need help with a linear algebra problem.

[stomlin108]

Given that (c1, c2, c3 ) is a solution of the following homogeneous system of equations:

a11x1 + a12x2 + a13x2 = 0
a21x1 + a22x2 + a23x2 = 0
a31x1 + a32x2 + a33x2 = 0

Demonstrate that for any real number k (k1, k2, k3 ) is also a solution.

I am not sure how to demonstrate this. I am guessing that unless C1, C2, C3 is zero, that there would be infinitely many solutions. Do I set the second two problems up as paramtetric solution, such as

c1=-S-T
c2=S
c3=T
for all real numbers S and T

then plug in k1, k2, k3 to show that it is also a solution?

As you can see, I am confused! I can do the work with numbers, but am confused with the letters

 

[Deleted member 4993]

Given that (c1, c2, c3 ) is a solution of the following homogeneous system of equations:

a11x1 + a12x2 + a13x2 = 0
a21x1 + a22x2 + a23x2 = 0
a31x1 + a32x2 + a33x2 = 0

Demonstrate that for any real number k (k1, k2, k3 ) is also a solution.

I am not sure how to demonstrate this. I am guessing that unless C1, C2, C3 is zero, that there would be infinitely many solutions. Do I set the second two problems up as paramtetric solution, such as

c1=-S-T
c2=S
c3=T
for all real numbers S and T

then plug in k1, k2, k3 to show that it is also a solution?

As you can see, I am confused! I can do the work with numbers, but am confused with the letters

For a non-trivial solution,

\(\displaystyle |A_{ij}|\ \ = \ \ 0\)

Then any set of real numbers will be a solution.

 

[soroban]

Hello, stomlin108!

Too many subscripts . . . I'll simplify the problem . . .

\(\displaystyle \text{Given that }\,(x_o,y_o,z_o)\text{ is a solution of the following homogeneous system of equations:}\)

. . . \(\displaystyle \begin{array}{cccc}a_1x + b_1y + c_1z &=& 0 & [1] \\ a_2x + b_2y + c_2z &=& 0 & [2] \\ a_3x + b_3y + c_3z &=& 0 & [3] \end{array}\)

\(\displaystyle \text{demonstrate that for any real number }k,\;k(x_o,y_o,z_o)\text{ is also a solution.}\)

\(\displaystyle \text{Since }(x_o,y_o,z_o)\text{ is a solution to equation [1], then: }\;a_1x_o + b_1y_o + c_1z_o \:=\:0\)


\(\displaystyle \text{Multiply by }k\!:\;\;k(a_1x_o + b_1y_o + c_1z_o) \:=\:k(0)\)

. . \(\displaystyle \text{and we have: }\;a_1(kz_o) + b_1(ky_o) + c_1(kz_o) \:=\:0\)

\(\displaystyle \text{Hence, }(kx_o,ky_o,kz_o) \:=\:k(x_o,y_o,z_o)\text{ is a solution of equation [1].}\)


\(\displaystyle \text{In a similar fashion, show that }k(x_o,y_o,z_o)\text{ is a solution of equations [2] and [3].}\)

 

[stomlin108]

thanks! That really helped