# Find the unique value of K for which the system has infinitely many solutions

So I've been working through this linear algebra problem. Find the unique value of k for which the following system has infinitely many solutions

\begin{align} 2x_1 + 2x_2 - x_3 + x_4 &= 10 \\ 3x_1 + x_2 + x_3 - 2x_4 &= 1 \\ x_1 + 4x_2 - x_3 + 9x_4 &= 13 \\ x_1 + x_2 + x_3 + kx_4 &= -1 \end{align}

$$\begin{bmatrix} 2 & 2 & -1 & 1 & 10\\ 3 & 1 & 1 & -2 & 1\\ 1 & 4 & -1 & 9 & 13\\ 1 & 1 & 1 & k & -1 \end{bmatrix}$$

So, I've reduced this to

$$\begin{bmatrix} 1 & 0 & 0 & -2 & 1\\ 0 & 1 & 0 & 3 & 2\\ 0 & 0 & 1 & 1 & -4\\ 0 & 0 & 0 & k-2 & 0 \end{bmatrix}$$

which I think the value of $$k$$ should be $$2$$. But then this makes $$x_4$$ equal to $$0$$ which implies that there is a unique solution to the system? Can anyone help?
*edit - I've been consulting tutor who told me that k = 3 as this would make that entry 1 which gives an infinite solution. Thanks for all your help!

• $k=2$ is precisely the value which does not make $x_4$ equal to $0$! Commented Apr 18, 2023 at 21:26
• A (questionable) shortcut is compute the determinant of the original matrix, as a function of $~k~$ and then determine, which values of $~k~$ cause the determinant to equal $~0.~$ I refer to this shortcut as questionable, because it may actually require more time/effort than the approach taken in the posted question. Commented Apr 18, 2023 at 21:28
• Try to plug in $k=2$ into your last reduced matrix. Recall that this matrix corresponds to the coefficients. Commented Apr 18, 2023 at 21:28

We can use the Rouché–Capelli theorem that can be more readable in this form:

Given the linear system $$Ax = B$$ and the augmented matrix (A|B).

1. If rank(A) = rank(A|B) = the number of rows in x, then the system has a unique solution.

2. If rank(A) = rank(A|B) < the number of rows in x, then the system has Infinitely number of solutions.

3. If rank(A) < rank(A|B), then the system is inconsistent.

To prove that the system at hand has Infinitely number of solutions, try to show that there is at least one row that looks like this in the agumented matrix.

$$\begin{pmatrix}0&0&0&0&| 0\end{pmatrix}$$

We can reach this result by transforming the augmented matrix to row-echelon form to get:

$$\begin{pmatrix}2&2&-1&1&10\\ 0&-2&2.5&-3.5&-14\\ 0&0&3.25&3.35&-13\\ 0&0&0&k-2&0\end{pmatrix}$$

In this transformed version, when $$k=2$$ the system will have infinite number of solutions because the fourth variable (say, $$x_4$$) can assume any possible number without constraints, for each of those values, we can computer the corresponding value of the other variables.

Note that you must use the augmented matrix not just the coefficient matrix for this solution.

A second approach would be using this fact:

A nxn homogeneous system of linear equations has a unique solution (the trivial solution) if and only if its determinant is non-zero. If this determinant is zero, then the system has an infinite number of solutions.

In your case the determinant's value is $$k-2$$. When $$k=2$$, there would be an infinite number of solutions.

Note that you don't need to use the augmented matrix to find the determinant.

For infinitely many solutions to exist, you should be able to write eqn (4) as a linear combination of the first 3, i.e.

$$\text{eqn}(4) = a \times \text{eqn}(1) + b \times \text{eqn}(2) + c \times \text{eqn}(3)$$

Plugging in the numbers:

$$2a + 3b + c = -1$$

$$2a + b + 4c = -1$$

$$-a + b -c = -1$$

$$a - 2b + 9c = -k$$

From the first two equations,

$$2b-3c = 0$$, i.e. $$b = 3c/2$$

From the first and third equations,

$$5b-c = -3$$, i.e. $$15c/2-c = -3$$, i.e. $$c=-6/13$$, $$b = -9/13$$

From the third equation, $$a = b-c + 1 = 10/13$$

Finally, $$k = -a + 2b -9c = -10/13 -18/13 + 54/13 = 26/13 = 2$$