Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Determine real number(s)for $a,b$ such that the system has no solution, has a unique solution, and has more than one solution: $$\begin{align*} x-2y+az-t&=1\\ -x+y-z+t&=-1\\ (a+1)y-a^2z+at&=0\\ (b+1)y-abz-a^2t&=b \end{align*}$$

I could not transform the matrix into row reduced form.So that I could not find $a$ and $b$. I will be glad if someone could solve it.

share|cite|improve this question

Your matrix will have $a$ and $b$ indicated. You then have to be careful with them.

(I hope I didn't make any silly arithmetic or algebra mistakes; but even if I did, you should be able to see how one proceeds: doesn't matter what $a$ and $b$ are, so long as you are careful not to divide by an expression containing them, since you won't know ahead of time that what you are dividing by is not zero).

The augmented matrix for the system is: $$\left(\begin{array}{rrrr|r} 1 & -2 & a & -1 & 1\\ -1 & 1 & -1 & 1 & -1\\ 0 & a+1 & -a^2 & a & 0\\ 0 & b+1 & -ab & -a^2 & b \end{array}\right).$$ Proceed as usual. First, we add the first row to the second row: $$\left(\begin{array}{rrrr|r} 1 & -2 & a & -1 & 1\\ 0 & -1 & a-1 & 0 & 0\\ 0 & a+1 & -a^2 & a & 0\\ 0 & b+1 & -ab & -a^2 & b \end{array}\right).$$ Then, multiply the second row by $-1$: $$\left(\begin{array}{rrrr|r} 1 & -2 & a & -1 & 1\\ 0 & 1 & 1-a & 0 & 0\\ 0 & a+1 & -a^2 & a & 0\\ 0 & b+1 & -ab & -a^2 & b \end{array}\right).$$ Now, subtract $(a+1)$-times the second row from the third row; multiplying the second row by $-(a+1)$ gives $(0\quad -(a+1)\quad (a^2-1)\quad 0\>|\>0)$, so we have: $$\left(\begin{array}{rrrr|r} 1 & -2 & a & -1 & 1\\ 0 & 1 & 1-a & 0 & 0\\ 0 & 0 & -1 & a & 0\\ 0 & b+1 & -ab & -a^2 & b \end{array}\right).$$ Multiply the third row by $-1$: $$\left(\begin{array}{rrrr|r} 1 & -2 & a & -1 & 1\\ 0 & 1 & 1-a & 0 & 0\\ 0 & 0 & 1 & -a & 0\\ 0 & b+1 & -ab & -a^2 & b \end{array}\right).$$ Now subtract $(b+1)$-times the second row from the fourth row. Multiplying the second row by $-(b+1)$ gives $(0\quad (b+1)\quad (ab+a-b-1)\quad 0\>|\>0)$, so we get: $$\left(\begin{array}{rrrr|r} 1 & -2 & a & -1 & 1\\ 0 & 1 & 1-a & 0 & 0\\ 0 & 0 & 1 & -a & 0\\ 0 & 0 & a-b-1 & -a^2 & b \end{array}\right).$$ Now, subtract $(a-b-1)$-times the third row from the fourth row. Multiplying the third row by $-(a-b-1)$ gives $(0\quad 0\quad (1+b-a)\quad (a^2-ab-a)\>|\>0)$, so we get: $$\left(\begin{array}{rrrr|r} 1 & -2 & a & -1 & 1\\ 0 & 1 & 1-a & 0 & 0\\ 0 & 0 & 1 & -a & 0\\ 0 & 0 & 0 & -(ab+a) & b \end{array}\right).$$ Now we can figure out the rank easily enough.

If $-(ab+a) = -a(b+1)\neq 0$, then the coefficient matrix has rank $4$ and the augmented matrix has rank $4$, so the system has exactly one solution.

If, however, $-a(b+1)=0$ and $b\neq 0$, then the coefficient matrix would have rank $3$ and the augmented matrix would have rank $4$, so that would be no solutions.

And, finally, if $-a(b+1)=0$ and $b=0$, then both the coefficient and augmented matrices have rank $3$, so the system would have an infinite number of solutions.

So now, it's just a matter of determining what are the conditions on $a$ and $b$ that give you those three possibilities. Suggestion: Deal with them in the inverse order I listed them: first the case where $-a(b+1)=0$ and $b=0$, because you'll be able to say exactly what $a$ and $b$ are; then consider the middle case; and finally the first mentioned case.

share|cite|improve this answer
    
Thank you very much for your effort.I was confused with a and b's when I was doing row reductions.Thank you again for the each steps that you have done and for your clear explanations. – Piril Jun 18 '11 at 20:59
    
@Piril: Arturo's answer is compeletly right. :) – Babak S. Jun 20 '11 at 10:07

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.