Creative way to solve a linear system

Question

Does anyone know a method for solving the following linear system:

Here, $\pi_0, \, \pi_1, \, \pi_2$ are the variables.

I tried solving by setting up a simple matrix and row-reducing, but it quickly became a nightmare. Can anyone see a simpler way?

Answer 1

Try adding the second equation to the first one. Then you are left with $\pi_2-b\pi_1=0$ which is just the third equation with a negative sign. As you have a zero on the right side, you see that equation 3 is redundant and you can go on using a 3x3 system. I hope that helps

Answer 2

Hint

If you had the first and second equations, you get the third. So, ignore the first one and consider only the other.

If you don't want to use matrix, eliminate successivel variables until you end withe one equation for one unknown. Take care : may be there is a problem somewhere !

I am sure that you can take from here.

Answer 3

Try $\pi_0=\frac{1-b}{ab+a-b+1}$, $\pi_1=\frac{a}{ab+a-b+1}$ and $\pi_2=\frac{ab}{ab+a-b+1}$, for $ab+a-b+1\neq 0$ to check your answer. Write your system as $Ax=b$ with $A\in M_3(K)$ and $b=(0,0,1)^t$. You should compute $\det(A)$. (Hint: it is equal to $ab+a-b+1$). If it is nonzero, then there is a unique solution given by $x=A^{-1}b$. For $\det(A)=0$ compute $\ker (A)$.

Answer 4

Matrix method doesn't look so badly for me: \begin{matrix} a & -1 & 1 & | & 0\\ -a & 1-b & 0 &| & 0\\ 1 & 1 & 1 & | & 1 \end{matrix}

Add the first row to the second and the first divided by $-a$ to the third: \begin{matrix} a & -1 & 1 & | & 0\\ 0 & -b & 1 &| & 0\\ 0 & \frac{a+1}{a} & \frac{a-1}{a} & | & 1 \end{matrix}

Add the second row multiplied by $\frac{a+1}{ab}$ to the third: \begin{matrix} a & -1 & 1 & | & 0\\ 0 & -b & 1 &| & 0\\ 0 & 0 & \frac{a-b+ab+1}{ab} & | & 1 \end{matrix}

Find $\pi_2$: \begin{matrix} a & -1 & 1 & | & 0\\ 0 & -b & 1 &| & 0\\ 0 & 0 & 1 & | & \frac{ab}{a-b+ab+1} \end{matrix}

Then $\pi_1$: \begin{matrix} a & -1 & 1 & | & 0\\ 0 & 1 & 0 &| & \frac{a}{a-b+a+1}\\ 0 & 0 & 1 & | & \frac{ab}{a-b+ab+1} \end{matrix}

And $\pi_0$: \begin{matrix} 1 & 0 & 0 & | & \frac{1-b}{a-b+ab+1}\\ 0 & 1 & 0 &| & \frac{a}{a-b+ab+1}\\ 0 & 0 & 1 & | & \frac{ab}{a-b+ab+1} \end{matrix}

I'm sorry for such detailed calculations, I just wanted to show you that regardless of method you use, you can't avoid these operations with fractions, matrix method is just a simplified representation of other solution methods.

Answer 5

Part of your confusion is probably from the fact that you have 4 equations and only 3 unknowns. Either, one of your equations is redundant, or there is no solution to your statement. So really, the question comes down to: which equation to eliminate?

There might be no clear way to answer that, as any 3 of the equations may be linearly independent. Try picking 3 that are 'likely' to be linearly independent, and solving for the results. Once you have the results, see if the results are consistent with the last equation (i.e., plug in the results into the last equation). If you find from picking 3 of the equation that you end up with a row of 0s at the bottom, then just replace the last row with your 4th equation.

If you are looking for purely alternative ideas, you could use substitution to reduce the total variable count. However, this will not change the overall complexity of the process. Still, It may make each step easier to understand, so it is worth a try.

Try for example, solving for π2 in the third equation to get:

π2 = bπ1

Now, replace π2 with bπ1 in equations 2 and 4. Your new equations become:

-aπ0 + (1-b)π1 = 0
aπ0 - π1 + (bπ1) = 0
π0 + π1 + (bπ1) = 1

Notice a couple things here: First, there is no need to rewrite the third equation, since it is just going to tell you how π2 relates π1. Second, notice that the first two equations are redundant, and fortunately, they are consistent. Therefore, you can just drop one of them.

You are left with 2 equations with 2 unknowns:

aπ0 + (b-1)π1 = 0
π0 + (b+1)π1 = 1

Solving the rest from here can now either be done with matrix methods (having eliminated the extra equation), or again using substitution. Having solved for π0 and π1, you can then find π2 based on equation 3.