Matrix rank condition for consistency of a system of linear equations.

Question

I am stuck in this problem and I want to know if there's anything wrong in the approach. Along with the correct solution, kindly point out what is wrong in my approach!

Show that there are two values of k for which the equations, $$\begin{align} kx+3y+2z & =1 \\ x+(k-1)y& = 4 \\ 10y+3z& =-2\\ 2x-ky-z& =5 \end{align}$$ are consistent. Find their common solution for that value of k which is an integer.

My attempt

Writing the given system of equations in matrix form, \begin{align} \begin{bmatrix} k &3 &2 \\ 1 &k-1 &0 \\ 0 & 10 & 3\\ 2 & -k &-1 \end{bmatrix}\begin{bmatrix} x\\ y\\ z \end{bmatrix}& = \begin{bmatrix} 1\\ 4\\ -2\\ 5 \end{bmatrix} \\\\i.e., AX & = B \end{align}

Now we'll construct the augmented matrix A|B and reduce it to echelon form (to comapre rank later): \begin{align} \therefore A|B & = \begin{bmatrix} k &3 &2 &|1\\ 1 &k-1 &0 &|4\\ 0 & 10 & 3&|-2\\ 2 & -k &-1 &|5 \end{bmatrix} \\ \\R_{2} \mapsto R_{2}-\frac{R_{1}}{k} & = \begin{bmatrix} k &3 &2 &|1\\ 0 &k-1-\frac{3}{k} &\frac{-2}{k} &|4-\frac{1}{k}\\ 0 & 10 & 3&|-2\\ 2 & -k &-1 &|5 \end{bmatrix}= \begin{bmatrix} k &3 &2 &|1\\ 0 &\frac{k^{2}-k-3}{k} &\frac{-2}{k} &|\frac{4k-1}{k}\\ 0 & 10 & 3&|-2\\ 2 & -k &-1 &|5 \end{bmatrix} \\ \\R_{3} \mapsto R_{3}-\frac{10kR_{2}}{k^{2}-k-3} & = \begin{bmatrix} k &3 &2 &|1\\ 0 &\frac{k^{2}-k-3}{k} &\frac{-2}{k} &|\frac{4k-1}{k}\\ 0 & 0 & \frac{3k^{2}-3k+11}{k^{2}-k-3}&|\frac{-2k^{2}-38k+16}{k^{3}-k^2-3k}\\ 2 & -k &-1 &|5 \end{bmatrix} \\ \\R_{4} \mapsto R_{4}-\frac{2R_{1}}{k} & = \begin{bmatrix} k &3 &2 &|1\\ 0 &\frac{k^{2}-k-3}{k} &\frac{-2}{k} &|\frac{4k-1}{k}\\ 0 & 0 & \frac{3k^{2}-3k+11}{k^{2}-k-3}&|\frac{-2k^{2}-38k+16}{k^{3}-k^2-3k}\\ 0 & \frac{-k^{2}-6}{k} &\frac{-k-4}{k} &|\frac{5k-2}{k} \end{bmatrix} \end{align}

Now, how do I proceed from here. I want to check the condition that rank of A|B = rank of A for consistency. But even before that, for the matrix to be in echelon form, all the terms of last row have to be zero which is not possible for any k.

Correct answer is $k=3$ or $\frac{-1}{25}$

Source: This book.

Answer 1

As @xbh pointed out, the procedure needs to be continued two steps further and then equate the very last entry to 0. Unfortunately this leads to be a polynomial of k in 4th degree and there's probably a slip somewhere in the calculations.

Alternately, a necessary (although not sufficient) condition for consistency would be,

$$\begin{align*}\left |A|B \right | = 0 \\ \\ \begin{vmatrix} k & 3 & 2 & 1\\ 1 & k-1 & 0 & 4\\ 0 & 10 & 3 & -2\\ 2 & -k & -1 & 5 \end{vmatrix} = 0 \\ \\ 25k^{2}-74k-3 = 0 \\ \\ \therefore k = 3 \ \ or \ \ \frac{-1}{25} \end{align*} $$

We have to now put back each of these k in the system of equations to check that they're consistent indeed.

For the integer value of $k=3$, we get $x=2, y=1, z=-4$

Answer 2

xbh’s suggestion to swap rows should be followed aggressively. This will allow you to avoid dividing by expressions that involve $k$ for as long as possible and then, when you can’t avoid it any longer, to divide by the lowest-degree polynomials.

So, the first step is to swap rows 1 and 2. After clearing the first column, you should have $$\begin{bmatrix}1 & k-1 & 0 & 4 \\ 0 & -k^2+k+3 & 2 & -4k+1 \\ 0 & 10 & 3 & 2 \\ 0 & -3k+2 & -1 & -3\end{bmatrix}.$$ Dividing by the quadratic term at the (2,2) position looks unpleasant, so swap the second and third rows and clear the column below the pivot to get $$\begin{bmatrix}1 & k-1 & 0 & 4 \\ 0 & 10 & 3 & 2 \\ 0&0 & \frac1{10}(3k^2-3k+11) & -\frac15(k^2+19k-8) \\ 0&0 & \frac1{10}(9k-16) & -\frac15(3k+13) \end{bmatrix}.$$ You can’t avoid dividing by a polynomial in $k$ at this point, but you can at least swap the last two rows so that the quadratic is in the numerator. You will, of course, need to check the case $9k-16=0$ separately, since otherwise you’d have divided by zero. In any case, by aggressively swapping rows like this, the worst thing that you’re faced with at the end is solving a quadratic equation or two.