Matrices. Please help :) For what value of x will the following matrix A be of rank 1) equal to 3 2) less than 3
where is the matrix?
\[A=\left[\begin{matrix}3-x & 2 & 2 \\ 1 &4-x & 0 \\ -2 & -4 & 1-x\end{matrix}\right]\]
really, you need to find x so that the determinant is non-zero which will make the matrix have full rank (3). and then find x so that the determinant is 0 then the matrix will not have full rank which means the rank will be less than 3
how do i do that
do you know how to calculate the determinant?
should i equate the determinant to zero?
once you get the expression for the determinant, you should have a 3rd degree polynomial. if you find the roots of that polynomial, then when x is the value of anoy of the roots, the determinant will be 0 and the second part of your question will be answered. if you pick any value that is not a root, then the determinant will be non-zero and you will have answered the first part of the question.
oh okay-tysm
do you understand? the dots i'm trying to help you connect?
yep
the first part is actually the easier of the two because you can use trial and error and there are an infinite number of values that will work. but for part 2, a maximum of 3 values work, and only one value if the polynomial has a pair of complex roots!
if you are doing by hand, you can use the rational roots theorem to find the roots.
did you find the answers?
nope
did you get the polynomial?
you there?
did you want help finding the answers?
the rank is 3 when the null space is trivial, i.e. only zero; this means that the entire codomain is mapped to. so we want \(\det A\ne 0\).
if the null space is not trivial, then we will invariably have incomplete rank and so for rank less than 3 we want to solve \(\det A=0\)
Join our real-time social learning platform and learn together with your friends!