On an Exercise in finding the rank of a matrix.

Question

The exercise states

Given $A = \left( \begin{array}{ccc} a & b & c \\ 3 & 0 & 2 \\ 1 & 1 & 1 \end{array} \right)$ knowing that $\det(A) = 1$ find the rank of $N = \left( \begin{array}{ccc} a-1 & b-1 & c-1 & 0 \\ 4 & 1 & 3 & a \\ 1 & 1 & 1 & 1\end{array} \right) $

The fact that $\det(A) = 1$ tells us that $ +3c-2a-b = 1$ but I kind seem to figure out how to use this.

Answer 1

$\require{cancel}$Look at the subdeterminant: $$ \begin{align*}\begin{vmatrix} a-1 & b-1 & c-1 \\ 4 & 1 & 3 \\ 1 & 1 & 1\end{vmatrix} &= \begin{vmatrix} a & b & c \\ 3+1 & 0+1 & 2+1 \\ 1 & 1 & 1\end{vmatrix} - \cancelto{0}{\begin{vmatrix} 1 & 1 & 1 \\ 4 & 1 & 3 \\ 1 & 1 & 1\end{vmatrix}} \\ &= \begin{vmatrix} a & b & c \\ 3 & 0 & 2 \\ 1 & 1 & 1\end{vmatrix} + \cancelto{0}{\begin{vmatrix} a & b & c \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{vmatrix}} \\ &= 1 \neq 0.\end{align*}$$Since we have a $3\times 3$ subdeterminant which is non-zero in a $3\times 4$ matrix, follows that ${\rm rank}(N) = 3$.

Answer 2

Since $A$ is a $3\times 3$ matrix and $\det A\neq 0$ it follows that rank$(A)=3$, now, by performing elementary operations on the rows of $N$ we get \begin{align} \begin{pmatrix} a-1 & b-1 & c-1 & 0 \\ 4 & 1 & 3 & a \\ 1 & 1 & 1 & 1 \end{pmatrix}\sim \begin{pmatrix} a & b & c & 1 \\ 3 & 0 & 2 & a-1 \\ 1 & 1 & 1 & 1 \end{pmatrix} \end{align} Where we have added row $3$ to the row $1$ and subtracted row $3$ from the row $2$ of $N$.

It follows rank$(N)=$ rank$(A)=3$.