Video: Determinants and Invertibility

Given that the matrix [7, 1 and βˆ’7, π‘Ž] is invertible, what must be true of π‘Ž?

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Video Transcript

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

Let’s recap. For a two-by-two matrix with elements π‘Ž, 𝑏, 𝑐, 𝑑, its inverse is given by one over the determinant of 𝐴 multiplied by 𝑑, negative 𝑏, negative 𝑐, π‘Ž, where the determinant is found by multiplying the elements π‘Ž and 𝑑 and then subtracting the product of the elements 𝑏 and 𝑐.

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

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

The determinant of 𝐴 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. That’s seven π‘Ž minus negative seven, which simplifies to seven π‘Ž plus seven.

We said that this matrix is invertible, so its determinant cannot be equal to zero. So we can say that seven π‘Ž 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‘Ž is not equal to negative seven.

Next, we’ll divide both sides by seven. And that gives us π‘Ž is not equal to negative one.

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

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