Any one can solve it step by step

Question

into echelon form and find the rank

*Find a parametric solution for the system

5_w ^_ 2_x + y ^_ z = 0

2_w + x + y + z = ^_1

^_w + 3x ^_ y + 2_z = 3

Answer 1

Gaussian Elimination:

5 -2 1 -1 | 0

2 1 1 1 | -1

-1 3 -1 2 | 3

2*R3+R2; 5*R3+R1; Then -R3

============================

0 13 -4 9 | 15

0 7 -1 5 | 5

1 -3 1 -2 | -3

7*R1 and 13 * R2

===================

0 91 -28 63 | 105

0 91 -13 65 | 65

1 -3 1 -2 | -3

-R1 + R2; then R1/7

====================

0 13 -4 9 | 15

0 0 15 2 |-40

1 -3 1 -2 | -3

There is one degree of freedom;

Declares Free Variable Z

Per Row #2: 15y + 2z = -40

15y = -2z - 40

15y = -2(z+20)

y = (-2/15)(z+20)

Per Row #1:

13x - 4y + 9z = 15

13x = 15 - 9z + 4y

13x = 15 - 9z + 4( -2/15)(z+20)

= 15 - 9z + (-8/15)(z+20)

= 15 - 9z + (-8/15)z + (-160/15)

= (225/15) - 9z + (-8/15)z + (-160/15)

= (225/15) - (135/15)z + (-8/15)z + (-160/15)

= (-143/15) z + 65/15

dividing by 13:

x = (-11/15)z + 5/15 = (-11/15)z + 1/3

= 1/3 - (11/15)z

= (5-11z)/15

Per original first equation:

5w - 2x + y - z = 0

5w = z - y + 2x

= z + (2/15)(z+20) + 2(5-11z)/15

= (15/15) z + (2/15)z + (8/3) + (2/3) - (22/15)z

= (-5/15)z + 10/3

= (-5/15)z + 50/15

= (50 - 5z)/15

= 5( 10 -z)/15

dividing by 5: (10-z)/15

==================================================================

Algebraic solution:

EQ1 - EQ2: 3w - 3x - 2z = 1 <-- Equation ALPHA

EQ2 + EQ3: w + 4x + 3z = 2 <-- Equation BETA

ALPHA - 3 * BETA : -15x - 11z = -5

15x + 11z = 5

15x = 5 - 11z

x = (5-11z)/15 <--- agrees with solution above

Per Equation BETA: w + 4x + 3z = 2

w = 2 - 3z - 4x

= 2 - 3z - 4( 5 - 11z)/15

= (30 - 45z - 4(5-11z))/15

= (30 - 45z - 20 + 44z)/15

= (10 -z)/15 <--- agrees with solution above

per equation #1:

5w - 2x + y - z = 0

y = z + 2x - 5w

= z + 2(5-11z)/15 - 5(10-z)/15

= (15z + 2(5-11z) - 5(10-z))/15

= (15z + 10 - 22z - 50 + 5z)/15

= (-2z -40)/15

= (-2/15)z - 40/15

= (-2/15)(z+20) <-- agrees with solution above

============================================================================================

Equation #1:

-------------

5w - 2x + y - z = 5((10-z)/15) - 2 ((5-11z)/15) + (-2/15)(z+20) - z

= (5(10-z) -2(5-11z) + -2(z+20) - 15z)/15

= ( 50 -5z - 10 + 22z + -2z - 40 - 15z) /15

= 0

Equation #1 checks out!!!

Equation #2:

---------------

2w +x+y+z = 2( (10-z)/15) + (5-11z)/15 + (-2/15)(z+20) + z =

(2(10-z) + (5-11z) + -2(z+20) + 15z) / 15 =

( 20 - 2z + 5 - 11z + -2z + -40 + 15z) / 15 =

-15/15 = -1

Equation #2 checks out!!!

Equation #3:

--------------

-w + 3x - y + 2z = (z-10)/15 + 3(5-11z)/15 + (2/15)(z+20) + 2z =

((z-10) + 3(5-11z) + 2(z+20) + 30z) / 15 =

( z - 10 + 15 - 33z + 2z + 40 + 30z) / 15=

45/15 =

3

Equation #3 checks out!!!

Answer 2

Okay, if this the system of algebraic equation if the form

5w ^_ 2x + y ^_ z = 0

2w + x + y + z = ^_1

^_w + 3x ^_ y + 2z = 3

then it can be solved in many way. But you have to make sure you have written everything correctly.

There are 4 unknowns and thee equations so we can solve for 3 unknowns w.r.t to 4th one. Rewrite the system as

^_ 2x + y ^_ z = -5w

x + y + z = ^_1-2w

3x ^_ y + 2z = 3+w

I just replaced equations :

x + y + z = ^_1-2w

^_ 2x + y ^_ z = -5w

3x ^_ y + 2z = 3+w

STEP1 leave the first equation and exclude x from the second one. To do that we leave first equation unchanged

x + y + z = ^_1-2w

now multiple this equation by 2, we get 2x +2 y +2 z = ^_2-4w and add this to the second equation:

2x +2 y +2 z = ^_2-4w

^_ 2x + y ^_ z = -5w

------------------------------------

3y+z=2-9w

Now, this time we multiple the first equation by -3: ^_ 3x -3y-3 z = 3+6w and add to the third equation:

^_ 3x -3y-3 z = 3+6w

3x ^_ y + 2z = 3+w

---------------------------

-4y-z=6+7w

So we have

x + y + z = ^_1-2w

3y+z=2-9w

-4y-z=6+7w

So we excluded x, in the next step, we exclude y or z. Since it is easier to eliminate z, we eliminate z

x + y + z = ^_1-2w

3y+z=2-9w

-y=8+2w

Now find y from the last equation, then z from the second and x from the first equation