### Video Transcript

Given that the matrix seven, one, negative seven, π is invertible, what must be true of π?

Letβs recap. For a two-by-two matrix with elements π, π, π, π, its inverse is given by one over the determinant of π΄ multiplied by π, negative π, negative π, π, where the determinant is found by multiplying the elements π and π and then subtracting the product of the elements π and π.

Notice this means that if the determinant of π΄ is zero, then π΄ can have no multiplicative inverse, since one over the determinant of π΄ is one over zero, which is undefined.

We are told that the matrix seven, one, negative seven, π is invertible. This means its determinant cannot be zero. Letβs set up an equation to this effect.

The determinant of π΄ is found by multiplying the top left element with the bottom right element, then subtracting the product of the top right element and the bottom left. Thatβs seven π minus negative seven, which simplifies to seven π plus seven.

We said that this matrix is invertible, so its determinant cannot be equal to zero. So we can say that seven π plus seven is not equal to zero. Then, we can solve this inequation by subtracting seven from both sides. And we get that seven π is not equal to negative seven.

Next, weβll divide both sides by seven. And that gives us π is not equal to negative one.

For the matrix seven, one, negative seven, π to be invertible, π cannot be equal to negative one.