Finding number of solutions in system of three equations with four variables and one parameter

Question

As a continuation (my previous question regarding this topic) of reminding myself Algebra I would like you to check my calculations, and if it possible - to suggest better $U$ matrix transformations.

There is a system of equations like this:

$\begin{cases} -3x-2y-pz-4t=-1\\ 2x+1y-1z+(1-p)t=1 \\ 1x+py+2z+4t=p+5 \end{cases}$

As you can see there are three equations, four variables and one parameter $p$.

I'd like to discuss how many there are solutions depending on value of parameter $p$.

Using Kronecker–Capelli theorem I can be assured that for any $p$ value there won't be only one solution (because $rA < n$ and $rU < n$, where $n$ is number of variables equal $4$).

So, my $A$ matrix is: \begin{align} A = &\left[\!\! \begin{array}{cccc} -3&-2&-p&-4 \\ 2&1&-1&1-p \\ 1&p&2&4 \end{array} \!\!\right] \end{align}

And $U$ matrix is: \begin{align} U = &\left[\!\! \begin{array}{cccc|c} -3&-2&-p&-4&-1 \\ 2&1&-1&1-p&1 \\ 1&p&2&4&p+5 \end{array} \!\!\right] \end{align}

Now, I transformed $U$ matrix to $REF$ form. Any suggestions how to make it quicker or without such ugly denominators would be beneficial.

\begin{align} &\left[\!\! \begin{array}{cccc|c} -3&-2&-p&-4&-1 \\ 2&1&-1&1-p&1 \\ 1&p&2&4&p+5 \end{array} \!\!\right]\rightsquigarrow \left[\!\! \begin{array}{cccc|c} 1&p&2&4&p+5 \\ 2&1&-1&1-p&1 \\ -3&-2&-p&-4&-1 \end{array} \!\!\right]\rightsquigarrow \left[\!\! \begin{array}{cccc|c} 1&p&2&4&p+5 \\ 0&1-2p&-5&-p-7&-2p-9 \\ 0&-2+3p&-p+6&8&3p+14 \end{array} \!\!\right]\rightsquigarrow \left[\!\! \begin{array}{cccc|c} 1&p&2&4&p+5 \\ 0&1&-5\over{1-2p}&-p-7\over{1-2p}&-2p-9\over{1-2p} \\ 0&-2+3p&-p+6&8&3p+14 \end{array} \!\!\right], p \neq {1\over{2}} \rightsquigarrow \left[\!\! \begin{array}{cccc|c} 1&p&2&4&p+5 \\ 0&1&-5\over{1-2p}&-p-7\over{1-2p}&-2p-9\over{1-2p} \\ 0&1&-p+6\over{-2+3p}&8\over{-2+3p}&3p+14\over{-2+3p} \end{array} \!\!\right], p \neq {2\over{3}} \rightsquigarrow \left[\!\! \begin{array}{cccc|c} 1&p&2&4&p+5 \\ 0&1&-5\over{1-2p}&-p-7\over{1-2p}&-2p-9\over{1-2p} \\ 0&1&-p+6\over{-2+3p}&8\over{-2+3p}&3p+14\over{-2+3p} \end{array} \!\!\right] \rightsquigarrow \left[\!\! \begin{array}{cccc|c} 1&p&2&4&p+5 \\ 0&1&-5\over{1-2p}&-p-7\over{1-2p}&-2p-9\over{1-2p} \\ 0&0&2(p+2)(p-1)\over{(1-2p)(3p-2)}&3(p+2)(p-1)\over{(1-2p)(3p-2)}&-2(p+2)\over{(1-2p)(3p-2)} \end{array} \!\!\right] \end{align}

The transformation steps are as follows:

1. Swap $w_1$ and $w_3$
2. Subtract $2w_1$ from $w_2$ and add $3w_1$ to $w_3$
3. Divide $w_2$ by $1-2p$
4. Divide $w_3$ by $-2+3p$
5. Subtract $w_2$ from $w_3$

As I see, I should discuss the following $p$ parameter values:

1. $p = -2$ (would make numerator of fraction zero)
2. $p = 1$ (would make numerator of fraction zero)
3. $p = {1\over{2}}$ (because while the transformation was made I assumed it to be able to divide)
4. $p = {2\over{3}}$ (because while the transformation was made I assumed it to be able to divide)
5. $p = R \setminus \{-2, {1\over{2}}, {2\over{3}}, 1\}$ (others)

Firstly, let's deal with $p = {1\over{2}}$ and $p = {2\over{3}}$. If I count a minor of $A$ by taking $I = J = {1, 2, 3}$ (first, second, third row & column), the determinant is equal to:

$$\begin{vmatrix} -3&-2&-p\\ 2&1&-1\\ 1&p&2 \end{vmatrix}=-6-2p^2+2+p-3p+8=-2p^2-2p+4$$

Either if I put ${1\over{2}}$ or ${2\over{3}}$ for $p$, the $det \neq 0$, so the rank of $A$ and $U$ when $p = \{{1\over{2}}, {2\over{3}}\}$ is $3$.

What about $p = -2$? If $p = -2$ then $rA = 2 = rU$.

What about $p = 1$? If $p = 1$ then $rA = 2 \neq rU = 3$.

What about others? If $p = R \setminus \{-2, {1\over{2}}, {2\over{3}}, 1\}$ then $rA = 3 = rU$.

So the final answer is:

If $p = -2$ - infinitely many solutions (with two parameters).

If $p = 1$ - no solutions.

If $p = R \setminus \{-2, 1\}$ - infinitely many solutions (with one parameter).

Is the resolution of this exercise ok?

Answer 1

Your solution is okay, but you could have prevented the ugly denominators and special cases $p=1/2,2/3$ by avoiding the first step, $w_1\leftrightarrow w_2$:

$\begin{bmatrix} \begin{array}{cccc|c} -3&-2&-p&-4&-1\\ 2&1&-1&1-p&1 \\ 1&p&2&4&p+5 \end{array} \end{bmatrix}\xrightarrow[w_3\to w_3+w_1/3]{w_2\to w_2+2w_1/3} \begin{bmatrix} \begin{array}{cccc|c} -3&-2&-p&-4&-1\\ 0&-1/3&-1-2p/3&-5/3-p&1/3\\ 0&p-2/3&2-p/3&8/3&p+14/3 \end{array} \end{bmatrix}\\\xrightarrow{w_3\to w_3+(3p-2)w_2} \begin{bmatrix} \begin{array}{cccc|c} -3&-2&-p&-4&-1\\ 0&-1/3&-1-2p/3&-5/3-p&1/3\\ 0&0&-2(p-1)(p+2)&-3(p-1)(p+2)&2(p+2) \end{array} \end{bmatrix}$

From here, it can be easily seen that the system has:

$\begin{cases}\text{no solution},&p=1\\\text{infinitely many solutions with 2 free variables},&p=-2\\\text{infinitely many solutions with 1 free variable},&\text{otherwise}\end{cases}$