### Video Transcript

Using matrix inverses, solve the
following for π. Thatβs π multiplied by the matrix
with elements negative three, two, four, negative three is equal to the matrix with
elements zero, negative two, three, zero.

Weβre given a matrix equation and
asked to solve for π. We know, in fact, that π is a
matrix since if π were a scalar, weβd be able to find the value for π satisfying
the equation shown. And by equality of matrices, this
would mean that π would satisfy all four of the shown equations. π is equal to zero satisfies the
first and fourth equations. However, for the second equation,
the only value of π satisfying this is π is negative one. And for the third equation, the
only value of π satisfying this is π is three over four. Since a scalar quantity cannot be
three things at once, our π must be a matrix.

Now we know that for a matrix
multiplication to be possible, if a matrix π΄ has size π by π and a matrix π΅ has
size π by π, then their product must have the size π by π, that is, π rows and
π columns. If we call the resulted matrix πΆ,
in our case, πΆ is a two-by-two matrix. This means both π and π are
two. So if our matrix π is labeled π΄
and it multiplies in matrix π΅, we know π΅ is a two-by-two matrix; therefore, π is
also two in which case our matrix π, corresponding to π΄, is also a two-by-two
matrix. So letβs let our matrix π have
elements π one, π two, π three, and π four.

Now weβre asked to use matrix
inverses to solve the equation. And we recall that for an π-by-π
invertible matrix π, its inverse π minus one is an π-by-π matrix such that π
multiplied by π inverse is equal to π inverse multiplied by π. And thatβs equal to the identity,
where the identity matrix has all of its elements equal to zero except those on the
leading diagonal, which are all equal to one.

Now, remember, for a two-by-two
matrix π with elements π, π, π and π, the inverse of π is one over the
determinant of π multiplied by the matrix where π and π have been swapped and we
take negative π and negative π. And remember, of course, that the
determinant of a two-by-two matrix is ππ minus ππ.

To solve the equation, weβre going
to isolate π on the left-hand side. If we call our two other matrices
π΄ and π΅, respectively, our equation says π multiplied by π΄ is equal to π΅. Weβre going to use the inverse of
the matrix π΄ to eliminate π΄ on the left-hand side. So if we multiply both sides on the
right by inverse π΄, by the inverse law, we know that π΄ multiplied by π΄ inverse is
equal to the identity matrix. So we now have π multiplied by the
identity matrix is π΅ multiplied by π΄ inverse. Any matrix multiplied by the
identity matrix is itself. So we have π is equal to π΅
multiplied by π΄ inverse. So now we need to find the inverse
of the matrix π΄.

Making a little room, our matrix π΄
has elements negative three, two, four, and negative three. For the inverse of π΄, weβre going
to need the determinant. And the determinant is ππ minus
ππ. And thatβs negative three
multiplied by negative three minus two multiplied by four. And thatβs nine minus eight, which
is equal to one. The inverse is given by swapping π
and π and taking the negatives of π and π within our matrix and multiplying by
the determinant. In our case, π and π are both
negative three. So swapping those leaves them the
same. And taking the negative of π and
π, where π is two and π is four, gives us negative two, negative four. The inverse of π΄ is therefore the
matrix with elements negative three, negative two, negative four, and negative
three.

We can easily check that this is
actually the inverse of the matrix π΄ by multiplying the two together. Our result should be the identity
matrix. Using matrix multiplication, we
have negative three multiplied by negative three plus two multiplied by negative
four, negative three multiplied by negative two plus two multiplied by negative
three, and similarly for the remaining two elements. This evaluates to nine minus eight,
six minus six, negative 12 plus 12, and negative eight plus nine, which does indeed
give us the two-by-two identity matrix. We do therefore have the correct
inverse matrix for the matrix π΄.

And now inserting this into our
equation for the inverse of π΄, performing matrix multiplication should then give us
the matrix π. We have zero times negative three
plus negative two times negative four, zero times negative two plus negative two
times negative three, three times negative three plus zero times negative four, and
finally three times negative two plus zero times negative three.

Zero times negative three is
zero. Negative two times negative four is
positive eight. Zero times negative two is
zero. And negative two times negative
three is six. Three times negative three is
negative nine. Zero times negative four is
zero. Three times negative two is
negative six. And zero times negative three is
zero. Our result is therefore the matrix
with elements eight, six, negative nine, and negative six.

Equating this with our matrix π,
our elements are therefore π one is equal to eight, π two is equal to six, π
three is equal to negative nine, and π four is equal to negative six. And this is our solution π.