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Consider following system of equations:

$\begin{bmatrix} 1 &2 &3 &4 \\ 5&6 &7 &8 \\ a&9 &b &10 \\ 6&8 &10 & 13 \end{bmatrix}$$\begin{bmatrix} x1\\ x2\\ x3\\ x4 \end{bmatrix}$=$\begin{bmatrix} 0\\ 0\\ 0\\ 0 \end{bmatrix}$

The locus of all $(a,b)\in\mathbb{R}^{2}$ such that this system has at least two distinct solution for ($x_{1},x_{2},x_{3},x_{4}$) is

  1. a parabola
  2. a straight line
  3. entire $\mathbb{R}^{2}$
  4. a point
in Linear Algebra
edited by
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0
straight line

here equation will be $ax_{1}+bx_{2}+cx_{3}+dx_{4}=0$

here no polynomial degree of $x$

and also there are atleast 2 distinct point

and the line passes through these points

So, this must be a straight line
0
can you please explain in detail
0
$\begin{pmatrix} 1& 2&3 &4\\ 5& 6 &7 &8 \\ a&9 &b &10 \\ 6&8 &10 &13 \end{pmatrix}$ $\rightarrow$$\begin{pmatrix} 1& 2&3 &4\\ 0& -4&-8&-12 \\ a-k&9-2k &b-3k &10 -4k\\ 0&-4 &-8 &-11 \end{pmatrix}$$\rightarrow$

 

$\begin{pmatrix} 1& 2&3 &4\\ 0& -4&-8&-12 \\ a-k&9-2k-4l &b-3k-8l &10 -4k-12l\\ 0&0 &0 &1 \end{pmatrix}$

so for getting atleast two distinct solution

a-k=0  $\Rightarrow$ a=k ........(1)

9-2k-4l=0 $\Rightarrow$ 2k+4l=9........(2)

b-3k-8l=0 $\Rightarrow$ 3k+8l=b .........(3)

10-4k-12l=0 $\Rightarrow$ 4k+12l=10 .....(4)

solving 2 & 4 getting l = -2,k = $\frac{17}{2}$

so a = $\frac{17}{2}$ and b= $\frac{19}{2}$

i.e it is a point
0
@Sambit

a and b not solution

Here $\left ( x_{1},y_{1},z_{1},w_{1} \right )$ is one solution

Similarly $\left ( x_{2},y_{2},z_{2},w_{2} \right )$ another solution
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@srestha at least two solution
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srestha they asked to evaluate all values of (a,b) if the system needs to have atleast two solutions

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@Tesla

yes more than two also possible

but equation will be always linear

No parabolic or circular

So, should be straight line

if it is $x^{2}+y^{2}=a^{2}$

$y^{2}=4ax$ then it can form circular figure

but here equation is linear
0
No I am saying that there are 3 different solutions for 3 difference a,b now they have ask locus of a,b so 3 different value of a,b can stastify a parabola, pair of straight lines ?  For straight line all 3 a,b must be colinear that is stastify so y=MX+c

1 Answer

1 vote
The determinant of the matrix is easily calculated: $4a+4b-72$.  Now there are more than one solution iff that determinant equals zero. Clearly the locus is a straight line!

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