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## Finding the inverse of a matrix using its determinant

Current time:0:00Total duration:2:48

# Finding inverse of a 2x2 matrix using determinant & adjugate

## Video transcript

Let's attempt to take the
inverse of this 2 by 2 matrix. And you'll see the
2 by 2 matrices are about the only
size of matrices that it's somewhat pleasant
to take the inverse of. Anything larger than that,
it becomes very unpleasant. So the inverse of
a 2 by 2 matrix is going to be equal to 1 over
the determinant of the matrix times the adjugate
of the matrix, which sounds like a very fancy word. But we'll see for by a 2 by 2
matrix, it's not too involved. So first let's think about what
the determinant of this matrix is. Well, we've seen this before. We just look along
the two diagonals. It's 3 times 2 minus
negative 7 times 5. So this is going to be equal to
1 over 3 times 2 minus negative 7 times 5. And then the adjugate
of A-- and here I'm really just teaching
you the mechanics of it. And it's a little unfortunate
that in a typical algebra II class you kind of just go
into the mechanics of it. But at least this will get
us to where we need to go. So the adjugate of
A, you literally just need to swap the two
elements on this diagonal. So put the 2 where the 3 is
and the 3 where the 2 is. So this element right
over here, this 3 will go right over there. This 2 will go right over here. And then these two elements, you
just take the negative of them. So the negative-- let
me do a new color. Actually, I'm running
out of colors. The negative of
that is negative 5. The negative of
that is positive 7. So we are left
with-- this is going to be equal to 1
over-- 3 times 2 is 6. Negative 7 times
5 is negative 35. But then we have this
positive over here. So this whole thing
becomes plus 35. So 6 plus 35 is 41. So the determinant
of our matrix is 41. We're going to take 1
over the determinant and multiply it times our
adjugate, times 2, negative 5, 7, and 3. So we get -- so this is the
drum roll part -- 2 over 41, negative 5 over 41. I'm just multiplying each of
these elements times 1 over 41. 7 over 41, and 3 over 41. And we are done.