I need to find the conditions for $a$ and $b$ such that $$3x-2y+z=b$$ $$5x-8y+9z=3$$ $$2x+y+az=-1$$ has (i) unique solution, (ii) no solution (iii) infinitely many solutions

Here's what I did. Firstly, I put the situation into an augmented matrix form, and tried reducing to row echelon form to get the rank. Let C be the augmented matrix $C=[A:B]$ for the system of equations $AX=B$

$$ C= \left[ \begin{array}{ccc|c} 3&-2&1&b\\ 5&-8&9&3\\ 2&1& a & -1\\ \end{array} \right] $$

The operations that I used, in order, are:

$$R_1 \rightarrow R_1-R_2$$

$$R_1\rightarrow \frac{-R1}{2}$$

$$R_2\rightarrow R_2-5R_1$$

$$R_3\rightarrow R_3-2R_1$$

$$R_3 \rightarrow R_3-R_2$$ Reducing using these operations to row echelon form, I got:

$$ C= \left[ \begin{array}{ccc|c} 1&-3&4&\frac{3-b}{2}\\ 0&7&-11&3+5(\frac{3-b}{2})\\ 0&0& a+3 & \frac{7b-29}{2}\\ \end{array} \right] $$

I know, for unique solution, the ranks of A and C must be equal, and they should be equal to 3 (the number of unknowns), i.e.,

$($rank of the matrix A is being denoted by $\rho(A)$$)$

$\implies a+3\neq0$ and, $b \in \mathbb{R}$

This is the correct answer.

However, for no solution, I know that $\rho(A)\neq\rho(C)$

That is possible when, $a+3=0$ and $\frac{7b-29}{2} \neq 0$ $\implies b \neq \frac{29}{7}$

But, the answer says the conditions are $a = -3 $ and $b \neq \frac{1}{3}$ for the no solution part. What is my mistake here? I don't know where I went wrong, any help would be nice. If someone can point out the error in my steps, it would be very helpful

  • $\begingroup$ Is it not $27b-9$ instead of $7b-29$? $\endgroup$ Jan 15 at 16:18
  • $\begingroup$ According to the transformations that I used, no, that isn't the case. I'll edit the post with the transformations I used $\endgroup$
    – Techie5879
    Jan 15 at 16:20
  • 1
    $\begingroup$ Your solution seems correct, but I didn't check your work in reducing the matrix, and I believe this is probably where the mistake is. This is what I would check. $\endgroup$
    – GSofer
    Jan 15 at 16:23
  • $\begingroup$ I obtain for $a=-3$ that $b=1/3$ and $x=(5(3z - 1))/21$ , $y=(11(3z - 1))/21$. $\endgroup$ Jan 15 at 16:24
  • $\begingroup$ As can be checked using WolframAlpha, you must have made a mistake in your row operations. $\endgroup$
    – Christoph
    Jan 15 at 16:25

It seemed to be a calculation error. I have now found that the previous row-echelon form that I got was incorrect


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