right?

0 votes

Let $x$ and $y$ be two vectors in a $3$ dimensional space and $<x,y>$ denote their

dot product. Then the determinant

$det\begin{bmatrix}<x,x> & <x,y>\\ <y,x> & <y,y>\end{bmatrix}$

- is zero when $x$ and $y$ are linearly independent
- is positive when $x$ and $y$ are linearly independent
- is non-zero for all non-zero $x$ and $y$
- is zero only when either $x$ or $y$ is zero

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2 vectors are linearly independent when we can't represent one vector to another vector with some constant multiple ..for example if we have 2 vectors **x **and **y **..then these are linearly independent if **x ** ≠ c.**y **..where c = constant..and these 2 vectors will be linearly dependent when **x = **c.**y **...Now here in question ,if we take any example of 2 linearly independent vectors and put in the determinant then result can be anything ie. zero , positive or negative number...

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ur logic is not correct @ankit

linear dependence is same as linear independence, where one vector is is subset of another vector

So, if one contains 0, other also contains 0

linear dependence is same as linear independence, where one vector is is subset of another vector

So, if one contains 0, other also contains 0

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please check here

http://cs.brown.edu/courses/cs053/current/slides/10-13-2017.pdf

It is clearly told linearly dependendence is another name of linearly (in)dependence

Can u show it with example (difference between linear dependence and independence)

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@Srestha Ma'am ..sorry I didn't find where it is written that "linearly dependence is another name of linearly independence" ..

For example , please refer 1) https://www.khanacademy.org/math/linear-algebra/vectors-and-spaces/linear-independence/v/linear-algebra-introduction-to-linear-independence

2)

1 vote

< x,x > = x$^{2}$

< y,y > = y$^{2}$

< x,y > = <y,x > = xycos$\Theta$

--------------------------------------------------------------------

Now, determinant = x$^{2}$y$^{2}$ - (xycos$\Theta$)$^{2}$

= x$^{2}$y$^{2}$ - x$^{2}$y$^{2}$cos$^{2}$$\Theta$

= x$^{2}$y$^{2}$(1-cos$^{2}$$\Theta$)

= **x$^{2}$y$^{2}$(sin$^{2}$$\Theta$)**

So from above equation determinant will be either zero or positive:

- Determinant will be zero when x and y are linearly dependent (i.e $\Theta$ = n$\pi$) or if either of x and y is zero.
- Determinant will be positive when x and y are linearly independent (i.e $\Theta$ $\neq$ n$\pi$)

So** option B** satisfies the condition.