Determining the values of $\lambda$ for which the matrix is invertible

Question

I'm working on a homework problem and am a little stuck. The question is: Determine the values of $\lambda$ for which the matrix $$\begin{pmatrix} \lambda &-1&0\\ -1&\lambda&-1\\ 0&-1&\lambda\\ \end{pmatrix}$$ is invertible.

I looked around and one site suggested finding the determinant, which I did: $|A| = \lambda^3$

What I don't understand is what I'm supposed to do next, now that I have the determinant.

Answer 1

The key principle here is that a matrix is invertible if an only if its determinant is non-zero.

We apply this principle here as follows:

Set

$A(\lambda) = \begin{bmatrix} \lambda &-1 & 0 \\ -1 & \lambda & -1 \\ 0 & -1 & \lambda\end{bmatrix}; \tag{1}$

then we have

$\det A(\lambda) = \lambda^3 - 2\lambda; \tag{2}$

thus $\det A(\lambda) = 0$ precisely when

$0 = \lambda^3 - 2 \lambda = \lambda(\lambda^2 - 2); \tag{3}$

the roots of this equation are easily seen to be

$\lambda = 0, \pm \sqrt 2; \tag{4}$

$A(\lambda)$ is thus invertible for all other values of $\lambda$.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!

Answer 2

Hint: $A$ is invertible iff it has a non-zero determinant.

Answer 3

Suppose the determinant of this matrix is zero, then compute $\lambda$ that solves this equation. Then all $\lambda$ that don't solve this equation are what you are looking for.