### Video Transcript

Find the values of π₯ and π¦ given
the following: the matrix one, three, negative two, one multiplied by the matrix
two, zero, π₯, π¦ equals the matrix eight, negative nine, negative two, negative
three.

To solve this for π₯ and π¦, we can
consider how we obtain some of the components and the resultant matrix. Letβs start with the top left
component, which is eight. We know that the eight mustβve been
obtained by the multiplication of the top row of the first matrix with the left-hand
column of the second matrix. That is one times two add three
times π₯ equals eight or two add three π₯ equals eight. We then obtain that three π₯ equals
six by subtracting two from both sides. And we find that π₯ must be equal
to two.

Now if we think about how the
negative nine was obtained, as this is the top right element, this is the
multiplication of the top row of the first matrix with the right-hand column of the
second matrix. That is, one times zero add three
times π¦ must equal negative nine. This simplifies to three π¦ equals
negative nine. Therefore, π¦ must be equal to
negative three. Weβre then able to verify our
answer by checking that these values of π₯ and π¦ work for the bottom two values in
the resultant matrix.

To get the bottom left value of the
resultant matrix negative two, we must do the bottom row of the first matrix
multiplied by the left-hand column of the second matrix. That is negative two times two add
one times π₯, which is two, equals negative two. This gives us negative four add two
equals negative two, which is true. So our value of π₯ is correct. And we can check the bottom
right-hand component of the resultant matrix by multiplying the bottom row of the
first matrix with the right-hand column of the second matrix. That is negative two times zero add
one times negative three equals negative three. That is zero add negative three
equals negative three, which is true. So we know our value for π¦ is also
correct. Itβs always good when possible to
confirm the values that you found for π₯ and π¦.