Find $x$ for which the rank is as minimal/maximal as possible

Question

Find an $x$ in $\Bbb R$ for which rank of the matrix $$A=\begin{bmatrix}1 & 1&1&1 \\1 & -1&-1&1\\1 &-3 &-3 &x \end{bmatrix}$$ is as minimal/maximal as possible.

I was thinking of row reducing the matrix and then count the pivot.

Answer 1

The rank of the first $3$ columns is $2$ since the second column is equal to the third but the first is not a multiple of them. Thus the minimal possible rank is $2$, which is obtained by choosing $x=2$, in which case the 4th column equals the first and therefore does not increase the rank of the matrix.

On the other hand the maximal possible rank is $3$ in the case that $x$ can be choosen so that the 4th column is linearly independent of the first two columns. (The third column can be ignored since it is equal to the second and does not add to the rank of the matrix). In order to find such an $x$ you can calculate the determinant of $$A'=\begin{bmatrix}1 & 1&1 \\1 & -1&1\\1 &-3 &x \end{bmatrix}$$ and set it different from $0$. If there exists such an $x$ then the maximal rank of A is $3$ otherwise it is $2$.

---

Indeed $$det(A')=(-x+3)-(x-1)+(-3+1)=-x+3-x+1-3+1=-2x+2$$ where $$-2x+2\neq0 \implies x\neq 1$$ Hence for every $x\neq1$ the rank of the matrix is $3$.

Answer 2

The most general method to compute the rank is to apply the Gauss elimination algorithm: $$A=\begin{bmatrix}1 & 1&1&1 \\1 & -1&-1&1\\1 &-3 &-3 &x \end{bmatrix} \\\to \begin{bmatrix}1 & 1&1&1 \\0 & -2&-2&0\\0 &-4 &-4 &x-1 \end{bmatrix} \\\to \begin{bmatrix}1 & 1&1&1 \\0 & -2&-2&0\\0 &0 &0 &x-1 \end{bmatrix} $$ The rank is $3$ unless $x=1$ in which case it is $2$.