Question Video: Solving Determinants Using Properties Mathematics

Without expanding, find the value of the determinant [8, −3, −2 and 7, 1, −8 and 24, −9, −6].

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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.

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