Find the value of the determinant of the three-by-three matrix negative six, negative two, seven, negative six, nine, four, negative two, zero, one times the determinant of the matrix negative two, one, negative two, zero.
It should be quite clear that in order to answer this problem, we will need to begin by working out the determinant of each of our matrices. It’s somewhat simpler to calculate the determinant of a two-by-two matrix. And this definition will help us with the determinant of a three-by-three matrix in a moment. So we’ll begin with this two-by-two matrix. Suppose we have a two-by-two matrix whose elements are 𝑎, 𝑏, 𝑐, 𝑑. Its determinant is 𝑎𝑑 minus 𝑏𝑐. We find the product of the elements in the top left and bottom right and subtract the product of the elements in the top right and bottom left.
So for our matrix then, we find the product of negative two and zero and then subtract the product of one and negative two. Negative two times zero is zero, and negative one times negative two is two. So the determinant is zero plus two, which is simply two. Now that we know the determinant of our two-by-two matrix, let’s move on to calculating the determinant of the three by three.
Now let’s suppose that our matrix 𝐴 is the three-by-three matrix with elements 𝑎, 𝑏, 𝑐, 𝑑, 𝑒, 𝑓, 𝑔, ℎ, 𝑖. It looks somewhat complicated, but there is a pattern to finding its determinant. It’s 𝑎 times 𝑒𝑖 minus 𝑓ℎ. In other words, it’s the first element in the first row times the determinant of the matrix 𝑒, 𝑓, ℎ, 𝑖, essentially, the determinant of the elements that are not in 𝑎’s row or column. Then we subtract 𝑏 multiplied by the determinant of the matrix 𝑑, 𝑓, 𝑔, 𝑖. That’s 𝑑𝑖 minus 𝑓𝑔. Finally, we add 𝑐 times the determinant of the matrix 𝑑, 𝑒, 𝑔, ℎ, so times 𝑑ℎ minus 𝑒𝑔.
So what does this mean for the determinant of our three-by-three matrix? We multiply the first element by the determinant of the two-by-two matrix not in this row or column, so by the determinant of the matrix nine, four, zero, one. The determinant of this matrix is nine times one minus four times zero. And we’re multiplying this by negative six. Then we subtract negative two times the determinant of the matrix not in this row or column. So that’s the determinant of the matrix negative six, four, negative two, one, which is negative six times one minus four times negative two.
Finally, we add seven times the determinant of the two-by-two matrix not in this row or column. So that’s seven times negative six times zero minus nine times negative two. Evaluating this expression will give us the determinant of this matrix. It’s negative 54 minus negative four plus 126, which is equal to 76. So the value of the expression in our question is the product of our two determinants. It’s the product of 76 and two, which is 152. The answer is 152.