Find m so that $(m+1,1,1)$ , $(1,-m,-1)$ , $(m,1-m,2)$ are linearly dependent I formed an augmented matrix    $$\left(\begin{array}{ccc|c}m+1&1&m&0\\1&-m&1-m&0\\1&-1&2&0\end{array}\right)$$
I now that we do reduced row echelon form for the augmented matrix. But do we do the same in this case ? I failed to get the reduced row echelon of this. How can we get the value of m ?
 A: Hint: Via the invertible matrix theorem
Given a set of $n$ vectors from $\Bbb R^n$, forming a square matrix $A$ using those vectors as the columns we have:
The columns of $A$ are linearly independent $\iff$ $\det(A)\neq 0$
via contrapositive, we have the columns of $A$ are linearly dependent $\iff \det(A)=0$
Calculating the determinant of the matrix we have:

 $-2m^2-4m$.  By setting that equal to zero it implies...

A: By row-reducing the matrix, we get
$$ \begin{pmatrix} m + 1 & 1 & m \\ 1 & -m & 1 - m \\ 1 & -1 & 2 \end{pmatrix} \xrightarrow{R_1 \leftrightarrow R_3} 
\begin{pmatrix} 1 & -1 & 2 \\ 1 & -m & 1 - m \\ m + 1 & 1 & m \end{pmatrix} \xrightarrow{R_2 = R_2 - R_1} \\
\begin{pmatrix} 1 & -1 & 2 \\ 0 & -m + 1 & -1 - m \\ m + 1 & 1 & m \end{pmatrix} \xrightarrow{R_3 = R_3 - (m+1)R_1} 
\begin{pmatrix} 1 & -1 & 2 \\ 0 & -m + 1 & -1 - m \\ 0 & m + 2 & -2 - m \end{pmatrix} \xrightarrow{R_2 = R_2 + R_3} \\
\begin{pmatrix} 1 & -1 & 2 \\ 0 & 3 & -3 - 2m \\ 0 & m + 2 & -2 - m \end{pmatrix} \xrightarrow{R_3 = R_3 - \frac{m+2}{3}R_2}
\begin{pmatrix} 1 & -1 & 2 \\ 0 & 3 & -3 - 2m \\ 0 & 0 & \frac{2}{3}m(m+2) \end{pmatrix}. $$
Hence, the rows/columns will be linearly independent if and only if $m \neq 0, -2$.
