# Rank and Solution Type of a System of Linear Equations, dependent on parameter.

Analyse the linear system of equations:

$$(2-\lambda)x + y +2z=0$$

$$x+(4-\lambda)y-z=0$$

$$2x-y+(2-\lambda)z=0$$

where $$\lambda$$ is an arbitrary real parameter. (a) What is the rank of the associated coefficient matrix, A, depending on the value of $$\lambda$$?

(b) Based on the rank result in (a) what is your expectation for the qualitative structure of the solution?

(c) Confirm your expectation by an explicit calculation.

My approach:

Perform row operations as indicated:

$$r_1 \leftrightarrow r_3$$

$$r_2-\frac{1}{2}r_1 \rightarrow r_2$$

$$r_3 - (1-\frac{1}{2}\lambda)r_1 \rightarrow r_3$$

$$r_3-\frac{4}{9}r_2 \rightarrow r_3$$

to obtain the matrix: $$\begin{bmatrix} 2 & -1 & 2-x \\ 0 & \frac{9}{2}-x & -(2-\frac{1}{2}x) \\ 0 & -1/18x & (x+\frac{4}{9})(2-\frac{1}{2}x) \end{bmatrix}$$

(using $$x$$ to represent $$\lambda$$)

Then the only case I can think of where the rank is 2, is if $$x=4$$, then the $$A_{23},A_{33}=0$$, making column 1 and column 3 linearly dependent.

Since the bottom right hand 2x2 matrix can otherwise never have two sets of zeros, I believe the rank is 3 in all other cases.

(note: we haven't done determinants and eigenvalues yet so if you could share a solution that doesn't include these then that would be most helpful. Of course, I didn't simply want to take the determinant and solve a cubic)

(b) The nature of the solutions resembles the nullspace: if rank=3 => dimKerA=0=> solution is a point. rank = 2 => dimKer=1 => solution is a line. similarly a plane if dimKer=2, but this cannot occur.

(c) I explicitly test $$\lambda=2$$ and get inconsistent equations meaning there is no solution? I was not expecting this. I would have thought there will always be solutions as RHS is [0,0,0]

$$\lambda=4 \implies$$ [x,y,z]=[0,0,0], which I wasn't expecting either. Contradicts my solution to (b)

*Edit: I retried the question stopping 1 row operation earlier:

A-> $$\begin{bmatrix} 2 & -1 & 2-\lambda \\ 0 & \frac{9}{2}-\lambda & \lambda-2 \\ 0 & 2-\frac{1}{2}\lambda & \lambda(2-\frac{1}{2}\lambda) \end{bmatrix}$$

I then re-analysed cases of this matrix: $$\lambda=4 \implies$$ rank=2 => solution will be a line.

$$\lambda=9/2 \implies$$ rank=3 solution will be a point.

$$\lambda=0 \implies$$ rank=3 solution is a point

$$\lambda=2 \implies$$ rank=3 solution is a point.

Otherwise: rank also 3, since row2 independent to row3 => solution is a point.

In all cases, solutions only exist if the equations are consistent? How do I find whether or not the solution exists?

• Are you sure of the last component $-(2-\lambda)$ ?Were it instead $(2-\lambda)$, the whole issue would have been an eigenvalue issue... Apr 9, 2022 at 21:57
• Sorry which component do you mean? Apr 9, 2022 at 22:00
• I mean the very last term of your initial system : $-(2-\lambda)z$ Apr 9, 2022 at 22:03
• @JeanMarie, not a typo- I performed a row operation to switch r1 and r3. Apr 9, 2022 at 22:39
• Ok Thanks for your help @JeanMarie. GMT here (midnight). If you are able, perhaps another time, to describe the nature of the solutions now that we have the eigenvalues that would be immensely helpful. Apr 9, 2022 at 23:11

Your parametric system can be written:

$$(M-\lambda I_3)X=0 \tag{1}$$

with

$$M=\begin{pmatrix}2&1&2\\1&4&-1\\2&-1&2\end{pmatrix}\tag{2}$$

and where $$I_3$$ is the $$3 \times 3$$ identity matrix, $$X$$ is the column vector with entries $$x,y,z$$, and $$0$$ the null vector.

In fact, (1) can be recognized as the eigenvalue issue for matrix $$M$$.

The characteristic polynomial of $$M$$ is the determinant of system (1) ; it can be factored in the following way:

$$\det(M-\lambda I_3)=-(\lambda-4)(\lambda^2-4\lambda-2)\tag{3}$$

The roots of this polynomial are the eigenvalues of $$M$$:

$$2+\sqrt{6}, \ \ 4, \ \ 2-\sqrt{6}\tag{4}$$

Therefore it is not a surprise that

• if $$\lambda$$ is none of these values, the determinant of system (1) is non zero; therefore, this "homogeneous" system is invertible with unique solution:

$$(x,y,z)=(0,0,0)$$

• otherwise, if $$\lambda$$ is equal to one of the values in (4), it means that the determinant of system (1) is $$0$$ ; you have to treat each of the 3 cases in a separate way (the system becomes a rank-2 system, meaning practically that you can suppress for example the last equation.
• Thanks, I understand this solution as this was what I studied in high school, but my linear algebra course in college wants us to solve this problem using just row operations looking at the resultant upper-echelon matrix to find the nature of solutions. (I believe our next chapter is determinants and eigenvalues, which makes this kind of question trivial) Apr 9, 2022 at 22:41