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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
in Linear Algebra by Junior (553 points)
edited by | 410 views
answer will be A)

Answer will be D)
ok,but linearly independent set makes determinant 0

but not necessarily x,y needs to be 0

It is a property


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...

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

@ Sukanya Das

have u any reference?

I think A) will be corrct


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


I think  ankitgupta.1729 reference is right


please check here

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)

@ ankitgupta.1729


@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)


2 Answers

+2 votes

Answer :- D

by Boss (17.6k points)
+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:

  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.


by Active (1.1k points)
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