Linear Algebra | Use pure elimination to solve

Question

X	Y	Z	W
1	-2	4	1	16
4	-2	2	1	8
2	1	4	1	36
-2	-1	-6	1	-54
3	2	1	1	23

Answer 1

A typical guess, seeing that there are 5 equations in 4 unknowns, would be that there is no solution. But in working through it, we see that there is a single solution. That means that one equation is some combination of the others. I didn't take the time to see what that was.

There are tons of ways to solve these types of equations. It depends only upon your imagination. Dayv O., above, showed one method. I chose to multiply row 1 by -1 and add it to the other 4 equation - giving me new equations for Row 2, Row 3, Row 4, and Row 5. I then added Row 2 and 4 to get:

y - 12z = -78

Then I multiplied Row 2 by -2 and added the result to Row 5 to get:

-2y - 3z = -33

Then I multiplied y - 12z = -78 by 2 and added the result to -2y - 3z = -33 giving z = 7

Then I began back substituting giving the values of x, y, and w. Finally, I double checked my work because of my suspicions of a "no solution" result to ensure the values gave me true results for all 5 equations.

But you can do this problem the way it makes sense for you.

Answer 2

answer: x=2,y=6,z=7,w=-2

four variables, only need four equations, I used 1,2,3, and 5

1,-2,4,1=16

4,-2,2,1=8

2,1,4,1=36

3,2,1,1=23

1,-2,4,1=16

-3,0,2,0=8,,,,,,,,,,,-3x+z=8

1,3,0,0=20,,,,,,,,,,,,x+3y=20

-2,-4,3,0=-7

-9x+6z=24,,,4x+8y-6z=14

-5x+8y=38

x+3y=20

5x+15y=100

23y=138

y=6,x=2,z=7,w=-2

1,-2,4,1=16

-3,0,2,0=8,,,,,,,,,,,-3x+z=8

1,3,0,0=20,,,,,,,,,,,,x+3y=20

-2,-4,14,0=-7

1,-2,4,1=16

-3,0,2,0=8,,,,,,,,,,,-3x+2z=8

1,3,0,0=20,,,,,,,,,,,,x+3y=20

-2,-4,3,0=-7

1,-2,4,1=16

-3,0,2,0=8

1,3,0,0=20,,,,,,,,,,9,0,-6,0=-24,,,-4,-8,6,0=-14

5,-8,0,0=-38,,,,,,,,,,,5x+15y=100,,,,-5x+8y+38

y=6

x=2

z=7

w=-2

x=2,y=6,z=7,w=-2

Answer 3

The number of unknowns is 4 and we have 5 equations, so for there to be a consistent solution, you should be able to pick up any four equations, solve the system, and ensure those values also solve the 5th one. Otherwise, the answer would be that there is no solution to the system of equations.

You can discard any one row for this exercise, and I am going to discard the first row (for no particular reason).

Eq.	X	Y	Z	W	Result
(A)	1	-2	4	1	16
(B)	4	-2	2	1	8
(C)	2	1	4	1	36
(D)	-2	-1	-6	1	-54
(E)	3	2	1	1	23

While there are many ways to eliminate, the most efficient way is to look for common coefficient values. Interestingly the coefficient of W is 1 for all equations. As a first step in reduction, I will subtract equation (B) from each of equations (C), (D), and (E), resulting in the following where variable W is eliminated:

Eq.	X	Y	Z	W	Result
(C1)	-2	3	2	0	28
(D1)	-6	1	-8	0	-62
(E1)	-1	4	-1	0	15

Next, let’s proceed to eliminate variable Z. Let’s take the following actions on equations D1 and E1 to create their new equivalent versions:

D2 = D1 plus 4 times C1

E2 = E1 plus 0.5 times C1

Eq.	X	Y	Z	W	Result
(D2)	-14	13	0	0	50
(E2)	-2	5.5	0	0	29

Reduce one last time to eliminate X.

D3 = D2 minus 7 times E2

Eq.	X	Y	Z	W	Result
(D3)	0	-25.5	0	0	-153

(D3) gives us y = -153 / -25.5 => y = 6

Stepping back through our steps gives us the other variables.

(E2) gives us: -2x + 5.5 * 6 = 29 => x = 2

(C1) gives us: -2 * 2 + 3 * 6 + 2z = 28 => z = 7

Finally, (B) gives us: 4 * 2 – 2 * 6 + 2 * 7 + w = 8 => w = -2

We are not done until we put these values into the equation we discarded in the beginning to check if the solution holds. As a recap, the equation (A) we discarded is below:

Eq.	X	Y	Z	W	Result
(A)	1	-2	4	1	16

Let’s check if our solution holds true.

Put x = 2, y = 6, z = 7 and w = -2 into (A):

LHS = 2 – 2 * 6 + 4 * 7 – 2 LHS = 2 – 12 + 28 – 2 = 16

The result is 16, which matches the RHS.

So our solution holds!

Answer 4

(x,y,z,w) = (2,6,7,2)

x =w = 2, y=6, z =7