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A challenging system of equations


Active member
Jan 27, 2012
Hello all,

I am working on this system of equations, and I do not get the same results as they appear in the final solution in the book, need your assistance here...this is the question:

Discuss the solutions of the equation system:

\[\begin{matrix} ax_{1}+bx_{2}+2x_{3}=1\\ ax_{1}+(2b-1)x_{2}+3x_{3}=1\\ ax_{1}+bx_{2}+(b+3)x_{3}=2b-1 \end{matrix}\]

I have applied elementary row operations R2->R2-R1 and R3->R3-R1 and got this matrix:

\[\begin{pmatrix} a &b &2 &1 \\ 0 &b-1 &1 &0 \\ 0 &0 &b+1 &2b-2 \end{pmatrix}\]

However, in the book they say the matrix after elementary row operations is:

\[\begin{pmatrix} a &1 &1 &1 \\ 0 &b-1 &1 &0 \\ 0 &0 &b+1 &2(b-1) \end{pmatrix}\]

which is odd since I see no reason to touch the first row.

This is not the end of the troubles, the solution to the problem according to the book is:

"There are six cases:

b=1: infinite solution
b=5, a=0: infinite solution
b=5, a~=0: unique solution
b=-1: no solution
b~=+1 or -1 or 5, a~=0: unique solution
b~=1 or 5, a=0: no solution

(~= means not equal, gave up figuring it out in latex)

Where did they get the 5 from and where and how shall I see this in the matrix ?

Thanks !


Well-known member
MHB Math Scholar
Feb 15, 2012
First off, the difference in matrices should not bother you, they just did one more step, subtracting row 2 from row 1.

If one actually calculates what we get from back substitution, we have:

$x_1 = \dfrac{-(b+1)(b - 5)}{a(b^2 - 1)}$

$x_2 = \dfrac{-2(b-1)}{b^2-1}$

$x_3 = \dfrac{2(b-1)}{b+1}$.

This tells us straight off the bat we need to look at the cases where $a = 0$ and $b = -1,1$ on their own.

Clearly, if $b = -1$ we can have no solution, as this means:

$0 = (b+1)x_3 = 2(b - 1) = -4$, from row 3.

So we have to look at the following possibilities (all of which assume $b \neq -1$):

$a = 0, b = 1$
$a \neq 0, b = 1$
$a = 0, b \neq 1,-1$
$a \neq 0, b \neq 1,-1$.

Let's look at the case $b = 1, a\neq 0$ first. Here we see that we must have $x_3 = 0$ and that we have the infinite number of solutions:


Now in the case $b = 1, a = 0$, we find that $x_1$ can be anything, and we get the infinite number of solutions:

$(t,1,0)$. So if $b = 1$, we get an infinite number of solutions. So far we have found that we agree with 2 of the 6 cases of your text.

Now let's turn to the case $a \neq 0, b \neq 1,-1$.

This gives the unique solution above (at the beginning of this post), which actually covers cases 3 and 5 of your text (which could be combined). That's 4 out of 6 so far.

If we have $a = 0, b \neq 1,-1$ clearly if we have ANY solution at all, we have infinitely many, since $a = 0$ drops $x_1$ right out of the system.


$x_3 = \dfrac{2(b - 1)}{b+1}$ (from row 3) poses no problem at all, and:

$x_2 = \dfrac{-2}{b+1}$ (from row 2 and row 3) doesn't pose any problem, either.

But now row 1 tells us that:

$\dfrac{2(b - 1)}{b+1} + \dfrac{-2}{b+1} = 1$ that is:

$2b - 4 = b+1 \implies b = 5$

(which is case 2), so if $b \neq 5$ there is no solution (which is case 6, sort of, there's no need to distinguish between the sets:

a)$b \neq 1,-1,5, a = 0$ and
b)$b \neq 1,5, a = 0$ and $b \neq -1$

because if $b = -1$ we already know there is no solution from case 4).

So your text is correct, they just arranged the answers differently than you might have been thinking about them.

(by the way, the latex for "not equals" is \neq)