### Video Transcript

Use matrices to solve the following
system of equations.

Remember, if we can write our
system of equations in the form π΄π± equals π, where π΄ is the coefficient matrix,
π± is the column vector that contains all of the variables, and π is the constant
vector, then we can multiply both sides of this equation by the inverse of π΄. And when we do, we find that the
variable vector π± is equal to the inverse of π΄ times π.

And so to solve our system of
equations, we need to set it up in this form and then find the inverse of our
coefficient matrix. We have three equations with three
variables. And so vector π΄ is going to be
three by three. Weβre going to list the
coefficients of our variables in order from our first equation. Itβs negative one, eight, and
negative three. These are simply the coefficients
shown.

In our second equation, theyβre
four, negative three, and eight. And in our third equation, thereβre
six, negative 12, and 19. Then the vector π± contains all of
our variables. Itβs π₯, π¦, π§. And of course our constant vector
contains our constants negative 10, 12, and 18. So our first job is to find the
inverse of the matrix π΄.

There are of course a number of
ways to do this. One is to augment it with the
identity matrix and perform elementary row operations. We might even choose to use a
calculator. But letβs recall how we can use the
minors, cofactors, and adjugate method. We begin by finding the matrix of
minors. We ignore the values on the row and
column weβre looking at and calculate the determinant of the remaining values. So for the first element on the
first row, weβre going to do negative three times 19 minus eight times negative 12,
which is 39. We then ignore all elements on the
first row and second column. And we work out four times 19 minus
eight times six, which is 28. Finally, we ignore elements on the
first row and third column. And we work out four times negative
12 minus negative three times six. And we get negative 30.

Continuing the process, and the
matrix of minors is as shown. We then find the matrix of
cofactors using that sort of checkerboard pattern. This involves changing the sign of
the second element in the first row, the first and third elements in our second row,
and the second element in our third row. Next, we find the adjugate. The elements on a diagonal stay the
same, and then we simply swap the positions of all of the rest. We go directly across that
diagonal. So we swap negative 116 and
negative 28, 55 and negative 30, and negative four and 36.

Then our final step is to multiply
this by one over the determinant of π΄. We can use the values of the matrix
of minors to do so. Weβre going to multiply each of the
elements in the top row of our original matrix by the cofactor for the same
location. So we multiply negative one by
39. We then subtract the product of
eight and 28. And then we add the product of
negative three and negative 30. And so the determinant of π΄ is
negative 173. And so we find the inverse of π΄ is
negative one over 173 times that adjugate.

Letβs clear some space, and weβre
now able to solve our system of equations. π± is therefore equal to the
inverse of π΄ times π, which is as shown. Now, we could multiply each of the
elements inside the inverse of π΄ by negative one over 173. Or we can do that right at the
end. Letβs begin by finding the dot
product of the elements in the first row of our inverse and the vector π. When we do, we get negative one
over 173 times 39 times negative 10 minus 116 times 12 plus 55 times 18, which is
negative 792. So π₯ is negative one over 173
times negative 792, which is the same as 792 over 173.

We then do the same for π¦, this
time finding the dot product of the elements on this row with our vector, giving us
negative one over 173 times 196, which is negative 196 over 173. Letβs do this for π§ by finding the
dot product of the elements on the third row of the inverse and the column vector
π. And this time, we get negative one
over 173 times 210 or negative 210 over 173.

And so weβve found the values of
π₯, π¦, and π§. Letβs put them back into vector
form. It is in fact easier to do so by
taking out that constant factor of one over 173. And so weβve solved our system of
equations. In vector form, we can say that π₯,
π¦, π§ is equal to one over 173 times the vector 792, negative 196, negative
210.