Analyse the linear system of equations:

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



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?

  • $\begingroup$ Are you sure of the last component $-(2-\lambda)$ ?Were it instead $(2-\lambda)$, the whole issue would have been an eigenvalue issue... $\endgroup$
    – Jean Marie
    Apr 9, 2022 at 21:57
  • $\begingroup$ Sorry which component do you mean? $\endgroup$
    – Game Code
    Apr 9, 2022 at 22:00
  • $\begingroup$ I mean the very last term of your initial system : $-(2-\lambda)z$ $\endgroup$
    – Jean Marie
    Apr 9, 2022 at 22:03
  • 1
    $\begingroup$ @JeanMarie, not a typo- I performed a row operation to switch r1 and r3. $\endgroup$
    – Game Code
    Apr 9, 2022 at 22:39
  • 1
    $\begingroup$ 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. $\endgroup$
    – Game Code
    Apr 9, 2022 at 23:11

1 Answer 1


Your parametric system can be written:

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



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:


  • 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.
  • 1
    $\begingroup$ 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) $\endgroup$
    – Game Code
    Apr 9, 2022 at 22:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .