# Find constant a such that the linear equation system has two different solutions?

I have the following equation system shown in its augmented matrix form: However, this is the one solution for $$a = \pm \sqrt{2}$$ (from what I can see). So is there really any $$a$$ such that there are two different solutions? If you a smart enough to solve it - please take me through the process, like how you engage the problem :)

• WolframAlpha says that you computed the solution incorrectly, can you share your work, please? – Viktor Glombik Apr 30 '19 at 9:13
• Thanks for the input, Viktor. However, it is correct :) Try substituting the solution back into the equations and you will see it :)) – Dip Apr 30 '19 at 10:21
• Why did you delete the matrices? Your question makes no sense now. – amd Apr 30 '19 at 20:51

There cannot be TWO solutions, as the set of solutions $$S$$ is either empty (when the ranks of the matrix and of the augmented matrix are different), or an affine subspace, of codimension this common rank.

If you only mean (at least) two different solutions, it is equivalent to $$\dim S >0$$, i.e. the codimension (or the common rank of the matrix and the augmented matrix) is $$<3$$.

Anyway, when you divide by some coefficient in a row operation, it is always under the assumption this coefficient is $$\ne 0$$, so when it is $$0$$, you should examine the linear system before dividing.

• They don't say exactly two solutions. Just two different solutions. – Yves Daoust Apr 30 '19 at 9:34
• @YvesDaoust: As the O.P. wrote two in upper case, I interpreted it as meaning ‘exacrtly two’. Anyway, the end of the first sentence implicitly gives the answer. Probably I should make things quite clear. – Bernard Apr 30 '19 at 9:41
• you are right, these uppercase are ambiguous. – Yves Daoust Apr 30 '19 at 9:44
• Yea, I see the problem with emphasizing two. – Dip Apr 30 '19 at 10:26

The determinant of the system is $$a(a-3)$$.

With $$a=0$$,

$$\left[\begin{array}{rrr|r} 2&3&2&2\\ 1&0&1&0\\ 0&2&0&0 \end{array}\right]$$ is impossible ($$2\ne2\cdot0+\dfrac32\cdot0$$) and with $$a=3$$, $$\left[\begin{array}{rrr|r} 2&6&2&5\\ 1&3&1&3\\ 3&2&6&0 \end{array}\right]$$ as well ($$5\ne2\cdot3$$).

Hence for other values of $$a$$, the solution is always unique.

By Gaussian elimination:

We subtract the second row

$$\left[\begin{array}{rrr|r} 2&3+a&2&2+a\\ 1&a&1&a\\ a&2&2a&0 \end{array}\right]$$ to obtain a scrambled echelon form $$\left[\begin{array}{rrr|r} 0&3-a&0&2-a\\ 1&a&1&a\\ 0&2-a^2&a&-a^2 \end{array}\right].$$

which we reorder as

$$\left[\begin{array}{rrr|r} 1&a&a&1\\ 0&a&2-a^2&-a^2\\ 0&0&3-a&2-a\\ \end{array}\right].$$

• Wow - thanks! Great. So to answer the question: "Find constant a such that the linear equation system has two different solutions": it does not depend on a, but two different solution sets are obtainable. Is that correct, sir? – Dip Apr 30 '19 at 10:13
• Because your Guassian part did not seem to make any assumptions about a – Dip Apr 30 '19 at 10:14
• @Dip: sorry, can't make sense of your comments. – Yves Daoust Apr 30 '19 at 10:41
• Alright, let me try to rephrase it. Try to answer the question "Find constant a such that the linear equation system has two different solutions" directly. You seem to make no assumptions about a in your answer but found "another" solution set. So would "t does not depend on a, but two different solution sets are obtainable" be a correct answer to "Find constant a such that the linear equation system has two different solutions"? – Dip Apr 30 '19 at 10:49
• The main question is about finding a value of a such that there are two different solutions. And since it seems like you make no assumptions about a in your Gaussian part, I am wondering what your direct answer to ""Find constant a such that the linear equation system has two different solutions" would be. – Dip Apr 30 '19 at 10:50