  y  z  5
x yz2 x  y  z  2
by putting in echelon form. To solve this system, we need to transform it to an equivalent system. By equivalent, I mean another system of equations that has the same solutions. The system we’ll transform it to has the form x  ay  bz  c
y  dz  e z f
where a, b, c,d, e, and f are numbers. A system in this form is said to be in echelon form. To get the system into echelon form, we have three tools at our disposal. The following operations can be applied to a system of equations to give an equivalent system: 1. interchange any two equations in the system; 2. multiply both sides of an equation in a system by any nonzero number; 3. replace any equation in the system with a multiple of one equation added to a multiple of another equation. Step 1: The first step in solving this system is to make the coefficient on the first variable in the first equation a 1. It is currently a 2. There are two options: we can multiply the equation by ½ to make the coefficient a 1 or interchange the first equation with another to make the coefficient a 1. Multiplying would introduce fractions into the process which we’ll try to avoid. The best option is to interchange the first and second equations (E1  E2): x yz 2
2x  y  z  5 x  y  z  2
Step 2: Now that there is a 1 in front of the x in the first equation, we want to use that equation to eliminate x from both of the equations below it. To eliminate x from the second equation, multiply the first equation by ‐2 and add it to the second equation. Put this new equation in place of the second equation (‐2E1 + E2  E2): 2 x  2 y  2 z  4
2  x  y  z   2  2 
2 x  y  z  5  2 x  y  z  5  y  3z  1
Inserting the sum in the system of equations yields x y z  2
 y  3z  1 x  y  z  2
To eliminate x from the third equation, multiply the first equation by ‐1 and add it to the third equation (‐3E1 +E3  E3):  x  y  z  2
1 x  y  z   1 2 
x  y  z   2  x  y  z  2  2y
 4
The system of equations is now x y z  2
 y  3z  1 2 y
 4
Step 3: Next we need to make the coefficient on the y term in the second equation a 1. The easiest way to do this is to multiply the equation by ‐1 (‐1E2  E2): 1  y  3 z   11  y  3z  1 With this change, the system becomes x y z  2
y  3z   1 2 y
 4
Step 4: Now we need to eliminate y below the 1 just created. To eliminate y in the third equation, multiply the second equation by 2 and add it to the third equation (2E2 + E3  E3): 2 y  6 z  2
2  y  3 z   2  1
 4 2 y
  4  2 y
6 z  6
The system of equations becomes x y z  2
y  3z   1 6 z  6
Step 5: To put the matrix in echelon form, we need to make the coefficient in front of z in the third term a 1. To do this, multiply the third equation by 1/6 (1/6 E3  E3): 1
6
 6 z   16  6   z  1 The system in echelon form is x y z  2
y  3z   1 z  1
Step 6: Substitute the value for z into the second equation and solve for y. This gives y  3  1  1
y  3  1 y2
Now take the values for y and z and put them into the first equation and solve for x: x  2   1  2
x 1  2 x 1
The solution is  x, y, z   1, 2, 1 .