# Question Video: Solving Determinants Using Properties Mathematics

Evaluate [−6, 1 and 1, 1] + [−5, 1 and 1, 1] + [−4, 1 and 1, 1] + ... + [10, 1 and 1, 1].

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### Video Transcript

Evaluate the determinant of the matrix negative six, one, one, one plus the determinant of the matrix negative five, one, one, one plus the determinant of the matrix negative four, one, one, one. And we keep adding determinants of matrices of this form up to the determinant of the matrix 10, one, one, one.

In this question, we’re asked to evaluate an expression. And we can see in this expression every term is the determinant of a two-by-two matrix. So, one way to answer this question would just be to evaluate all of these determinants and add them together. However, we can simplify this expression first by using the properties of determinants.

To do this, we need to notice something interesting about all of the matrices we’re given. All of these matrices have exactly the same second row. And one of our properties of determinants tells us how to add two matrices together which only have one row or column which differ. We recall that we can add the determinants of two matrices together which have all of the same elements except for the same row or column by combining these into one matrix where we add the corresponding entries in the row or column that differs.

Now, we could use this property to just add two of the determinants together. However, we can notice, in this case, all of our matrices are of the same form. So instead, we’ll just apply this property to all of the terms. So, to use this property on this sum, we first know that the second row of this matrix is going to be one, one. Next, the entry in row one, column one of our matrix is going to be the sum of all of the entries in row one, column one of our terms. This will, of course, be negative six plus negative five. And we keep adding integers of this form up to 10.

We would then do the same for the entries in row one, column two. However, we can see that all of these entries are equal to one. So, when we add these together, we would just get the number of terms. And the number of terms from negative six to 10 is equal to 17. Therefore, by using the properties of determinants, we were able to rewrite the sum of determinants as one determinant.

Now, to evaluate the determinant of this two-by-two matrix, we’re going to need to evaluate the sum we have in its first row and first column. And there’s several different ways we can do this. For example, this is an arithmetic sequence with first term negative six and common difference one. We could then use the formula for the sum of a finite arithmetic sequence. However, there is a second way we can do this. We can notice the term negative six will cancel with six, negative five will cancel with five, and this will continue all the way down to negative one canceling with one. And of course, adding zero doesn’t change its value. So, this simplifies to give us seven plus eight plus nine plus 10. And if we evaluate this, we see it’s equal to 34. Therefore, the sum of these determinants is equal to the determinant of the two-by-two matrix 34, 17, one, one.

Finally, we can evaluate the determinant of this matrix by finding the difference in the products of its diagonals. That’s 34 times one minus 17 times one, which is equal to 17. Therefore, we were able to evaluate the sum of these determinants by using our properties of determinants. We were able to show this sum was equal to 17.