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System of linear equations with a parameter

Yankel

Active member
Jan 27, 2012
398
Hello all,

I need your help with this tricky problem.

Determine the value of \[\lambda\] so that the following linear equation system of three unknowns has only the zero solution:

\[\begin{pmatrix} 1 &1 &1 &0 \\ 1 &\lambda &1 &0 \\ \lambda &1 &1 &0 \end{pmatrix}\]

I have started working on the system to bring it to the form in which I can answer the question, and after one stage I got here:

\[\begin{pmatrix} 1 &1 &1 &0 \\ 0 &\lambda-1 &0 &0 \\ 0 &1-\lambda &1-\lambda &0 \end{pmatrix}\]

Now here is where I am stuck, I don't know how to proceed, knowing that any division by \[\lambda-1\] or \[1-\lambda\] is not allowed since it can be a division by 0.

Will appreciate your guidance. Thanks.
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
Can't you replace the last line by the sum of the last two?

In general, if it makes sense to divide by an expression that may be equal to zero, you can go ahead and divide and note that later you have to go back and examine what happens when the expression does equal zero.

Also, am I missing something or there is a single value of $\lambda$ where the system has infinitely many solutions, while for all other values of $\lambda$ the system has a single zero solution?
 

Yankel

Active member
Jan 27, 2012
398
No, you are not missing anything, this is indeed the answer, and yes, I could have just sum them, a big Oops from my side ! :eek:

But generally speaking, like you said, so it is legal to divide, as long as I go and check it later ? So in this example:

\[\begin{pmatrix} 1 &k &1 &0 \\ 0 &1-k^{2} &0 &0 \\ 0 &-k^{2}-k-1 &-k &0 \end{pmatrix}\]

I can divide the second row by 1-k^2, then do

\[R_{3}x \mapsto R_{3}-(-k^{2}-k-1)R_{2}\]

?

If to so, I need to check what happen if 1-k^2 is 0. In this case k=1,-1.

The solutions in this case, are that for k=0,1,-1 there is an infinite number of solutions. Getting that directly, means that I have nothing left to check ? Am I correct with my solution ?

Thanks !


Edit:

Where should I check the values which comes from the division (that yield division by 0) ? In the original equations ?
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
If you want to avoid the mental gymnastics that go along with dividing by quantities that contain a variable, I would merely bring the (3x3 portion of the) matrix to upper-triangular form and use "back-substitution". So starting with:

$\begin{bmatrix}1&1&1&0\\0&\lambda - 1&0&0\\0&1 - \lambda&1 - \lambda&0 \end{bmatrix}$

by adding row 2 to row 3 we get:

$\begin{bmatrix}1&1&1&0\\0&\lambda - 1&0&0\\0&0&1 - \lambda&0 \end{bmatrix}$

The last row tells us that:

$(1 - \lambda)z = 0$

If we want the ONLY possible solution to be $z = 0$, clearly we can take any $\lambda \neq 1$.

Similarly, the second row tells us:

$(\lambda - 1)y = 0$

and here, again any $\lambda \neq 1$ forces $y= 0$.

Finally, the top row tells us:

$x + y + z = 0$.

If $\lambda \neq 1$, we have $y = z = 0$, which then forces $x = 0$.

Here is another approach:

The system will have a unique solution (the 0-vector, since it is a homogeneous system) iff the determinant of your original 3x3 matrix is non-zero. So let's take the determinant, and see what we discover:

$\begin{vmatrix}1&1&1\\1&\lambda&1\\ \lambda&1&1 \end{vmatrix} = \lambda + \lambda + 1 - \lambda^2 - 1 - 1 = -1 + 2\lambda - \lambda^2$

Thus this is non-zero when:

$\lambda^2 - 2\lambda + 1 \neq 0$

That is, when:

$(\lambda - 1)^2 \neq 0 \implies \lambda - 1 \neq 0 \implies \lambda \neq 1$