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In this section we’re going to come back and take one more
look at solving systems of equations. In
this section we’re actually going to be able to get a general solution to
certain systems of equations. It won’t
work on all systems of equations and as we’ll see if the system is too large it
will probably be quicker to use one of the other methods that we’ve got for
solving systems of equations.
So, let’s jump into the method.
Proof : The proof
to this is actually pretty simple.
First, because we know that A
is invertible then we know that the inverse exists and that 
. We also know that the solution to the system
can be given by,
From the section on cofactors we know how to define the
inverse in terms of the adjoint
of A. Using this gives us,
Recall that is the cofactor of . Also note that the subscripts on the
cofactors above appear to be backwards but they are correctly placed. Recall that we get the adjoint by first
forming a matrix with in the ith
row and jth column and
then taking the transpose to get the adjoint.
Now, multiply out the matrices to get,
The entry in the ith
row of x, which is in the solution, is
Next let’s define,
So, is the matrix we get by replacing the ith column of A with b. Now, if we were to
compute the determinate of by expanding along the ith column the products would be one of the ’s times the appropriate cofactor. Notice however that since the only difference
between and A
is the ith column and so
the cofactors we get by expanding along the ith
column will be exactly the same as the cofactors we would get by expanding A along the ith column.
Therefore, the determinant of is given be,
where is the cofactor of from the matrix A. Note however that this is
exactly the numerator of and so we have,
as we wanted to prove.
Let’s work a quick example to illustrate the method.
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Example 1 Use
Cramer’s Rule to determine the solution to the following system of equations.
Solution
First let’s put the system into matrix form and verify
that the coefficient matrix is invertible.
So, the coefficient matrix is invertible and Cramer’s Rule
can be used on the system. We’ll also
need det(A) in a bit so it’s good
that we now have it. Let’s now write
down the formulas for the solution to this system.
where is the matrix formed by replacing the 1st
column of A with b, is the matrix formed by replacing the 2nd
column of A with b, and is the matrix formed by replacing the 3rd
column of A with b.
We’ll leave it to you to verify the following
determinants.
The solution to the system is then,
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Now, the solution to this system had some somewhat messy
solutions and that would have made the row reduction method prone to
mistake. However, since this solution
required us to compute 4 determinants as you can see if your system gets too
large this would be a very time consuming method to use. For example a system with 5 equations and 5
unknowns would require us to compute 6 determinants.
At that point, regardless of how messy the final answers are there is a
good chance that the row reduction method would be easier.