Find unknown coefficient of a matrix with known rank. Let there be a matrix A = $\begin{bmatrix} 8 & 7 & 5 \\ 4 & 5 & 6 \\ 7 & 8 & λ \end{bmatrix}$.
Find the λ value, for which matrix A has a rank(A) = 2.
First I need to do the reduced row echelon form (I guess). Should I start by dividing the first row by 8 or by substracting the third row from the first? Will the result be the same? How should I proceed with the information rank(A) = 2 given?
 A: Hint:  You need row $3$ to be a linear combination of rows $1$ and $2$.  Write $a(8,7,5)+b(4,5,6)=(7,8,\lambda) $.
Solve the system  $\begin{cases}{8a+4b=7 \\7a+5b=8}\end{cases}$ for $a$ and $b$.
Then $\lambda =5a+6b$.
If we let $M=\begin{pmatrix}8&4\\7&5\end{pmatrix}$, then $M^{-1}=\frac1{12}\begin{pmatrix}5&-4\\-7&8\end{pmatrix}$.
Thus 

$\begin{pmatrix} a\\b\end{pmatrix}=M^{-1}\begin{pmatrix}7\\8\end{pmatrix}=\begin{pmatrix}\frac14\\\frac54\end{pmatrix}\,\therefore \lambda =\frac{35}4$

A: You want to process by Gauss-Jordan elimination.
So you want to find the set of elementary operations to find the rank.
I recommend starting with the following operations:


*

*Line 2 becomes line 2 - half of line 1

*Line 3 becomes line 3 - (7/8) of line 1


See more here: https://en.wikipedia.org/wiki/Gaussian_elimination
A: The two first rows are linearly independent, therefore the rank is 2 or 3. To have the rank 2, one can solve $\det A = 0.$ It is a linear equation in $\lambda$ with unique solution.
