Find all values of h such that rank(A) = 2

Question

Find all values of $h$ such that rank($A$) = $2$.

$A$ = $\begin{bmatrix} 1 & h & -1\\ 3 & -1 & 0\\ -4 & 1 & 3 \end{bmatrix} $

I used row transformations to get

$A$ = $\begin{bmatrix} 1 & h & -1\\ 0 & -1-3h & 3\\ 0 & 1+4h & -1 \end{bmatrix} $

But how do I solve to get the rank? I know the general idea is that rank($A$) = $2$ when dim(col($A$)) = dim(row($A$)) = $2$

Answer 1

You know that the first row is independent, but the second and third rows must be dependent. In other words, you can write $$-1-3h=C(1+4h)\\3=C(-1)$$ Therefore $C=-3$. Plug it into the first equation, to find $h$

Answer 2

Well, you know that the column space has at least two linearly independent vectors. So, you just want the middle vector in your transformed matrix to be in the span of the other 2. Well, you can always add the left vector in the column space ($\hat{i}$, you might call it) so that the top component of the middle vector will correspond to some linear combination of the other two columns. So you need the ratio of the second two elements in the middle column to correspond to the ratio of the second two elements in the right column. That is, you want $$\frac{-1-3h}{1+4h} = \frac{3}{-1} $$ This gives $$ 3h+1 = 12h+3$$ thus $h= \frac{-2}{9}$.

Notice, also, that this equation only holds for $h=-\frac{2}{9}$. If you put in another value of $h$, this equation won't hold and so the projection of the middle vector to the $yz$ plane, that is, the second two components, won't be some scalar multiple of any two components of the vector in the right column. So if I take a linear combination of the vector in the left column and the right column, and look at the $y$ and $z$ components (the projection to the $yz$ plane), the left vector scaled will have no contribution, since it doesn't have a $y$ or $z$ term, and the $yz$ component won't be a scalar multiple of the same component for the vector on the right, because the $h$ is wrong the ratio won't work out for that to happen. So This is the only such $h$.

So then you get the middle column will look like $\begin{bmatrix} \frac{-2}{9} & \frac{-1}{3} & {\frac{1}{9}}\end{bmatrix}^\top$. You can see that this vector is simply $-\frac{1}{9}$ the vector in the right column plus $\frac{1}{9}$ times the vector on the left. Which works out as we hope.

So it's $$\boxed{h = - \frac{2}{9}}$$

Answer 3

Notice the plane $9x+y+3z=0$ is precisely the span of the first and third column of your matrix. All you need to do is find $h$ so that $(h,-1,1)$ resides on this plane; this will guarantee that you matrix has rank two. So $9h-1+3=0$ implies $h=-2/9$ as desired.

Answer 4

It is easy to see that the first and the second row of $A$ are independent, hence

$$rank(A) \ge 2.$$

Furthermore, since $\det(A)=-2-9h,$

$$rank(A) =2 \iff \det(A)=0 \iff -2-9h=0.$$

Answer 5

There are several ways to approach this problem. Here are two approaches that I will not use in this answer. (1) Rank is preserved by both row operations and column operations, and so we can do column operations to simpliffy the matrix further. (2) The matrix has rank 3 if and only if it is invertible, and we can use the determinant to test for invertibility. However, because the first and third columns (or second and third rows) are linearly independent, we must have rank at least 2. Therefore, we only need to find h where $\det(A)=0$.

Let us observe that the first and third columns are linearly independent, and so we will have rank $2$ if and only if the second column is a linear combination of the first and third. Call the columns $v_1, v_2, v_3$. Because the second coordinate of $v_3=0$, if $v_2=av_1+bv_3$, then looking at the second coordinate, that tells us $a=-1/3$. Looking at the third coordinate, we get $(-4)/(-3)+3b=1$, so $b=-1/9$. Looking at the first coordinate, this gives $h=1/(-3)+(-1)/(-9)=-2/9$. This is the only way that $v_2$ can be a linear combination of $v_1$ and $v_3$.