### Video Transcript

Find the value of the determinant
of the matrix given by negative three π₯ plus three, four π₯ squared plus four,
negative three π¦ plus four, negative π¦ squared plus three.

Remember, for a two-by-two matrix
π΄ with elements π, π, π, π, the determinant of this matrix can be found by
subtracting the product of elements π and π from the products of elements π and
π.

In the matrix in our question, π
is negative three π₯ plus three. π is four π₯ squared plus
four. π is negative three π¦ plus
four. And π is negative π¦ squared plus
three. Weβre then going to find the
product of elements π and π. Thatβs negative three π₯ plus three
multiplied by negative π¦ squared plus three. And from this, weβre going to
subtract the product of elements π and π. Thatβs four π₯ squared plus four
multiplied by negative three π¦ plus four.

We can use our preferred method to
expand each of these brackets. Iβm going to use the FOIL
method. Remember βFβ stands for first. We multiply the first term in the
first bracket by the first term in the second. Negative three π¦ multiplied by
negative π¦ squared is three π₯π¦ squared. We then multiply the outer
terms. Negative three π₯ multiplied by
three is negative nine π₯.

βIβ stands for inner. So weβre going to multiply the
inner terms. And three multiplied by negative π¦
squared is negative three π¦ squared. And finally, weβre going to
multiply the last term in each bracket. Three multiplied by three is
nine.

Letβs repeat this process for the
second set of brackets. Multiplying the first term from the
first bracket by the first term in the second bracket gives us negative 12π₯ squared
π¦. Four π₯ squared multiplied by four
is 16π₯ squared. Multiplying the inner two terms
gives us four multiplied by negative three π¦, which is negative12π¦. And finally, four multiplied by
four is 16.

Weβre going to simplify this
expression by collecting like terms where we can, noting that weβre going to be
subtracting everything in this second bracket. Our first three terms are three
π₯π¦ squared minus nine π₯ minus three π¦ squared. Then, nine minus 16 gives us
negative seven. Weβre subtracting negative12π₯
squared π¦. So it gives us plus 12π₯ squared
π¦. Weβre subtracting 16π₯ squared. And then, we subtract negative
12π¦. So thatβs the same as adding
12π¦. Itβs not entirely necessary, but we
can move the constant to the end.

And then we see the determinant of
the matrix in our question is three π₯π¦ squared minus nine π₯ minus three π¦
squared plus 12π₯ squared π¦ minus 16π₯ squared plus 12π¦ minus seven.