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Problem: Solve the following system in function of the parameter $b$:

\begin{align*} \begin{cases} -bx + 2y - (2+b^2)z + bu &= -2 \\ x -2y + bz -u &= 0 \\ x + (2b-4)y + (2-b)z + (b-1)u &= 2 \\ x -2by -(3b+2)z + (4b-5)u &= 2b-4 \end{cases} \end{align*}

Attempt at solution: We write down the augmented matrix of this system, and then apply the operations: $R_1 \leftrightarrow R_2, R_2 \rightarrow R_2 + b R_1$. This gives us the matrix: \begin{align*} \left(\begin{array}{cccc|c} 1 & -2 & b & -1 & 0 \\ 0 & (2-2b) & (2+b^2) & 0 & -2b \\ 1 & (2b-4) & (2-b) & (b-1) & 2 \\ 1 & -2b & -(3b+2) & (4b+5) & (2b-4) \end{array}\right) \end{align*} After that, we do $R_3 \rightarrow R_3 - R_1$ and $R_4 \rightarrow R_4 - R_1$: \begin{align*}\left(\begin{array}{cccc|c} 1 & -2 & b & -1 & 0 \\ 0 & (2-2b) & (2+2b^2) & 0 & -2b \\ 0 & (2b-2) & (2-2b) & b & 2 \\ 0 & (-2b+2) & (-4b-2) & (4b-4) & (2b-4) \end{array}\right) \end{align*} Now I want a leading $1$ at the position $a_{22}$.

Case 1. Let $b = 1$. Then our matrix reduces to \begin{align*} \left(\begin{array}{cccc|c} 1 & -2 & 1 & -1 & 0 \\ 0 & 0 & 4 & 0 & -2 \\ 0 & 0 & 0 & 1 & 2 \\ 0 & 0 & -6 & 1 & -2 \end{array}\right) \end{align*} From the second row we see than that $u =2$. Substituting this in the last equation gives $z = - \frac{2}{3}$. But from the second row $z = - \frac{1}{2}$, which is a contradiction. Does this mean I can conclude the system has no solutions in this case?

Then if $b \neq 1$, should I just proceed with Gauss-elimination untill I hit another case distinction?

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  • $\begingroup$ Do you really need a leading $1$? If the only thing you need is solving the system, then this would be an innecesary complication. However, you're going to need those cases once you finish the reduction. $\endgroup$ – Daniel May 3 '15 at 21:09
  • $\begingroup$ I don't know. How else could I solve it? $\endgroup$ – Kamil May 3 '15 at 21:13
  • $\begingroup$ I mean, you can let the $(2-2b)$ as it is, and keep going with the other rows. However, if $2-2b=0$, then the matrix wouldn't be reduced. Don't get confused, you're doing it great, and yes, you must work all the cases as they appear. $\endgroup$ – Daniel May 3 '15 at 21:14
  • $\begingroup$ Are the entries supposed to be real or complex numbers? If they are real numbers, you can divide row 2 by $a_{2,3}$: $2+2b^2$ because it can never be zero for real entries. By the way... it should probably be $2+b^2$ if you look closely on the equation system before. $\endgroup$ – mathreadler May 4 '15 at 5:53

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