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Let $x$ and $y$ be two vectors in a $3$ dimensional space and $<x,y>$ denote their
dot product. Then the determinant
$det\begin{bmatrix}<x,x> & <x,y>\\ <y,x> & <y,y>\end{bmatrix}$

1. is zero when $x$ and $y$ are linearly independent
2. is positive when $x$ and $y$ are linearly independent
3. is non-zero for all non-zero $x$ and $y$
4. is zero only when either $x$ or $y$ is zero

edited | 295 views
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right?
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ok,but linearly independent set makes determinant 0

rt?
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but not necessarily x,y needs to be 0

It is a property

rt?
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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
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have u any reference?

I think A) will be corrct

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I have given the definitions which I have learned in 12th..u can check it on wikipedia..PFA..

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I think  ankitgupta.1729 reference is right

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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" ..

2)

by Boss (14.3k points)

< 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:

1. 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.
2. Determinant will be positive when x and y are linearly independent (i.e $\Theta$ $\neq$ n$\pi$)

So option B satisfies the condition.

ago by Junior (935 points)

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+1 vote
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