0
$\begingroup$

\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 $v_2, v_3, v_4$?

$\endgroup$
  • $\begingroup$ @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. $\endgroup$ – César Rosendo Nov 7 '17 at 4:02
1
$\begingroup$

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 $v_1=v_2(-1/2)+v_4(-5/2)$

$\endgroup$
0
$\begingroup$

Yes, you will then have to solve for $a, b, c$.

By examining the first entry, we have $$a+b-c = 2$$

Do the same thing for the second and third entries as well.

Gaussian elimination might be helpful in obtaining $a,b,$ and $c$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.