### Video Transcript

Without expanding, find the value
of the determinant of the three-by-three matrix eight, negative three, negative two,
seven, one, negative eight, 24, negative nine, negative six.

In this question, we’re asked to
evaluate the determinant of a three-by-three matrix. We could do this by expanding over
any of its rows or columns. However, the question explicitly
asks us to do this without expanding. There’s a few different methods we
could use to evaluate determinants by using different properties of the
determinant. For example, we can recall that
we’re allowed to add and subtract linear multiples of the rows and columns to other
rows and columns of the matrix. And this won’t affect the
determinant of this matrix.

We could then use these to try and
rewrite our matrix as an upper triangular matrix. Then, the determinant of an upper
triangular matrix is the product of the entries on its leading diagonal. However, this is a complicated
process, so we should always check to see if there’s an easier method first. We can start by recalling that we
can take out shared factors among a row or column of the matrix. In particular, we can notice the
third row of this column all share a factor of three. Taking out the shared factor of
three from the row gives us three multiplied by the determinant of the
three-by-three matrix eight, negative three, negative two, seven, one, negative
eight, eight, negative three, negative two.

At this point, we might also notice
that in the third column of this matrix every single entry shares a factor of two or
negative two. And we could take this factor
outside of our matrix in exactly the same way. However, we can already evaluate
the determinant of this matrix by using a different property. We can instead notice that the
first row of this matrix is equal to its third row. We can then recall if a matrix has
a repeated row or column, then its determinant will be equal to zero. So, because the first and third row
of this matrix is repeated, we can conclude its determinant is zero, giving us three
times zero, which is equal to zero.

And before we finish with this
question, it’s worth noting we’ve shown a useful property about matrices. We’ve shown if one of the rows or
columns of this matrix is a scalar multiple of the other row or column of this
matrix, then its determinant is zero. In either case, we were able to
show the determinant of the three-by-three matrix eight, negative three, negative
two, seven, one, negative eight, 24, negative nine, negative six is equal to
zero.