### Video Transcript

Use determinants to solve the
system five 𝑥 equals negative two 𝑦 minus five plus three 𝑧, negative three 𝑥
minus 𝑦 plus one equals two 𝑧, and two 𝑦 minus 𝑧 equals negative five 𝑥 plus
three.

So, in a problem like this, the
first thing we want to do is actually rearrange our equations so that we have our
variables on the left-hand side. And then we have our answers on the
right-hand side, which are numerical values or constants. So our first equation rearranged is
gonna be five 𝑥 plus two 𝑦 minus three 𝑧 equals negative five. Then, for the second equation,
we’re gonna have negative three 𝑥 minus 𝑦 minus two 𝑧 equals negative one. And then finally, five 𝑥 plus two
𝑦 minus 𝑧 equals three.

Okay, great. We’ve got it like this, but why do
we want it in this form? We want it in this form so we can
set up a matrix equation. And when we do, what we get is the
matrix five, two, negative three, negative three, negative one, negative two, five,
two, negative one multiplied by the matrix for our variables, which is 𝑥, 𝑦,
𝑧. Then this is equal to our answer
matrix negative five, negative one, three. Okay, great. But how does this help us meet our
objective, which is to solve the system of equations using determinants? Well, what we’re gonna do is use
Cramer’s rule. And what Cramer’s rule tells us is
that we can find the variables or solutions to our system of equations using, for
example, 𝑥 is equal to, then we’ve got the determinant of the matrix Δ sub 𝑥 over
the determinant of the matrix Δ. And then this pattern continues for
𝑦 and 𝑧.

Okay, to use this then, what we
need to do is work out our determinants. The first determinant we’re gonna
work out is the determinant of Δ, which is gonna be our coefficient matrix. So what we’re gonna do is find out
the determinant of the matrix five, two, negative three, negative three, negative
one, negative two, five, two, negative one. So this is gonna be equal to five
multiplied by the determinant of the submatrix negative one, negative two, two,
negative one minus two multiplied by the submatrix negative three, negative two,
five, negative one minus three multiplied by the submatrix negative three, negative
one, five, two, remembering that when we find the determinants, the coefficients go
positive, negative, positive. And to find our submatrices, we
delete the column and row that our coefficient is in.

Okay, great. So now, we calculate this. And then remembering that when we
work out determinants of two-by-two matrices, what we do is cross multiply and then
subtract, we’re gonna get five multiplied by one plus four minus two multiplied by
three plus 10 minus three multiplied by negative six plus five which is equal to
two. So this is great cause it also
tells us that the matrix is nonsingular. So therefore, we know that there is
not gonna be an infinite number of solutions. And that’s because if there was,
then the determinant would be equal to zero. So now, what we’re gonna do is
clear a space and work out the other determinants we need to find.

So next, what we want to do is find
the determinant of Δ sub 𝑥. And the way we do that is by
substituting in the answer matrix values for the coefficient of 𝑥-value, so the
first column in our matrix. So now what we’re gonna want to do
is find the determinant of this matrix. And to do that, what we’re gonna do
is to use the same methods as we used for the previous determinant, which is gonna
give us a determinant value of negative 42. And you can see the working
there. Okay, great. So once again, we’re gonna clear a
bit of space and look at our next determinant.

So now what we’re gonna find is the
determinant of the matrix Δ sub 𝑦. And this is gonna be where we
substitute in our answer matrix for the 𝑦-coefficients in the matrix. So then once again, using the same
method to find the determinant, we’re gonna have a determinant of 112. And again, the working is shown
here. So once again, what we’re gonna do
is clear a space for the final determinant. So for the final one, it’s gonna be
the determinant of the matrix Δ sub 𝑧. So then once again, we go through
the same method to find the determinant of our three-by-three matrix. And what it gives us is a value of
eight.

So now what we have are all the
determinants we need to use Cramer’s rule to find out our variables 𝑥, 𝑦, and
𝑧. So, first of all, we’re gonna start
with 𝑥, which is gonna be equal to negative 42 over two. And we get that because it’s the
determinant of Δ sub 𝑥 over the determinant of Δ. So this is gonna give us a value of
𝑥 equals negative 21. And then for 𝑦, we’re gonna have
112 over two which is gonna give us a 𝑦-value of 56. And then finally, we’re gonna get
𝑧 is equal to eight over two, and this is gonna give us 𝑧 is equal four. So therefore, we can say the
solutions to our systems of equation are 𝑥 equals negative 21, 𝑦 equals 56, and 𝑧
equals four.