Solve the homogeneous linear system corresponding to the given coefficient matrix. (If there is no solution, enter NO SOLUTION. If the system has an infinite number of solutions, set x4 = t and x2 = s and solve for x1 and x3 in terms of t and s.)

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0 0 1 0
0 0 0 0
(x1, x2, x3, x4) =

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The solution of the system is { (-t. s. 0, t) : s,t ∈ R}

Step-by-step explanation:

Lets multiply the first row by (x1,x2,x3,x4) and equalize to 0

(1,0,0,1) * (x1,x2,x3,x4) = x1+x4

which is equal to 0 only when

x1+x4 = 0

x1 = -x4

Now we can replace x1 with -x4.

If we multiply the second row (0,0,1,0) with (x1,x2,x3,x4) we will obtain (0,0,1,0)*(x1,x2,x3,x4) = x3. Since we want that product to be equal to 0, then x3 = 0. If we multiply the third row of zeros with (x1,x2,x3,x4) we will obtain automatically 0.

As a consecuence, we have 2 conditions:

We call t = x4, s = x2, which can take any value. The solutions of the system is

{ (-t. s. 0, t) : s,t ∈ R}

answer: just subtact the units and then add the non negative : happy to

step-by-step explanation:

dc=6 square root of 3 and ac is 12 square root 3

c and d

step-by-step explanation:

square root of 3/2 = 18/c

cross multiply to get 18*2=√3*c

36/√3 * √3/√3 = c

c= 36*√3/3

c=12√3