# Video: Evaluating 3 × 3 Determinants

Evaluate |1, −9, −6 and −8, 4, 1 and 2, −1, 9|.

03:41

### Video Transcript

Evaluate the determinant of the three-by-three matrix one, negative nine, negative six, negative eight, four, one, two, negative one, nine.

Now when we’re looking to evaluate a determinant of a three-by-three matrix, then what we do if we take a look at the first row, and we use the first row as coefficients. And what we do is we multiply in turn each of these by the submatrix, so the two-by-two submatrix that’s formed when you delete the row and column that that value is in.

It’s also worth noting that our coefficients have to follow a pattern. So, our first column is positive, so the coefficient is multiplied by positive one. The second column is negative, so multiplied by negative one. Third column, positive. So, we’re gonna use that in a second when we put it altogether.

So, the first coefficient we’re looking at is the top-left term. So, it’s one because it’s in the first row. And we’re gonna multiply this by the determinant of this submatrix, which is formed, so the two-by-two submatrix when we delete the row and column that the one is in. So, it’s gonna give us one multiplied by the determinant of the two-by-two submatrix four, one, negative one, nine.

Then, next, we’re gonna have minus negative nine. And that’s cause, as we said, the second column has to be negative. And we’ve already got negative nine as our coefficient. Then, multiplied by the two-by-two submatrix, again, formed when you delete the row and column that the negative nine is in. So, it’s the determinant of the submatrix negative eight, one, two, nine. Then, finally, we’re gonna have minus six multiplied by the determinant of the two-by-two submatrix negative eight, four, two, negative one. And it stayed as negative six as the coefficient because, as we said, this third column is positive. So, the sign stays the same.

So, now the next stage is to know how to deal with the two-by-two determinant. Well, to find the two-by-two determinant, what we do is if we’ve got the two-by-two determinant of the matrix 𝑎, 𝑏, 𝑐, 𝑑, you multiply 𝑎 by 𝑑, so we diagonally multiply, and then subtract 𝑏 multiplied by 𝑐. So, first of all, we’re gonna have one multiplied by then we’ve got four multiplied by nine minus one multiplied by negative one. And then, we’ve got plus nine multiplied by negative eight multiplied by nine minus one multiplied by two. And we get that because we had minus negative nine. And if we subtract a negative, it turns positive.

And then, finally, you’ve got minus six multiplied by negative eight multiplied by negative one minus four multiplied by two. So, if we evaluate this, we’re gonna get 37. And that’s because we had one multiplied by then you’ve got 36 minus negative one, which is 36, add one, which is 37. Then, we’re gonna get minus 666. And that’s because we had nine multiplied by negative 74. That’s cause we had negative 72 minus two, which is negative 74, which gives us negative 666.

And then, minus zero. And we get that because we have negative eight multiplied by negative one, which is eight. Four multiplied by two, which is eight. Eight minus eight is just zero. So, this gives us a final answer of negative 629. So, therefore, we can say that if we evaluate the determinant of the three-by-three matrix one, negative nine, negative six, negative eight, four, one, two, negative one, nine, then the result is negative 629.