### Video Transcript

Find the set of real values of π₯
that make the two-by-two matrix π₯ minus three, eight, two, π₯ plus three
singular.

Letβs begin by defining the word
singular when it comes to matrices. We say that a matrix is singular if
itβs not invertible; it doesnβt have an inverse. We know that a matrix is invertible
if its determinant is not equal to zero, and the converse is also true. So, in other words, a matrix is
singular if its determinant is equal to zero. In this case then, we need to find
the set of real values of π₯ such that the determinant of our matrix is equal to
zero. The determinant of a two-by-two
matrix π, π, π, π is ππ minus ππ. We subtract the product of the
elements in the top right and bottom left from the product of those in the top left
and bottom right.

So, in this case, thatβs π₯ minus
three times π₯ plus three minus eight times two. If we distribute these parentheses,
we get π₯ times π₯, which is π₯ squared, plus three π₯ minus three π₯ minus three
times three, which is nine. That simplifies to π₯ squared minus
nine. And eight multiplied by two is
16. So the determinant of our matrix is
π₯ squared minus nine minus 16, which is π₯ squared minus 25. Weβre trying to find the set of
values of π₯ that make our matrix singular. In other words, which values of π₯
make the determinant zero? So, letβs set our expression for
the determinant equal to zero and solve for π₯. That is, π₯ squared minus 25 equals
zero.

Adding 25 to both sides of this
equation gives us π₯ squared equals 25. And then weβll take the square root
of both sides of our equations, remembering to take the positive and negative square
root of 25. That gives us π₯ is equal to
positive or negative five. We can use these squiggly brackets
to help us represent the set of values that make our matrix singular. They are negative five and
five. Note that at this stage, we could
check our solutions by substituting each value of π₯ into our original matrix and
then checking that the determinant is indeed equal to zero.