Are These Vectors Linearly Independent?

AI Thread Summary
The vectors a1 = (1; 2; 0), a2 = (2; 1; 3), and a3 = (0; 3; -3) are not linearly independent, as indicated by the zero determinant of the corresponding matrix. This implies that the matrix is singular, confirming the linear dependence of the vectors. Additionally, a dependence relation can be established, showing nontrivial solutions to the vector equation x_1a_1 + x_2a_2 + x_3a_3 = 0. Some participants suggest that understanding the concept of determinants should come after grasping simpler methods of determining linear dependence. Overall, the discussion emphasizes both determinant analysis and inspection methods for assessing linear independence.
Lynne
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Homework Statement


Given vectors:
a1 = (1; 2; 0),
a2 = (2; 1; 3),
a3 = (0; 3; -3).
Find out if these vectors are linearly independent.

Homework Equations



The Attempt at a Solution



<br /> \begin{cases}<br /> \lambda_1+2\lambda_2=0;\\<br /> 2\lambda+\lambda_2+3\lambda_3=0;\\<br /> 2\lambda_2-3\lambda_3=0;\\<br /> \end{cases}\\<br /> <br /> \lambda_1=\lambda_2=\lambda_3=0<br />

<br /> D=\begin{vmatrix} 1 &amp; 2 &amp; 0 \\ 2 &amp; 1 &amp; 3 \\ 0 &amp; 3 &amp; -3 \end{vmatrix}=0<br />

Vectors are not linearly independent because determinant is zero.
Am I correct?
 
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Yes, looks ok. It's also possible to do it by inspection since there are so few vectors involved. It shouldn't be too hard to spot that 2a1-a2=a3.
 
yes that is correct.
 
Defennder said:
Yes, looks ok. It's also possible to do it by inspection since there are so few vectors involved. It shouldn't be too hard to spot that 2a1-a2=a3.

Definitely agree with this approach. Determinants are thrown in too early without students understanding what they mean. It's best to seek a solution without them at an early level. Also seek Axler's paper :)
 
Lynne said:

Homework Statement


Given vectors:
a1 = (1; 2; 0),
a2 = (2; 1; 3),
a3 = (0; 3; -3).
Find out if these vectors are linearly independent.

Homework Equations



The Attempt at a Solution



<br /> \begin{cases}<br /> \lambda_1+2\lambda_2=0;\\<br /> 2\lambda+\lambda_2+3\lambda_3=0;\\<br /> 2\lambda_2-3\lambda_3=0;\\<br /> \end{cases}\\<br /> <br /> \lambda_1=\lambda_2=\lambda_3=0<br />

<br /> D=\begin{vmatrix} 1 &amp; 2 &amp; 0 \\ 2 &amp; 1 &amp; 3 \\ 0 &amp; 3 &amp; -3 \end{vmatrix}=0<br />

Vectors are not linearly independent because determinant is zero.
Am I correct?

Well yeah, since the determinant is zero, it means that the corresponding matrix is singular, so the column vectors of that matrix are linearly dependent, in which case your vectors actually consist of the columns of the matrix.
i.e
A=[a1,a2,a3] where a1,a2,a3 are column vectors you were given.

Another way of doing it is taking the dependence relation

x_1a_1+x_2a_2+x_3a_3=\bar 0 and solving this vector equation, and observing that there are nontrivial solutions to this vector equation, which actually meanst that the three vectors given are lin. dependent.
 
Thank you very much.
 
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