### Video Transcript

Consider the matrix equation. Find the values of π, π, π, and
π.

We can imagine this to be a bit
like a linear equation. And to solve, we can start by
subtracting the constant matrix from both sides. Thatβs the matrix five, seven,
negative 10, negative eight. Subtracting this matrix from both
sides, and weβre left with three, one, zero, one multiplied by π, π, π, π on the
left-hand side. For the right-hand side weβll need
to subtract five, seven, negative 10, negative eight from the matrix eight, negative
two, two, seven. And we could do this by subtracting
each of its individual elements.

Eight minus five is three. Negative two minus seven is
negative nine. Two minus negative 10 is 12. And seven minus negative eight is
15. And weβre left with three, one,
zero, one multiplied by π, π, π, π being equal to three, negative nine, 12,
15.

Now to solve for the matrix π, π,
π, π, we canβt divide by the matrix three, one, zero, one. Instead, weβll need to multiply
both sides by its multiplicative inverse. And the reason this works is
because when we multiply a matrix by its inverse, we get the identity matrix. So, if we multiply both sides by
the inverse of three, one, zero, one, weβll solve for π, π, π, π. For two by two matrix with elements
π, π, π, π, its inverse is found by one over the determinant of π multiplied by
π negative π negative ππ, where the determinant is the difference between the
product of the top-left and bottom-right elements and the product of the top-right
and bottom-left.

So, letβs begin by finding the
determinant of our matrix three, one, zero, one. π multiplied by π is the top-left
element multiplied by the bottom-right element. Thatβs three multiplied by one. And then we subtract the product of
the top-right and bottom-left element. Thatβs one multiplied by zero. And that gives us a determinant of
three.

Letβs use this information to find
the inverse of this matrix. One over the determinant is
one-third. We then swap the elements on the
top left and the bottom right. We swap one and three. And we multiply the elements on the
top right and the bottom left by negative one, essentially changing their signs. Then we can multiply each of the
elements by one-third and we get the inverse of the matrix to be a third, negative a
third, zero and one. And now we have this, we can
multiply both sides of our equation by the inverse.

On the left-hand side, we end up
with π, π, π, π. And on the right-hand side we end
up with a third, negative a third, zero, one multiplied by three, negative nine, 12,
15. Remember, order matters, so we must
complete this multiplication with the inverse in the front.

To multiply two matrices, we find
the dot product of the associated rows and columns. For π, we find the dot product of
the first row in the first matrix and the first column in the second. Thatβs negative three. For π, we find the dot product of
the first row in the first matrix and the second column in the second. Thatβs negative eight. For π, we find the dot product of
the second row in the first matrix and the first column in the second, which is
12. And for π, we find the dot product
of the second row in the first matrix and the second column in the second. Thatβs 15.

We have found π to be negative
three, π to be negative eight, π to be 12. and π to be 15. And in matrix form π, π, π, π
is negative three, negative eight, 12, 15.