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Find the rank of the following matrix depending on $\lambda\in\Bbb R$. $$A=\begin{pmatrix} 1&2&3&4\\ 2&\lambda&6&7\\ 3&6&8&9\\ 4&7&9&10 \end{pmatrix}$$

My attempt:

$$\begin{pmatrix} 1&2&3&4\\ 2&\lambda&6&7\\ 3&6&8&9\\ 4&7&9&10 \end{pmatrix}\sim\begin{pmatrix} 1&2&3&4\\ 0&\lambda-4&0&-1\\ 0&0&-1&-3\\ 0&-1&-3&-6 \end{pmatrix}\sim\begin{pmatrix} 1&0&-3&-8\\ 0&\lambda-4&0&-1\\ 0&0&-1&-3\\ 0&-1&-3&-6\\ \end{pmatrix}$$ For $\lambda=4$ we have: $$\begin{pmatrix} 1 &0&-3&-8\\ 0&0&0&-1\\ 0&0&-1&-3\\ 0&-1&-3&-6 \end{pmatrix}\sim\begin{pmatrix} 1&0&0&1\\ 0&0&0&-1\\ 0&0&-1&-3\\ 0&-1&0&3\end{pmatrix}\sim\begin{pmatrix} 1&0&0&0\\ 0&0&0&-1\\ 0&0&-1&0\\ 0&-1&0&0\\ \end{pmatrix}$$

$\Rightarrow r(A)=4$

For $\lambda\neq 4$ we have:

$$\begin{pmatrix} 1&0&0&1\\ 0&\lambda-4&0&-1\\ 0&0&-1&-3\\ 0&-1&0&3\\ \end{pmatrix}\sim\begin{pmatrix} 1&0&0&1\\ 0&0&0&3\lambda-13\\ 0&0&-1&-3\\ 0&-1&0&3 \end{pmatrix}$$

For$\lambda=\frac{13}{3}\Rightarrow r(A)=3$ and for $\lambda\neq \frac{13}{3} \Rightarrow r(A)=4$

Is this correct? Thanks!

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    $\begingroup$ It looks fine to me...! Yet, notice that after the very first step above, you could directly evaluate the matrix's determinant by the first column, and get that the determinant equals $\;-3\lambda+13\;$ , and from here to get when the rank isn't full. A little more work would give you rank equal to three in this case, and four in any other case. $\endgroup$ – DonAntonio Jan 31 '17 at 13:14
  • $\begingroup$ Other than possible arithmetic errors, your strategy looks sound, and it is reasonable for a parameterized matrix of the form you show to have full rank for all $\lambda$s except one and rank $n-1$ at that one value. (It wouldn't be possible for there to be exactly two $\lambda$s there the matrix is singular, for example, or for both $4$ and $2$ to be possible ranks, when there is only a single $\lambda$). So I see no direct signs that you're wrong. $\endgroup$ – Henning Makholm Jan 31 '17 at 13:17
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It looks fine to me!! Or, we have $A=\begin{pmatrix} 1&2&3&4\\ 2&\lambda&6&7\\ 3&6&8&9\\ 4&7&9&10 \end{pmatrix}$ with $\lambda \in \Bbb{R}$ and we know: $$r(A) \leq 4 \text{ and } \big[r(A)= 4 \leftrightarrow \det(A)\neq 0\big]$$

We have $\displaystyle\det(A)= 13 - 3 \lambda$ therefore: $$ r(A) =4\leftrightarrow \bigg(\det(A) \neq 0 \leftrightarrow 13 - 3 \lambda \neq 0 \leftrightarrow 13 \neq 3 \lambda \leftrightarrow \lambda \neq \frac{13}{3}\bigg)$$ $$4 \neq r(A) < 4\leftrightarrow\bigg(\det(A)=0 \leftrightarrow 13 - 3 \lambda = 0 \leftrightarrow 13 = 3 \leftrightarrow \lambda=\frac{13}{3} \bigg) $$

But if $\det(A)=0$ and exists a $T \in M(A)_{3}$ with $\det(T) \neq 0$ and such that for all $R\in M(A)_{3+1=4}$ which contain $T$ we have $\det(R)=0$ then $r(A)=3$ $(M(A)_{n}:=\{X| X \text{ is minor of order }n \text{ for }A\})$

And, with $\lambda=\frac{13}{3}$, we have a $T \in M(A)_{3}$ with $\det(T) \neq 0$: $$\det(T)=\det\begin{pmatrix} 2&\frac{13}{3}&6\\ 3&6&8\\ 4&7&9 \end{pmatrix}=\frac{-1}{3}\neq 0$$ and $M(A)_{3+1=4}=\{A\}$ and $\det(A)=0$ because $\lambda=\frac{13}{3}$ then $r(A)=3$

Summary:

$r(A)=4 \leftrightarrow \Bbb{R}\ni\lambda\neq \frac{13}{3}$

$r(A)=3 \leftrightarrow \Bbb{R}\ni\lambda =\frac{13}{3}$

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I will point out that you can also check the result in Wolfram Alpha.

Here is my computation - rather similar to yours.

$$\begin{pmatrix} 1 & 2 & 3 & 4\\ 2 &\lambda& 6 & 7\\ 3 & 6 & 8 & 9\\ 4 & 7 & 9 &10 \end{pmatrix}\overset{(1)}\sim \begin{pmatrix} 1 & 2 & 3 & 4\\ 2 &\lambda& 6 & 7\\ 3 & 6 & 8 & 9\\ 1 & 1 & 1 & 1 \end{pmatrix}\sim \begin{pmatrix} 0 & 1 & 2 & 3\\ 0 &\lambda-2& 4 & 5\\ 0 & 3 & 5 & 6\\ 1 & 1 & 1 & 1 \end{pmatrix}\overset{(2)}\sim \begin{pmatrix} 0 & 1 & 2 & 3\\ 0 &\lambda-2& 4 & 5\\ 0 & 1 & 1 & 0\\ 1 & 1 & 1 & 1 \end{pmatrix}\sim \begin{pmatrix} 0 & 1 & 2 & 3\\ 0 &\lambda-2& 4 & 5\\ 0 & 1 & 1 & 0\\ 1 & 1 & 1 & 1 \end{pmatrix}\sim \begin{pmatrix} 0 & 0 & 1 & 3\\ 0 & 0 & 6-\lambda & 5\\ 0 & 1 & 1 & 0\\ 1 & 1 & 1 & 1 \end{pmatrix}\sim \begin{pmatrix} 0 & 0 & 1 & 3\\ 0 & 0 & 0 & 3\lambda-13\\ 0 & 1 & 1 & 0\\ 1 & 1 & 1 & 1 \end{pmatrix}\sim \begin{pmatrix} 1 & 1 & 1 & 1\\ 0 & 1 & 1 & 0\\ 0 & 0 & 1 & 3\\ 0 & 0 & 0 & 3\lambda-13 \end{pmatrix}$$

$(1)$: Subtracted third row from the last one.
$(2)$: Subtracted twice the first row from the last one.
In both cases I did so because the resulting row seemed to be simple (only zeroes and ones), which made further row operations a bit easier.

From the final matrix we see that rank is three if $3\lambda-13=0$ and in all other cases rank in equal to four.

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