Lesson Video: Properties of Determinants | Nagwa Lesson Video: Properties of Determinants | Nagwa

Lesson Video: Properties of Determinants Mathematics • Third Year of Secondary School

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In this video, we will learn how to identify the properties of determinants and use them to simplify problems.

20:28

Video Transcript

Properties of Determinants

In this video, we will learn how to identify different properties of determinants and how to use these properties of determinants to simplify problems. Before we start listing the properties of determinants, we can recall we already know several different methods of calculating the determinant of different matrices.

For example, we know we can calculate the determinant of a square matrix by adding and subtracting the products of the diagonals. Or alternatively, we can calculate the determinant of a matrix by expanding over the first row. In fact, we can also show that we can expand over any row or column of the matrix. We will still get the same value for the determinant. What this means is we can choose any row or column of the matrix to expand over to calculate its determinant; we can choose the easiest row or column. We can use this result to prove a very useful property of determinants.

If we had a square matrix with an entire row or column filled with zeros, then we could calculate the determinant of this matrix by expanding over the zero row or column. In this calculation, every term would have a factor of zero. So we would be able to show the determinant of this matrix is zero. And it is worth reiterating here we can only calculate the determinant of square matrices. So, when we’re talking about the properties of determinants, we assume all of our matrices are square. Otherwise, we wouldn’t be able to calculate the determinant of this matrix.

Another useful property we can show is that taking the transpose of a matrix does not affect its determinant. In other words, for any square matrix 𝐴, the determinant of 𝐴 transpose is equal to determinant of 𝐴, where we recall to take the transpose of a matrix, we write the rows of the matrix as the corresponding columns of our new matrix. And before we continue, it’s worth pointing out here we can prove all of the properties of the determinant shown in this video. Most of the proofs of the properties will just involve using the definition of a determinant and checking the left-hand side of the equation and the right-hand side of the equation are equal. However, as we’ll see, there’s far too many properties of the determinant to prove in this video.

And before we move on to the next property, there is one useful thing. We can use the results about the determinant of a matrix being equal to the determinant of its transpose. If we wanted to prove one of our previous properties involving rows and columns of a matrix, we already know the determinant of a matrix is equal to the determinant of its transpose. And the rows of matrix 𝐴 are the columns of matrix 𝐴 transpose. So, we would only need to prove this property for rows and then use the transpose property to prove it for columns.

The next property we can show is that if all entries in a single row or column of a matrix share a common factor, then we can take out this shared factor outside of the determinant. And we can see an example of this. Consider the determinant of the two-by-two matrix two, one, five, 10. In this matrix, we can see the second row has a shared factor of five. Our determinant properties tell us we can take the shared factor of five outside of the calculation for the determinant. It’s equal to five times the determinant of the two-by-two matrix two, one, one, two. And we could calculate both sides of this equation separately.

In our first matrix, we take the difference in the products of the diagonal. That’s two times 10 minus five times one, which is equal to 15. And we would do the same in our second matrix. We take the difference in the products of the diagonals. That’s two times two minus one times one, which is equal to three. Then, we multiply this by a factor of five to see the right-hand side of the equation is also 15.

Another useful property we can show is if a square matrix has a repeated row or column, then its determinant will be equal to zero. And this result can be particularly useful for finding the determinants of large matrices. We should always check if there’s a repeated row or column in any matrix so we can determine that its determinant is zero.

The next useful property we can show is if we switch any two rows or columns of our matrix, then this will switch the sign of the determinant of that matrix. To see an example of this, I’d switch the two columns in our two-by-two matrix two, one, five, 10. Switching the first column, the second column gives us the two-by-two matrix one, two, 10, five. And we have notation for this: 𝑐 sub one is the first column of the matrix and 𝑐 sub two is the second column of the matrix. We use a double-sided arrow to show that we’re interchanging the two columns. If we then calculate the determinant of this matrix, we get one multiplied by five minus 10 multiplied by two, which we can calculate is negative 15, the negative of the original matrix. So, switching any two rows or columns of our matrix will switch the sign of our determinant.

