How do you analyse the rank of a matrix depending on a parameter

Question

I have the matrix $$ \begin{pmatrix} 3-t & 3 & 2t \\ -2 & 0 & -1 \\ 1 & 3 & 2+t \\ t+2 & 0 &t \end{pmatrix} $$ And I'm asked to evaluate the matrix's rank depending of $t$, using the determinants and minors of the matrix. However, the solution I see in the answer book is too short and only addresses one minor $$ \begin{pmatrix} 3-t & 3 & 2t \\ -2 & 0 & -1 \\ 1 & 3 & 2+t \end{pmatrix} $$ and its determinant equals $0$. But I've tried different minors and they dont give always the same answer. Is it because I'm doing it wrong or I should think of something else?

The answer book is this page 85 ex 30

Answer 1

The given matrix is $4\times 3$, hence there are four $3\times 3$ minors. What you should know from theory is that if all determinants of such minors are zero then the rank of the given matrix is less than 3, otherwise it is exactly $3$. Personally, I would start from the minor obtained by removing the first row whose determinant is very easy to compute: $$\begin{pmatrix} -2 & 0 & -1 \\ 1 & 3 & 2+t \\ t+2 & 0 &t \end{pmatrix} \implies \det \begin{pmatrix} -2 & 0 & -1 \\ 1 & 3 & 2+t \\ t+2 & 0 &t \end{pmatrix}=3(-2t+t+2)=3(-t+2).$$ So if $t\not=2$ then the rank is $3$, otherwise the rank is $\leq 2$. Plugging $t=2$ into the matrix we find $$\begin{pmatrix} 1 & 3 & 4 \\ -2 & 0 & -1 \\ 1 & 3 & 4 \\ 4 & 0 &2 \end{pmatrix}.$$ Are you able to find at least one $2\times 2$ minor (their number is 18) whose determinant is not zero? In that case the rank of the matrix is $2$ otherwise it is $1$.

Answer 2

You can use rank-nullity $$\begin{align}\ker(M) &= \{\vec v|M\vec v = 0\} \\ &= \left\{\vec v|\begin{pmatrix} 3-t & 3 & 2t \\ -2 & 0 & -1 \\ 1 & 3 & 2+t \\ t+2 & 0 &t \end{pmatrix}\vec v = 0\right\} \\ &= \left\{\vec v|\begin{pmatrix} 1 & 3/5 & 3t/5 \\ 0 & 1 & t-5/6 \\ 0 & 0 & 2-t \\ 0 & 0 & 0 \end{pmatrix}\vec v = 0\right\}\\ &= \begin{cases} \lambda\begin{pmatrix}-\frac{1}{2} \\ -\frac{7}{6} \\ 1 \end{pmatrix} & \text{if $t=2$}\\ \begin{pmatrix}0 \\ 0 \\ 0 \end{pmatrix} & \text{otherwise}\\ \end{cases} \end{align}$$

So: $$\operatorname{nullity}(M) = \begin{cases} 1 & \text{if $t=2$}\\ 0 & \text{otherwise} \end{cases}$$

and $$\begin{align}\operatorname{rank}(M) &= \dim(\operatorname{dom}(M))-\operatorname{nullity}(M)\\ &=\begin{cases} 2 & \text{if $t=2$}\\ 3 & \text{otherwise} \end{cases} \end{align}$$

Answer 3

By row operations, $$\begin{pmatrix} 3-t & 3 & 2t \\ -2 & 0 & -1 \\ 1 & 3 & 2+t \\ t+2 & 0 &t \end{pmatrix}\sim$$ $$\begin{pmatrix} 0 & 3t-6 & t^2+t-6 \\ 0 & 6 & 3+2t \\ 1 & 3 & 2+t \\ 0 & -3t-6 &-t^2-3t-4 \end{pmatrix}\sim$$ $$\begin{pmatrix} 0 & 0 & \frac32 t-3 \\ 0 & 6 & 3+2t \\ 1 & 3 & 2+t \\ 0 & 0 &\frac12 t-1 \end{pmatrix}$$ Now it is obvious that when $t=2$, the rank of the matrix is $2$ and otherwise rank is $3$.

Answer 4

The rest is (much) easier, requiring almost no calculation and no algebra.

Stare at this part of the matrix, which consists of the constant entries independent of $t:$

$$\begin{pmatrix} & 3 & \\ -2 & 0 & -1 \\ 1 & 3 & \\ & 0 \end{pmatrix}$$

By inspection,

• The always nonzero middle column shows the rank is at least 1.

• The coefficients on the second row then show the rank cannot be 1 (and therefore is at least 2), because neither the first nor the third columns can be nonzero multiples of the second column.

• The rank cannot exceed 3, the smaller of the numbers of rows and columns.

Equivalently, because you are asked to use determinants and minors, notice that the $(2,3)\times(1,2)$ minor,

$$\pmatrix{-2 & 0 \\ 1 & 3},$$

has nonzero determinant, whence the rank is always at least 2. Possibly the answer presumed you would notice this in due course (it's the only $2\times 2$ minor not explicitly depending on $t$) and that this doesn't involve any algebra to work through.

Consequently, the question comes down to characterizing the values of $t$ that reduce the rank of the matrix from its maximum of 3 to 2. The answer can be construed as finding some values of $t$ that definitely make the rank equal to 3.

The answer therefore is deficient insofar as it's conceivable that when the upper $3\times 3$ minor is of reduced rank, adjoining the bottom row increases the rank. This, I believe, is the motivation for your question, and it's legitimate. But the answer at least found that only $t=2$ can possibly reduce the rank. And indeed, plugging in that value yields a matrix with identical first and third rows, $(1,3,4),$ and whose fourth row $(4,0,2)$ is $-2$ times the second row, $(-2,0,-1):$ obviously of rank 2, QED.