### Video Transcript

Is the matrix three, one, negative three, one invertible?

For a two-by-two matrix π΄, where π΄ is equal to π, π, π, π, its inverse is given by the formula one over the determinant of π multiplied by π, negative π, negative π, π, where the determinant of π΄ is given by ππ minus ππ.

Notice that this means that if the determinant of the matrix π΄ is zero, the inverse does not exist, since one over the determinant of π΄ would be one over zero, which is undefined.

What we need to decide then is whether the determinant of the matrix three, one, negative three, one is equal to zero. That will tell us whether it has an inverse or not.

To find the determinant, we multiply the first element in the first row by the second element in the second row. Itβs three multiplied by one. Then we subtract the product of the second element in the first row and the first element in the second row. Thatβs one multiplied by negative three. That gives us three minus negative three, which is six.

Since the determinant of our matrix is not zero, it does have an inverse; itβs invertible. In fact, we can substitute the values we now know into our formula for the inverse of a matrix. It becomes one-sixth multiplied by one, negative one, three, three. Multiplying everything by one over the determinant gives us one-sixth, negative one-sixth, one-half, and one-half.