Another property we can show about determinants is that we can add or subtract scalar multiples of the rows or columns of the matrix to other rows or columns of the matrix without affecting the determinant. And to help us see what this means, let’s take a look at an example. If we wanted to calculate the determinant of the two-by-two matrix one, three, zero, two, we could do this directly, or we could rewrite our matrix by using this property. We can replace the second column of this matrix by subtracting three times the first column from this matrix. And remember, our property guarantees that this won’t affect the value of the determinant. We can write this as 𝑐 two is being replaced with 𝑐 two minus three 𝑐 one. Subtracting three lots of the first column from the second column, we get zero, two.

Therefore, the determinant of the two-by-two matrix one, three, zero, two is equal to the determinant of the two-by-two matrix one, zero, zero, two. This property is particularly useful for helping us simplify determinants of matrices of order three by three and higher.

There are many more properties of determinants we can show. So, let’s clear some space and go over a few more of these properties. Our next property that we can show is that if we multiply all of the entries of a row or column by the corresponding cofactors of a different row or column, then the sum of these values will be zero. This is quite a complicated property, so it’s much easier to see what this means in an example. Let’s start by applying this property to the three-by-three matrix one, two, three, four, five, six, seven, eight, nine. And we’ll use the first row of this matrix, but remember, we can use any row or column of this matrix.

We now want to multiply every entry of this row of the matrix by the corresponding cofactors of another row of this matrix. Then, when we sum these values, we should get zero. Let’s multiply these entries by the corresponding cofactors of the third row of the matrix; that’s the row seven, eight, nine. Let’s start by finding the matrix minor we get from the entry seven in our matrix. That’s the determinant of the two-by-two matrix we get by removing the first column and third row from the matrix. That’s the two-by-two matrix two, three, five, six. Remember, we need to check the parity of the sum of the row and column numbers of this entry, and it’s in the first column and third row. These sum to give four, which is even. So, this will be a positive minor.

Finally, in our property, we’re multiplying this by the corresponding entry in the first row. So, we multiply this by one. We can then do exactly the same for the next entry in this row. The matrix minor of this entry will be the determinant of the two-by-two matrix one, three, four, six. We need to take the negative of this value. And finally, we need to multiply this by the corresponding entry in the first row; that’s two.

We then do exactly the same for the third entry in this row. We get three times the determinant of the two-by-two matrix one, two, four, five. And our matrix property guarantees that this sum must be equal to zero. We can calculate this to check that this is true in this case. And there is another slightly different way we can think about this property. We could say that we’re calculating the determinant of this matrix by expanding over the third row of the matrix. However, we’re using the coefficients of the first row instead.

Another useful property of determinants we can show is the determinants of any square triangular matrix will be equal to the products of all of its entries on its main diagonal. And there’s a few things worth pointing out here. First, this works for both upper and lower triangular matrices. Similarly, since diagonal matrices are both upper and lower triangular matrices, this property will also work for diagonal matrices. Finally, if any of the entries on the main diagonal of our triangular matrix is zero, then this product will have a factor of zero. So, its determinant is zero, and therefore it’s not invertible. Let’s see a use of this property in an example.

Let’s say we want to calculate the determinant of the three-by-three matrix one, two, three, zero, four, five, zero, zero, six. We can see we’re calculating the determinant of an upper triangular matrix because all of the entries below the leading diagonal are zero. And therefore, our property tells us we can calculate the determinant of this matrix by just calculating the products of the main diagonals. That’s one times four times six, which is equal to 24.

The next property we’re going to discuss is true for a square matrix of any order and is true for any row or column of this matrix. However, if we wanted to include all of these possibilities in our statement, then the statement itself would get very complicated. So, we’ll just state this property for the first row of a three-by-three matrix. However, this does hold for any row, any column, and any-size square matrix.

The determinant of any square matrix where one of the rows or columns is written as a sum can be split as follows. It’s equal to the sum of the determinants of the following two matrices. And the only difference between these two matrices is we’ve split the row or column over which we had a sum in our original matrix. The first matrix took the first term in each of these sums, and the second matrix took the second term in each of these sums.

And finally, it’s worth reiterating although the statement we’ve written is only written for the first row of a three-by-three matrix, the statement holds for any square matrix and it works for any row or any column of the matrix. The second thing we might not notice about this statement is that it also works in the opposite direction. In other words, if we’re calculating the sum of the determinants of two or more matrices where the only difference involves one row or one column, then we can add these rows or columns together so we only need to calculate the determinant of one matrix.

