# Express $v_1$ as a linear combination of $v_2,v_3,v_4$

\begin{align}v_1 &= (2,-3,1) \\ v_2 &= (1,1,-2) \\ v_3 &= (1,1,1) \\ v_4 &= (-1,1,0) \end{align}

$$a(1,1,-2) + b(1,1,1) + c(-1,1,0) = (2,-3,1)$$

is this the correct way of expressing $v_1$ as a linear combination of $v_2, v_3, v_4$?

• @Soon_to_be_code_master Yes, that is the first step. You must find now the scalars a,b and c that satisfy that equation. Apply the definition of the product of a vector with a scalar, and vector addtion on the left side of your equation to basically get a system of linear equations when you make the entries of the vector on the left side equal to its correspondent entry on the right side. – César Rosendo Nov 7 '17 at 4:02

That is a linear combination of the three vectors, but I think the question is asking for the values of $a, b,$ and $c$. Using the definitions of scalar-vector multiplication and vector addition, you can re-write this equation as a system: $$a+b-c=2$$$$a+b+c=-3$$$$-2a+b+0c=1$$ This can be written as a matrix equation: $$\begin{bmatrix} 1&1&-1\\ 1&1&1\\ -2&1&0 \end{bmatrix}\begin{bmatrix}a\\b\\c\end{bmatrix}=\begin{bmatrix}2\\-3\\1\end{bmatrix}$$ Now, you can row reduce the augmented matrix down to: $$\begin{bmatrix}1&0&0&-(1/2)\\0&1&0&0\\0&0&1&-(5/2)\end{bmatrix}$$ So, $a=-1/2, b=0,$ and $c=-5/2$. Thus, $v_1$ can be written as the linear combination $v_1=v_2(-1/2)+v_4(-5/2)$
Yes, you will then have to solve for $a, b, c$.
By examining the first entry, we have $$a+b-c = 2$$
Gaussian elimination might be helpful in obtaining $a,b,$ and $c$.