welcome in this video we're going to solve
a two-part problem now the first part of
this problem is very much in the algebra
side of the course we're going to find
the eigenvalues and eigenvectors of this
matrix A and for the second part well
we're going to steal the problem from
the calculus side of the course and
solve it using our algebraic techniques
so firstly let's get this first part done
so the eigenvalues are the
solution to the characteristic equation
and the characteristic equation is the
determinant of A minus lambda i is equal
to zero so if we calculate that that is
determinant of 2 minus lambda 3 1 4
minus lambda if you do this calculation
you will end up with lambda squared
minus 6 lambda plus 5 and well the
solution to this are not too hard to see
because we can just factor it as lambda
minus 1 lambda minus 5 and so looking at
this of course the characteristic
equation is all of this equal to 0 and
the solution to this equation are quite
obvious they're given by lambda 1 is equal to 1
and lambda 2 is equal to 5 so we found
our eigen values now let's find the
corresponding eigen vectors so let's try
and find the eigenvector for our first
eigenvalue lambda 1 equals 1 so for
lambda 1 equals 1 we need to solve is
the equation A minus I times our vector
is equal to 0 and what is this as an
Augmented matrix well this is just the
system of equations 1 3 1 3 augmented is
0 0 and we can easily row reduce that let's
just take Row two and make it Row 2
minus Row 1 so that is the matrix 1 3 0 0 0 0
now we can
back substitute this so let's set the
second component
I'll say v2 equal to my parameter t and
when we solve this we get v1 is equal
to minus 3 times t so let's put
this into a vector so our eigen vector is
going to be t times the vector minus 3 1
although this isn't the complete story
because eigen vectors are never 0 vectors 
so I should restrict myself to the case t not equal to zero
now let's find our second eigenvector so
for lambda 2 equal to 5
we have to solve the equation A minus 5i
times our vector is 0 so again let's
take this and make an Augmented matrix
out of it so this will be the matrix
-3 3 1 -1 augmented with 0 0
again this is quite easy to solve let's
just do one row reduction so row 2 is
equal to row 2 plus 1/3 of row 1 and if we do
that we get the matrix -3 3 0 0
augmented with 0 0 and back substituting
let's set
second component of our vector equal to
a parameter so in this case I've been
making the parameter s and if we solved
for the first component we get v1 is
equal well it's also equal to s so
that gives us our second eigenvector
which I'll call u is s times the
vector the vector 1 1 and again vectors
eigen vectors are non zero vectors so
let's restrict ourselves to the case
where s is non zero so for the second
part of this question we're gonna solve
a calculus problem and we're not coming
to this calculus problem unarmed we're
actually given that this matrix A has
eigenvalues 5 and 1 and with respective
eigenvectors given over here but how
does this help us how do we relate this
matrix to the these equations well if
you take a look at these equations they
can be written in the following form so
let's take the derivative of the vector
containing x and y well you can actually
just write that as the matrix 2 3 1 4
times the vector containing x and y that
is it is just the matrix a times x y and
now problems of this form are quite easy
to solve since we know the eigenvectors
and eigenvalues of this matrix they're
just given by x y is equal to a constant
i'm going to call it alpha times the
vector first eigenvector 1 1 and the
exponential of the first eigenvalue
times t and now the second part which is
the second eigenvector times the exponential e
to the t so the second eigenvalue times
t and this is the general solution for
arbitrary constants alpha
and beta and thank you for watching
