We want to solve the matrix equation
AX equals B for matrix X given matrix A and matrix B.
So to solve the matrix equation AX equals B,
we'll be using the inverse of matrix A
to isolate matrix X.
Where once we find A inverse,
on the left we'll multiply
and we'll have A inverse times AX equals.
Now because matrix multiplication is not commutative,
on the right we must have A inverse times B,
not B times A inverse.
And then A inverse times A is equal to the identity matrix I
so we have I times X equals A inverse times B.
And the identity matrix times matrix X is just X,
so we have our solution, matrix X equals A inverse times B.
Now let's go ahead and set this up
using our given matrices.
So we'll have matrix A
times matrix X equals matrix B.
And now we need to find the inverse of matrix A.
Because matrix A is a three by three matrix,
there's no shortcut formula to find the inverse.
We'll find the inverse of matrix A
using an augmented matrix where if A is invertable,
meaning it has an inverse,
to find A inverse using an augmented matrix,
we want to write an augmented matrix
where on the left we have matrix A.
On the right we have a three by three identity matrix,
and then we'll perform our operations
to transform this augmented matrix
so instead of having A on the left,
we would have the three by three
identity matrix on the left.
Once we do this, the matrix on the right
will be A inverse.
We won't show all the steps on this due to time,
but the initial setup would look like this.
We'd have a three by six augmented matrix.
Sometimes you'll see a dashed line
cutting the matrix in half.
Where on the left we'd have matrix A.
On the right we'd have a three by three identity matrix.
And then from here we'd perform our operations
to obtain a three by three identity matrix here on the left.
When we do that, the identity matrix on the right
is transformed into A inverse.
So the end result,
again we're not going to show all the steps here.
We would perform our operations
so that we have a three by three
identity matrix on the left.
And by doing this, the matrix on the right will be A inverse
which would have a first row of one, negative four, three.
A second row of negative one, five, negative three.
And a third row of three, negative 17, ten.
So again, we started with this augmented matrix
where we had matrix A on the left
and the identity matrix on the right.
Perform our operations so that we had
the identity matrix on the left
and this matrix is now A inverse.
So now we're going to perform multiplication
with A inverse on both sides of the equation.
On the left we'll have A inverse times A times X.
So there's A inverse times matrix A which is here.
Times matrix X equals,
and then right we'll have A inverse
times matrix B.
So on the left, remember, this is A inverse
times A, which is equal to the identity matrix,
so we have identity matrix times matrix X equals
this product here, which notice how
because we have a three by three matrix
times a three by one matrix,
because number of columns in the first matrix
equal the number of rows in the second matrix,
multiplication is possible and the result
is going to be a three by one matrix.
So to find the element in row one column one of the product,
we'll multiply the first row of the first matrix
and the first column of the second matrix
which means A sub one comma one
is equal to one times negative 37
plus negative four times negative one
plus three times ten
so we have negative 37 plus four plus 30
which equals negative three.
And now to find the element in row two column one
we multiply row two of the first matrix
and column one of the second matrix.
So a sub two comma one is equal to
negative one times negative 37
plus five times negative one
plus negative three times ten.
So we have positive 37 plus negative five
plus negative 30 which is equal to positive two.
And finally the element in row three column one
is obtained by multiplying row three in the first matrix
by column one of the second matrix.
So we'd have three times negative 37
plus negative 17 times negative one
plus ten times ten.
And this is equal to positive six.
Well the identity matrix times X
is just equal to the matrix X,
and therefore we do have our solution.
So our solution is matrix X equals
this three by one matrix or this column matrix.
So again if you do need more help
on how to find the inverse of a three by three matrix
using an augmented matrix, I do have other videos
that show just how to find the inverse
of a three by three matrix.
I hope you found this helpful.