The rank of a matrix, dependent on the value of $t$

Question

I'm trying to analyse the rank of the following matrix, for $t\in \mathbb{R}$.

$$\begin{bmatrix} t+3 & 5 & 6 \\ -1 & t-3 & -6 \\ 1 & 1 & t+4 \end{bmatrix}$$

With $R_1\leftrightarrow R_3$, $-(t+3)R_1+R_3 \rightarrow R_3$, $R_1+R_2 \rightarrow R_2$, and finally $-R_2+R_3\rightarrow R_3$. I get

$$\begin{bmatrix} 1 & 1 & t-4 \\ 0 & t-2 & t-2 \\ 0 & 0 & -t^2-8t-4 \end{bmatrix}$$

And this makes that when $t=2$ or $t=2(-2 \pm \sqrt{3})$. However, these values are not what they are supposed to give(-2,2,4). Where did I make a mistake? Any help would be appreciated.

Answer 1

The problem is the last elementary operation. I should have done $R_2+R_3\rightarrow R_3$

Then I would have obtained $$\begin{bmatrix} 1 & 1 & t-4 \\ 0 & t-2 & t-2 \\ 0 & 0 & -t^2-6t-8 \end{bmatrix}$$ Giving the correct values for $t$.

Answer 2

\begin{align} \begin{bmatrix} t+3 & 5 & 6 \\ -1 & t-3 & -6 \\ 1 & 1 & t+4 \end{bmatrix} &\to \begin{bmatrix} 1 & 1 & t+4 \\ -1 & t-3 & -6 \\ t+3 & 5 & 6 \\ \end{bmatrix} &&R_1\leftrightarrow R_3 \\ &\to \begin{bmatrix} 1 & 1 & t+4 \\ 0 & t-2 & t-2 \\ t+3 & 5 & 6 \\ \end{bmatrix} &&R_2\gets R_2+R_1 \\ &\to \begin{bmatrix} 1 & 1 & t+4 \\ 0 & t-2 & t-2 \\ 0 & 2-t & -t^2-7t-6 \\ \end{bmatrix} &&R_3\gets R_3-(t+3)R_1 \\ &\to \begin{bmatrix} 1 & 1 & t+4 \\ 0 & t-2 & t-2 \\ 0 & 0 & -t^2-6t-8 \\ \end{bmatrix} &&R_3\gets R_3+R_1 \end{align} The roots of $-t^2-6t-8$ are $-2$ and $-4$.

• If $t\notin\{2,-2,-4\}$ the rank is $3$.

• If $t=2$, the rank is $2$.

• If $t=-2$ or $t=-4$, the rank is $2$.

The result you're given is wrong. The numbers $-2$, $2$ and $4$ are the eigenvalues of $$ A=\begin{bmatrix} 3 & 5 & 6 \\ -1 & -3 & -6 \\ 1 & 1 & 4 \end{bmatrix} $$ and your case is finding the eigenvalues of $-A$.