Find the value of K for which the given matrix has rank 2 
I think I've got the right intuition here. Row reduce and then by the knowlege that the matrix has a rank of 2, I know the third row will be one of zeros. Therefore, for the third entry in that third row, I will likely have some constant added or subtracted from K, which can be set equal to zero. 
I've row reduced down but I'm essentially at a point where I can't get a rational number for K. Hopefully this is purely a computational error but please let me know if I'm going wrong elsewhere. 

 A: In my opinion, the easy way to do this is to use the fact that if a $n×n$ matrix has a rank less than $n$, then the determinant is $0$. Thus, if a $3×3$ matrix has a rank of $2$, then the determinant is $0$. Therefore, we can just find the determinant in terms of $k$, set it to $0$ and solve.
$$169-13k=0 \implies k=13$$

However, your way is perfectly valid, but you had a computational error in your last step, so it should really be:
$$\begin{bmatrix}-5 & -6 & 4 \\ 0 & 13 & 13 \\ 0 & 41 & -24+5k\end{bmatrix}$$
Now, in order to eliminate the $41$ in the second component of the third row, multiply the second row by $41$, the third row by $-13$, and put in the third row:
$$\begin{bmatrix}-5 & -6 & 4 \\ 0 & 13 & 13 \\ 0 & 0 & 845-65k\end{bmatrix}$$
In order for this matrix to have a rank of $2$, the last row must be all zeroes, so:
$$845-65k=0 \implies k=13$$
A: Your approach is good, but you didn't finish doing row reduction. we can still go:
$\begin{pmatrix}
1 & \frac{6}{5} & -\frac{4}{5}\\
0 &1 & 1\\
0 & 0 & -65+5k\\\end{pmatrix}\\$
and then
$\begin{pmatrix}
1 & 0 & \frac{2}{5}\\
0 &1 & 1\\
0 & 0 & -65+5k\\\end{pmatrix}$.
Clearly the first two columns are linearly independent, so the rank is $2$ if and only if the last column is generated by the first two, which clearly happens if and only if $-65+k=0$
A: Hint

$$\begin{vmatrix} A \end{vmatrix}=0$$

A: The above given problem can be solved using 2 methods :
1.By echelon form
2.By using determinant
process of solving
process of solving
A: Since A is a 3 by 3 matrix of rank 2, all 3 by 3 minors, if any, must be of zero. Here the only 3 by 3 minor is det(A), so that det(A)= 0 will give -13k+169 = 0. Hence k = 13.
