Question

Problem 1.3 #4 - I don't understand how to find a combination that gives a zero vector?

Answer 1

of course the trivial solution is always available:\[x_1=x_2=x_3=0\]

Answer 2

Yes, that is true!

This particular matrix has a non-trivial (non-zero) solution as well.

Is this making any sense, @staciann?

Answer 3

I'm having trouble understanding this question too. Is A this matrix below instead?

\[\left[\begin{matrix}1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6&9\end{matrix}\right]\]
I simplified it down to x1 = x3 and x2 = -2(x3). I'm having trouble going from here. Thanks for your responses.

Answer 4

How about eliminating forward:
\[\left[\begin{matrix}1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9\end{matrix}\right]\rightarrow\left[\begin{matrix}1 & 4 & 7 \\ 0 & -3 & -6 \\ 0 & -6 & -12\end{matrix}\right]\rightarrow\left[\begin{matrix}1 & 4 & 7 \\ 0 & -3 & -6 \\ 0 & 0 & 0\end{matrix}\right]\]Would it make it easier to find a (non-zero) linear combinations of the three columns that result in all zeros ?

Answer 5

@linnyalgae, thank you. In my haste I transposed the columns to rows. I am sorry about that! Yes, your matrix A is correct.

And, hanson.char's method of reducing the rows is an excellent way to solve the problem.

So with that last echelon reduced matrix we have:
\[\left[\begin{matrix}1 & 4 & 7 \\ 0 & -3 & -6 \\ 0 & 0 & 0\end{matrix}\right] \left[\begin{matrix}x_{1}\\x_{2}\\x_{3}\end{matrix}\right] = \left[\begin{matrix}0\\0\\0\end{matrix}\right] \] or completing the multiplication:
\[\left[\begin{matrix}1x_{1}+ 4x_{2}+ 7x_{3} \\ -3x_{2} -6x_{3} \\ 0\end{matrix}\right] = \left[\begin{matrix}0\\0\\ 0\end{matrix}\right]\]

So with the second row of the matrix on the left, we see the ratio between x2 and x3. There are an infinite number of solutions to that problem, so to make things simple, choose a value of 1 for x3. Then you can solve for x2. Then plug the values you just found for x2 and x3 into the first row to solve what the value of x1 must be. And from there you will have the answer.

linnyalgae's solution is correct.