Rank of matrix with variable

Question

$$ \left( \begin{array} {ccc}1 & 1 & x \\ 1 & x & 1 \\ x & 1 & 1 \\ \end{array} \right) $$

Find the values of $x$ such that the matrix above has rank $1$, $2$ and $3$.

Can I get the matrix to reduced row echelon form, or is there another approach?

Answer 1

Hint. Compute the determinant by using Sarrus' rule. You will find a third degree polynomial which has $1$ as a root (for $x=1$ the matrix has rank $1$). If the determinant is not zero then the rank is $3$. Then consider the case when the determinant is zero.

P.S. The determinant is $3x-x^3-2=-(x-1)^2(x+2)$. Therefore for $x\not in \{1,-2\}$ the rank is $3$. If $x=1$ then the rank is $1$. For $x=-2$ the matrix is $$\left(\begin{array}{ccc}1 & 1 & 2 \\ 1 & 2 & 1 \\ 2 & 1 & 1 \\ \end{array} \right)$$ which has rank $2$ because the minor $\left(\begin{array}{cc}1 & 1\\ 1 & 2 \end{array} \right)$ has anon-zero determinat.

Answer 2

There are many methods, one is to compute its characteristic polynomial.

$$\chi(t) = -t^3 + (x+2) t^2 + (x-1)^2 t - (x-1)^2 (2 + x)$$ Then we see we have rank $3$ when $x \neq 1, -2$, rank $2$ when $x = -2$ and rank $1$ when $x = 1$.

Answer 3

Option one: row echelon form with contingencies: avoid doing anything which requires a decision, for instance dividing by a possibly zero scalar. You might get

\begin{equation} \begin{pmatrix} 1&1&x\\ 0&x-1&1-x\\ 0&0&-(x+2)(x-1)\\ \end{pmatrix} \end{equation}

From this, one may see that if $x\neq 1,-2$ then the matrix is full-rank. If $x=-2$, then it has rank $2$, and if if $x=1$, then it is rank $1$. For these latter deductions, just substitute and complete the row reduction.

Option two: nullity as Sylvester knew it. If the $(d-1)$th minors of the matrix all vanish, and at least one $d$th minor does not, then the nullity of the matrix is $d$, so then the matrix has rank $3-d$.

The $0$th minor is the determinant itself, $-(x-1)^2(x+2)$, which vanishes when $x=1,-2$, so the rank is $3$ otherwise (when $x\neq 1,-2$).

The $1$st minors are all either $\pm(1-x)$ or $1-x^2$, which all vanish exactly when $x=1$, so the rank is $2$ otherwise (when $x=-2$).

The $2$nd minors are the entries of the matrix, which never simultaneously vanish, so the rank is $1$ when $x=1$.