Making a matrix have a certain rank

Question

Find the values of $a$ and $b$ such that $$\mbox{rank} \begin{pmatrix} 3 & 2 & 5 \\ 1 & a & -1\\ 1 & 3 & b\\ \end{pmatrix} = 2$$

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I can only seem to get single values for $a$ and $b$. To get a matrix of rank 2 we need a matrix in row echelon form with one row of $0$'s at the bottom of the matrix. I.e. two linearly independent rows.

I am able to reduce the above matrix to:

$$\begin{pmatrix} 1 & 3 & b\\ 0 & -7 & 5-b\\ 0 & a-3 & -1-b\\ \end{pmatrix}$$

We get a row of $0$'s at the bottom when $b=-1$ and $a = 3$. Since the questions asks for an equation for a and b I feel like my solution is incomplete. Any advice?

Answer 1

Here is why your argument is incomplete. One way for your reduced matrix to have rank 2 is to have row three all zeros, as you said. HOWEVER, there is another possibility: row 2 and 3 being multiple of each other. (With row 1 this is not a question b/c row 1 cannot be multiple of either row 2 or row 3 since we have 1 versus 0s on first column.

So let us look at this second possibility: we will have rank 2 if $$ \frac{-7}{a-3} = \frac{5-b}{-1-b} \, . $$ Which hopefully is the same equation as the one in the answer below reached by computing determinant.

Answer 2

Hint: The determinant of the matrix is $$ \det(A)=3ab - 5a - 2b + 22. $$ So one necessary condition for rank two is that $\det(A)=0$. This is an equation in $a$ and $b$ as well. Indeed, for $(a,b)=(3,-1)$ the determinant is zero. Can it happen that the matrix has rank zero or one for any $(a,b)$?

Answer: $A$ has rank two if and only if $\det(A)=0$, i.e., $b= \frac{5a-22}{3a-2}$. For $3a-2=0$ the rank is three.