There are two more properties of determinants we need to show. First, the determinant of the product of two matrices is the product of their determinants, where, of course, both 𝐴 and 𝐵 need to be square matrices. Otherwise, we wouldn’t be able to calculate their determinants. And we can actually use this property to show one final useful property. The determinant of 𝐴 times 𝐴 will be equal to the determinant of 𝐴 multiplied by the determinant of 𝐴. In other words, the determinant of 𝐴 times 𝐴 is equal to the determinant of 𝐴 all squared.

We can continue this to show for any positive integer exponent 𝑛 the determinant of matrix 𝐴 to the 𝑛th power is equal to the determinant of matrix 𝐴 all raised to the 𝑛th power. Let’s now see some examples of how we can use these properties of determinants to simplify and evaluate problems involving the determinants of different matrices.

Find the value of the determinant of the three-by-three matrix four, one, negative eight, negative six, three, six, zero, zero, zero.

In this question, we’re asked to evaluate the determinant of a three-by-three matrix. And we might be tempted to jump in straightaway and evaluate this determinant by expanding over the first row. And this would work; we would get the correct answer. However, whenever we’re asked to evaluate a determinant, we can always check to see if we can simplify this problem by using properties of determinants. And in this case, we can notice the third row of this matrix is all zeros.

We can then recall the following fact about the determinants of matrices. If all of the entries of a row or column of a square matrix is zero, then its determinant will also be equal to zero. And it’s worth pointing out, in this case, this is exactly the same as saying that we’re going to calculate the determinant of our matrix by expanding over its third row, since then in our calculation of the determinant every single term would have a factor of zero. So, the determinant of this matrix would still be zero. In either case, we were able to show the determinant of the three-by-three matrix four, one, negative eight, negative six, three, six, zero, zero, zero is equal to zero because it has a row entirely made of zeros.

Let’s now see another example of calculating the determinant of a three-by-three matrix without expanding over the rows or columns.

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.

Let’s now see an example of combining multiple determinants into one determinant.

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.

Let’s now see one last example of using the properties of the determinants to answer questions.

Consider the determinant of the two-by-two matrix 𝑥, 𝑦, 𝑧, 𝑤 is equal to six. Find the value of the determinant of the two-by-two matrix 𝑥 minus 10𝑦, 𝑦, 𝑧 minus 10𝑤, 𝑤.

In this question, we’re given the determinant of a two-by-two matrix and asked to determine the determinant of a different two-by-two matrix. And both of these matrices contain four unknowns, so we can’t evaluate these directly. We might be tempted to expand the determinant in our first matrix and then expand the determinant in our second matrix and try to rewrite this expression in terms of our first determinant. However, there’s a much simpler method involving the properties of determinants.

To do this, we need to notice something interesting about the second matrix we’re given. In the first entry of the first column, we’re subtracting 10𝑦, and in the second entry of the second column, we’re subtracting 10𝑤. This is a scalar multiple of the second column of our first matrix. In fact, we’re subtracting this directly from the first column of this matrix. In other words, to generate the second matrix, we’re subtracting 10 lots of the second column from the first column of our first matrix. And we can recall adding and subtracting scalar multiples of one row or column to a different row or column will not affect the determinant. Therefore, the determinant of the second matrix will be equal to the determinant of the first matrix, which we’re told is equal to six.

Let’s now go over some of the key points of this video. First, we showed that we can use all of the many properties of determinants to simplify expressions involving determinants. This means whenever we’re given an expression involving determinants or asked to evaluate a determinant, we should always check to see if we can simplify this problem by using the properties of determinants. And there are many properties of the determinants, so many that we can’t list all of them in the key points.

A few of these are taking the transpose of a square matrix doesn’t affect its determinant. If all of the entries in a row or column of a square matrix are zero, then its determinant will be equal to zero. If a square matrix has a repeated row or column, then its determinant will be equal to zero. And the determinant of any triangular matrix will be the product of its main diagonals. And of course, this also extends to diagonal matrices.

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