How to deal with systems of linear equations with parameters

Question

Good morning everyone, I have with myself not a small-big problem - when checking whether a given system of equations is inconsistent / consistent / marked As long as I operate on the numbers themselves it is elegant, I check the conditions with Kronecker and it is elegant However, when the type task appears:

Specify for which $a \in \Bbb R $ the given system is inconsistent / consistent / marked $$\begin{cases} x+5y+a^2z = - 3 \\2x+2y-a^2z = 3a\\-4x + 3y + z = -5 \end{cases}$$ I stuck myself, every time I write a matrix, I come to some form like this: $$ \begin{bmatrix}0&-4&a^2&(-2-a)\\1&-1&0&(-1+a)\\0&-5&(a^2+1)&(-11+2a)\end{bmatrix}$$ I do not even know if the form I have brought is beneficial to me, I transform it until I intuitively see it, but in fact I do not know what to stop, what is the target form of martix, when I should stop transforming and start consider cases.

Then I act completely mindlessly - i.e. I check for which $a$, some elements are zeroing and I look what's going on for $$a = 0, a=1, a= \frac{11}{2}$$ Then there is the case when $$ a \in \Bbb R \setminus \left\{ 0,1,\frac{11}{2}\right\} $$ and I do not know how to behave, i.e. I have a problem with calculating it to the end, with the detection of some patterns of action in these cases. How to solve such systems of equations with parameters from the beginning to the end? Thanks for your time... PS: Please, don't mark this question as duplicate - I know that there were some problems like that, but nowhere I have seen a clearify "instruction" (I know it sounds stupid) how to deal with these tasks.

Answer 1

I think the problem here is that you don't have any formal algorithm for how to transform the equation and you are kind of just doing what feels right. Instead, you should focus on getting the determinant in reduced-row echelon form, or RREF. Please read this Wiki article to find out what RREF. In this answer, I will show you how to transform a matrix into RREF form. First, write out the augmented matrix:

$$\left[\begin{matrix}1 & 5 & a^2 & -3 \\ 2 & 2 & -a^2 & 3a \\ -4 & 3 & 1 & -5\end{matrix}\right]$$

Now, we have a $1$ in the top-left corner, which is good. Using this $1$, I will eliminate the $2$ and $-4$ in the first column using row addition. Subtract the second row by two times the first row and add the third row by four times the first row:

$$\left[\begin{matrix}1 & 5 & a^2 & -3 \\ 0 & -8 & -3a^2 & 3a+6 \\ 0 & 23 & 4a^2+1 & -17\end{matrix}\right]$$

Now, we have a $-8$ in the leading element of the second row, so we will divide by $-8$:

$$\left[\begin{matrix}1 & 5 & a^2 & -3 \\ 0 & 1 & \frac 3 8a^2 & -\frac 3 8 a-\frac 3 4 \\ 0 & 23 & 4a^2+1 & -17\end{matrix}\right]$$

Now, we can eliminate the $5$ and $23$ in the second column using row addition of the second row. Subtract the first row by five times the second row and subtract the third row by twenty-three times the second row:

$$\left[\begin{matrix}1 & 0 & -\frac 7 8a^2 & \frac {15} 8a+\frac 3 4 \\ 0 & 1 & \frac 3 8a^2 & -\frac 3 8 a-\frac 3 4 \\ 0 & 0 & -\frac{37}8a^2+1 & \frac{69}8a+\frac 1 4\end{matrix}\right]$$

Now, we have $-\frac{37}8a^2+1$ leading in the third row, so we need to divide the third row by $-\frac{37}8a^2+1$. However, before we do this, we need to make sure that we are not dividing by zero:

$$-\frac{37}8a^2+1\neq 0\rightarrow a\neq \pm 2\sqrt{\frac 2 {37}}$$

Thus, this system is consistent only for $a\neq \pm 2\sqrt{\frac 2 {37}}$ because otherwise, the last equation would turn into $0=\frac{69}8a+\frac 1 4$, which is clearly a contradiction. Using this knowledge, we can divide by $-\frac{37}8a^2+1$:

$$\left[\begin{matrix}1 & 0 & -\frac 7 8a^2 & \frac {15} 8a+\frac 3 4 \\ 0 & 1 & \frac 3 8a^2 & -\frac 3 8 a-\frac 3 4 \\ 0 & 0 & 1 & \frac{69a+2}{8-37a^2}\end{matrix}\right]$$

Finally, we can use row addition to eliminate the $-\frac 7 8a^2$ and $\frac 3 8a^2$ in the third column:

$$\left[\begin{matrix}1 & 0 & 0 & \frac{483a^3+14a^2}{64-296a^2}+\frac {15} 8a+\frac 3 4 \\ 0 & 1 & 0 & -\frac{207a^3+6a^2}{64-296a^2}-\frac 3 8 a-\frac 3 4 \\ 0 & 0 & 1 & \frac{69a+2}{8-37a^2}\end{matrix}\right]$$

This is the final RREF form of the original matrix. From this augmented matrix, we can finally see that the solution is as follows:

$$x=\frac{483a^3+14a^2}{64-296a^2}+\frac {15} 8a+\frac 3 4$$ $$y=-\frac{207a^3+6a^2}{64-296a^2}-\frac 3 8 a-\frac 3 4$$ $$z=\frac{69a+2}{8-37a^2}$$

Answer 2

From the first equation we get $$x=3-5y-a^2z$$ so we can eliminate $x$ in the second and third equation: $$-8y-3a^2z=3a-6$$ $$23y+4a^2z+z=17$$ From here we get $$y=-\frac{1}{8}(3a-6+3a^2z)$$ Now we get for $z$ $$z\left(-\frac{69}{8}a^2+4a^2+1\right)=17-\frac{138}{8}+\frac{69}{8}a$$ Can you finish?