### Video Transcript

If the matrix π΄ has rank two, what is the value of π?

In order to solve this question, we will use the fact that the question says that matrix π΄ has a rank of two. And since matrix π΄ is a three-by-three matrix, this tells us that the determinant of π΄ is equal to zero. And this is because if the determinant of π΄ is nonzero, then π΄ has a rank of three. And so we can use the fact that the determinant of π΄ is equal to zero to form an equation involving π, which can be solved to find the value of π.

Letβs now recall how to find the determinant of a three-by-three matrix. So given this matrix, we have that its determinant is equal to π lots of ππ minus πβ minus π lots of ππ minus ππ plus π lots of πβ minus ππ. Letβs quickly look at where this formula comes from.

When finding the determinant of a three-by-three matrix, we know that there will be three terms. And each of these terms will start with their respective entry in the first row of the matrix. And so in our case, thatβs π, π, and π. And we know that the signs of these terms will be positive and then negative and then positive. And so thatβs where we get the minus sign in front of the π from.

Now in order to work out what goes in the bracket of each term, we take the first part β so in the first term, thatβs π β and we cross out everything thatβs in the same row or column as π. And weβre left with a two-by-two matrix, which we then find the determinant of. And we multiply π by this determinant. So this gives us that ππ minus πβ in the brackets.

We then repeat this process for π. Crossing out everything in the same row and column as π leaves us with four entries, which form a two-by-two matrix. And we then find the determinant of this matrix and multiply it by π. And we then repeat this process for π. Crossing out the rows and columns, weβre left with this four-by-four matrix, which we find the determinant of, giving us then πβ minus ππ, which we multiply π by.

And so now weβre ready to use this formula to find the determinant of matrix π΄. And we get that the determinant of π΄ is equal to. We get six lots of 24 times minus 11 minus four times 18 minus negative nine lots of π timesed by negative 11 minus four times 15 plus one lot of π times 18 minus 24 times 15. And we can multiply inside the brackets to leave us with six lots of minus 264 minus 72 plus nine lots of minus 11 times π minus 60 plus 18π minus 360. And now we expand the brackets to leave us with. We get minus 1584 minus 432 minus 99π minus 540 plus 18π minus 360, which simplifies to minus 2916 minus 81π.

And now we have found an equation for the determinant of π΄. But we also have that the determinant of π΄ is equal to zero. So therefore, we can put our equation equal to zero. And then we can solve this for π. We obtain a solution of π is equal to negative 36.