Question

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Answer 1

To repeat, we have three vectors \[w_1=\begin{bmatrix}1\\2\\3\end{bmatrix}\quad w_2=\begin{bmatrix}4\\5\\6\end{bmatrix}\quad w_3=\begin{bmatrix}7\\8\\9\end{bmatrix}\]and for (a) we want to find \(c_1\), \(c_2\), and \(c_3\) such that \(c_1w_1+c_2w_2+c_3w_3 = 0\).

The way I typically do this sort of exercise is to try to find some way to express one of the vectors in terms of the other two; in this case I'll try to express \(w_3\) in terms of \(w_1\) and \(w_2\), since \(w_3\) is the largest vector. Note that this is equivalent to taking the equation \(c_1w_1+c_2w_2+c_3w_3 = 0\) and trying to solve for \(w_3\).

We can start with the first elements of \(w_1\) and \(w_2\), 1 and 4 respectively. Clearly 1 and 4 don't add up to 7, so \(w_3 \ne w_1+w_2\). If we try multiplying \(w_1\) by 2 and \(w_2\) by 1 we'd get \(2\cdot1 + 4 = 6\) as the first element of \(2w_1+w_2\), so clearly \(2w_1+w_2\ne w_3\).

What about multiplying \(w_2\) by 2? That gives us \(2\cdot4 =8\) as the first element of \(2w_2\), which is larger than the first element of \(w_3\). But... if we then subtracted the first element of \(w_1\) we'd get \(2\cdot4-1=8-1=7\) as the first element of \(2w_2-w_1\), and this is the same as the first element of \(w_3\). Trying this with the second elements of \(w_2\) and \(w_1\) we have \(2\cdot5-2=10-2=8\), which is the second element of \(w_3\). Finally, using the third elements of \(w_2\) and \(w_1\) we have \(2\cdot6-3=12-3=9\), which is the third element of \(w_3\).

So we have \(2w_2-w_1=w_3\). I'll leave it to you to put this in the form \(c_1w_1+c_2w_2+c_3w_3 = 0\).

As for the other two questions (b) and (c), start with the third question (c) and recall the definition of linear independence: Three vectors \(w_1\), \(w_2\), and \(w_3\) are linearly independent if \(c_1w_1+c_2w_2+c_3w_3 = 0\) only when \(c_1=c_2=c_3=0\); otherwise the vectors are linearly dependent. If you found the answer to question (a) above then you've got the answer to question (c).

As for the second question (b), remember that a matrix is invertible only if its columns (or rows, for that matter) are linearly independent. Otherwise the matrix is not invertible. Since the matrix here has the vectors \(w_1\), \(w_2\), and \(w_3\) as its columns, the answer to question (c) gives the answer to question (b).