Rank of matrix with a real parametric number

linear algebramatricesmatrix-rank

Cheers, I am given a matrix $ Α =
\begin{pmatrix}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 21+λ \\
\end{pmatrix}
$ and I am being asked to find its rank. I calculate the row echelon form of the matrix, and I end up with $\begin{pmatrix} 1 & 2 & 3 \\ 0 & -6 & λ \\ 0 & 0 & -\frac{λ}{2}-6 \\ \end{pmatrix}$. Now for the rank, shouldn't I take into account the two possibilities of the last row? So, if $λ \neq -12$, rank(A) = 3, AND if $λ=-12$ rank(A)=2 ? Is my thinking correct? Thanks

Best Answer

This is correct. Another point of view would be the following :

First, you see that the two first rows are linearly independant (of course they are not proportional), so this implies that the rank is $\geq 2$, i.e. it is $2$ or $3$.

Moreover, the determinant of the matrix is clearly a one-degree polynomial in $\lambda$, hence there is only one value of $\lambda$ that makes the matrix singular, i.e. that makes the rank equal to $2$.

Finally, to see which is the value, just "notice" that for the last coefficient being equal to $9$, the rows are in arithmetic progression (you have $R_3-R_2=R_2-R_1$), hence the matrix is not invertible : so the only value of $\lambda$ that makes the rank equal to $2$ must satisfy $21+\lambda=9$, so it is $\lambda=-12$.

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