### Video Transcript

Consider the following matrix
equation: the three-by-three matrix with elements negative six, three, negative one,
four, zero, negative six, two, two, one multiplied by the column matrix with
elements ๐ฅ, ๐ฆ, and ๐ง is equal to the column matrix with elements three, ๐, and
five. Find the value of ๐ that results
in ๐ฆ is equal to 31 over 16.

The given matrix equation
represents a system of three linear equations, which we would normally want to solve
for the unknowns ๐ฅ, ๐ฆ, and ๐ง. However, in this case, weโre given
a value for ๐ฆ; thatโs ๐ฆ is equal to 31 over 16. And weโre asked to find the value
of ๐. We still therefore have three
equations and three unknowns. And to solve these, we can use the
matrix inverse method.

If we label our three matrices ๐ด,
๐ฎ, and ๐ฏ to begin with, weโll see how this method works on our equation ๐ด
multiplied by ๐ฎ is equal to ๐ฏ. Assuming that ๐ด is invertible,
that is, its inverse exists, then multiplying our equation through on the left by ๐ด
inverse, we have ๐ด inverse ๐ด multiplied by ๐ฎ is equal to ๐ด inverse multiplied by
๐ฏ. Now we know that for an ๐-by-๐
invertible matrix ๐ด, ๐ด multiplied by ๐ด inverse is ๐ด inverse multiplied by ๐ด is
the identity matrix. Thatโs the ๐-by-๐ matrix with all
elements equal to zero except the leading diagonal with elements all equal to one
and which for a three-by-three matrix is as shown.

Then, on the left-hand side of our
equation, we have ๐ผ multiplied by ๐ฎ. And we know that ๐ผ multiplied by
๐ฎ is simply ๐ฎ. So, this isolates the matrix ๐ฎ
with elements ๐ฅ, ๐ฆ, and ๐ง on the left-hand side. And if we then multiply out the
right-hand side, weโll find our solution. And to multiply out the right-hand
side, we need to find the inverse of the matrix ๐ด. Now making some room, we recall
that for an ๐-by-๐ invertible matrix ๐ด, ๐ด inverse is given by one over the
determinant of ๐ด multiplied by the adjoint matrix of ๐ด. So, weโll need to find both the
determinant of ๐ด and the adjoint matrix of ๐ด, remembering that the adjoint matrix
of ๐ด is the transpose of the matrix of cofactors. So, letโs begin by finding the
determinant of the matrix ๐ด.

Since the second row of our matrix
๐ด has a zero element, letโs expand along this row. And remember, to find the
determinant of a matrix by expanding along the ๐ผth row, we use the formula the sum
from ๐ is one to ๐ multiplied by the element ๐ ๐๐ multiplied by negative one
raised to the power ๐ plus ๐ multiplied by the determinant of the matrix minor ๐ด
๐๐. Remember, the matrix minor ๐ด ๐๐
of a three-by-three matrix is the two-by-two matrix obtained by removing row ๐ and
column ๐. In our case then, we have the
determinant of ๐ด is equal to negative the element ๐ two one multiplied by the
determinant of the matrix minor ๐ด two one and so on.

Note that itโs very important to
get the signs correct at this point. And these are given by negative one
raised to the power ๐ plus ๐. So, for example, in our first term,
we have negative one raised to the power two plus one, which is negative one raised
to the power three, and thatโs negative one. Our first term is then four
multiplied by negative one multiplied by the determinant of the two-by-two matrix
obtained by removing row two and column one. That is the two-by-two matrix with
elements three, negative one, two, and one. Similarly, we obtain our second
term and our third term. And we can evaluate our
determinants by recalling that for a two-by-two matrix ๐ with elements ๐, ๐, ๐,
๐, its determinant is ๐๐ minus ๐๐.

For our first term, this then gives
us negative four multiplied by three multiplied by one minus negative one multiplied
by two and similarly for our second and third terms. Evaluating then gives us negative
four times five plus zero times negative four plus six times negative 18. That is negative 20 minus 108,
which is negative 128. The determinant of our matrix ๐ด is
then negative 128. So, we have our determinant. Letโs now find the adjoint matrix
of ๐ด.

Recalling that the adjoint of a
matrix ๐ด is the transpose of the matrix of cofactors and a cofactor ๐ถ ๐๐ is
negative one raised to the power ๐ plus ๐ multiplied by the determinant of the
matrix minor ๐ด ๐๐. The term negative one raised to the
power ๐ plus ๐ gives us the sign or parity of the cofactor. And for a three-by-three matrix,
this is as shown. In fact, weโve already worked out
three of the cofactors, those of the second row. But letโs write them all out in
full. Our nine cofactors are as
shown.

And so, now, we work out the
two-by-two determinants of the matrix minors. So, for example, for our cofactor
๐ถ one one, we have the determinant zero multiplied by one minus negative six
multiplied by two, and thatโs equal to 12. Our cofactors are then 12, negative
16, eight, negative five, negative four, 18, and negative 18, negative 40, negative
12.

Now, writing out our matrix of
cofactors, the adjoint of our matrix ๐ด is then the transpose of the matrix of
cofactors. That is the matrix where the rows
and columns are interchanged. Now, remember, weโre trying to find
the inverse of the matrix ๐ด. And thatโs given by one over the
determinant times the adjoint of matrix ๐ด. Our determinant is negative 128 and
our adjoint is as shown. So, the inverse of our matrix ๐ด is
negative one over 128 multiplied by the adjoint of matrix ๐ด. So, now, we can substitute this
into our original equation. That is ๐ฎ is equal to ๐ด inverse
multiplied by ๐ฏ.

Now, remember, weโre to find the
value of ๐ that results in ๐ฆ is equal to 31 over 16. So, before solving our equation,
letโs substitute this value in for ๐ฆ. We can solve this by multiplying
out the right-hand side. And using matrix multiplication,
for example, in our first row, we have 12 times three plus negative five times ๐
plus negative 18 times five. And evaluating each of our rows, we
have a first row of negative 54 minus five ๐, a second row of negative 248 minus
four ๐, and a third row of negative 36 plus 18๐ all multiplied by negative one
over 128. Making some room and rewriting, we
can now equate our right- and left-hand sides. And so, by equality of matrices, we
have the three equations shown.

Since we want to find the value of
๐, we use the second equation because ๐ is the only variable in this equation. So, letโs begin by multiplying both
sides by 128. Dividing numerator and denominator
by 128 on the right-hand side gives us one. And dividing numerator and
denominator on the left-hand side by 16 gives us eight multiplied by 31 divided by
one. So, on the left, we have 248 and,
on the right, 248 plus four ๐. Subtracting 248 from both sides
gives us zero is equal to four ๐. Hence, ๐ is equal to zero.

This means that the value of ๐
that results in ๐ฆ is equal to 31 over 16 in the given matrix equation is ๐ is
equal to zero. And now, for completionโs sake, we
could use this value to solve for ๐ฅ and for ๐ง. ๐ฅ is then 54 divided by 128, and
๐ง is then 36 divided by 128. And evaluating these gives us ๐ฅ is
equal to 27 over 64 and ๐ง is nine over 32.