# How to calculate a linear combination for a matrix' column?

I have a very weak understanding of linear dependency and linear combination, so I figured I'd check out some exercise about it:

$$A = \begin{bmatrix} 4 & 0 & 1\\ 2 & 3 & 6\\ 6 &-3 & -4 \end{bmatrix} \ \ , \ \ \left| \ A \ \right| = 0$$

a) Without additional calculations, determine whether the 3 columns of the matrix are linearly independent or not.

b) Write the vector that corresponds t othe third column of the matrix as a linear combination of the other two columns.

My book is not particularly clear about those concepts, but, from what I gather, I would say that for the first question the answer is that the 3 columns are dependent... why? Well, I think it is because since $\left| \ A \ \right| \neq 0$, the matrix must have infinite solutions that depend on a certain parameter... so yeah (ok what I said doesn't make much sense)...

As for the second question, I don't really even get it.

Can you better explain to me how to create a linear combination of a column? (I don't really need the answers to those questions)

• @Amzoti: Well, the answer I wrote is 'no', because there are infinite solutions depending on some parameter. Why, is that wrong? – Zol Tun Kul Apr 9 '14 at 3:38
• @Amzoti: So that's a linear combination? I need to find two numbers $a$ and $b$ that multiplied by the other two columns respectively will produce the third one? – Zol Tun Kul Apr 9 '14 at 3:43
• @Amzoti: $a = \frac{1}{4}$ and $b = \frac{11}{6}$. Do I have to write them in some specific format for the answer? – Zol Tun Kul Apr 9 '14 at 3:58
• @Amzoti: Oh the answer would be $$C_3 = \frac{1}{4}C_1 + \frac{11}{6}C_2$$ I think? – Zol Tun Kul Apr 9 '14 at 4:00
• @Amzoti: Ok, so in short: A system's columns are dependent if the rank is not full. Can the same be said for the rows? Actually, what if I'm asked if a specific column is dependent or not? Does it suffice simply checking the rank, or do I have to check if it has a leading $1$ from the reduced row echelon form? – Zol Tun Kul Apr 9 '14 at 4:05