Please help me in slicing this question.Given a matrix and need to find The locus of all (a,b) € R2 such that this system has atleast TWO DISTINCT solutions for (x1,x2,x3,x4) is ____
Options are Parabola; straight line ; entire plane of R2 ; a point
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1Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? – 5xum Feb 19 '18 at 08:20
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Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote. – 5xum Feb 19 '18 at 08:21
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1Hint: observe that your question is equivalent to ask "for what values of $;a,,b;$ is the given matrix singular ?" – DonAntonio Feb 19 '18 at 08:23
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I got an other expression for minor M 14 as 3a-8b=114.if we do all the 3x3 determinants of minors as zero.we will get some set of a,b values.how to know the locus. – sindhu Feb 19 '18 at 09:26
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if we get two singular solutions ,what does it mean..i dont know how to find this locus. – sindhu Feb 19 '18 at 09:39
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As commented, your homogeneous system has at least two different solutions iff the coefficients' matrix is singular iff its determinant is zero:
$$\begin{vmatrix}1&2&3&4\\ 5&6&7&8\\ a&9&b&10\\ 6&8&10&13\end{vmatrix}\stackrel{\begin{cases}R_2-5R_1\\R_3-aR_1\\R_4-6R_1\\{}&{}\end{cases}}=\begin{vmatrix}1&2&3&4\\ 0&-4&-8&-12\\ 0&9-2a&b-3a&10-4a\\ 0&-4&-8&-11\end{vmatrix}=$$$${}$$
$$=\begin{vmatrix} -4&-8&-12\\ 9-2a&b-3a&10-4a\\ -4&-8&-11\end{vmatrix}\stackrel{R_3-R_1}=\begin{vmatrix} -4&-8&-12\\ 9-2a&b-3a&10-4a\\ 0&0&1\end{vmatrix}=$$$${}$$
$$=\begin{vmatrix} -4&-8\\ 9-2a&b-3a&\end{vmatrix}=12a-4b+72-16a\stackrel?=0\iff a+b=18$$
DonAntonio
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