Lesson Video: Properties of Determinants Mathematics

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

18:03

Video Transcript

Properties of Determinants

In this video, we will learn how to identify the properties of determinants and use these properties to solve problems. We will focus on only four properties, and all of the properties listed will hold on square matrices of any order. However, we will only focus on matrices of order two by two and three by three. Before we begin with these properties, let’s start by recalling how we calculate the determinants of a two-by-two matrix.

We recall the determinant of a two-by-two matrix is the difference in the product of the diagonals. In this matrix, that’s π‘Ž one one times π‘Ž two two minus π‘Ž one two multiplied by π‘Ž two one. We can use this definition to consider our first property. Let’s start by calling this matrix 𝐴. We want to see if we can construct a formula for the determinant of a scalar multiple of 𝐴. And of course, in this case, we’re only working with two-by-two matrices. We can do this directly from the definition of a determinant and scalar multiplication. Remember, when we multiply a matrix by a scalar, we multiply every entry of our matrix by our scalar.

We can then evaluate the determinant of this matrix by taking the difference in the products of the diagonals. This then gives us the determinant of matrix π‘˜ times 𝐴 is equal to π‘˜ times π‘Ž one one multiplied by π‘˜ times π‘Ž two two minus π‘˜ times π‘Ž one two multiplied by π‘˜ times π‘Ž two one. And we can simplify this expression. We can notice the first term has a factor of π‘˜ squared and the second term also has a factor of π‘˜ squared. Taking out this factor of π‘˜ squared gives us the following expression. We can see it’s π‘˜ squared multiplied by the difference in the product of the diagonals. It’s the determinant of 𝐴. Therefore, we’ve shown for any scalar multiple π‘˜ and two-by-two matrix 𝐴, the determinant of π‘˜ times 𝐴 is π‘˜ squared multiplied by the determinant of 𝐴.

We could show a very similar result for three-by-three matrices by using the definition of a determinant, the biggest difference being when calculating a determinant of a three-by-three matrix, we have three factors. So, we’ll have factors of π‘˜ cubed instead of π‘˜ squared. And in fact, this property is true in general, although we’ve only proven it for two-by-two matrices. We have for any square matrix 𝐴 of order 𝑛 by 𝑛 and any scalar value of π‘˜, the determinant of π‘˜ times 𝐴 is equal to π‘˜ to the 𝑛th power multiplied by the determinant of 𝐴. When calculating the determinant of a matrix, we can take out a scalar factor of π‘˜. We just need to raise it to the 𝑛th power where 𝑛 is the order of matrix 𝐴. There are other operations we can apply to square matrices. So, let’s see how these affect the determinant of our matrix.

One of these operations is the transpose of a matrix. Remember, this means we write the rows as the columns of the matrix and the columns as the row of the matrix. Let’s see how the determinant of 𝐴 transposed compares of the determinant of 𝐴 for our two-by-two matrix. First, it’s worth reiterating when we take the transpose of a matrix, we write the rows of the matrix as the columns of our new matrix. We write the first row of matrix 𝐴 as the first column of matrix 𝐴 transpose. And we can see the only difference between matrix 𝐴 and matrix 𝐴 transpose is the element in row one, column two is switched with the element in row two, column one. We can calculate the determinant of this matrix by taking the difference in the products of the diagonals. We get π‘Ž one one times π‘Ž two two minus π‘Ž two one multiplied by π‘Ž one two. Then, we can write this in terms of the determinant of 𝐴 by using the commutativity of scalar multiplication. This expression is just equal to the determinant of 𝐴.

Therefore, we’ve proven for any two-by-two matrix the determinant of 𝐴 transpose is equal the determinant of 𝐴. We can prove the exact same result is true for three-by-three matrices or, in fact, any square matrix. The easiest way to do this is to expand the determinant of 𝐴 over the first row and then expand the determinant of 𝐴 transpose over the first column. This then gives us our second property. For any square matrix 𝐴, the determinant of 𝐴 transpose is just equal to the determinant of 𝐴. Let’s now move on to properties involving the determinants of different ways of combining matrices.

The first way we might think to combine matrices is to add them together. If we have two two-by-two matrices 𝐴 and 𝐡, what is the determinant of their sum? However, it turns out this is very difficult to do. And for any order matrices, this becomes impossible. So instead, let’s consider the product of these two matrices. What is the determinant of matrix 𝐴 times matrix 𝐡?

We’ll start by constructing an expression for the determinant of 𝐴 times 𝐡. Remember, when we find the product of two matrices, the elements in row 𝑖 and column 𝑗 of their products will be the sum of the products of the entries in row 𝑖 of matrix 𝐴 and column 𝑗 of matrix 𝐡. So, for example, the entry in row one, column one of matrix 𝐴 times 𝐡 is π‘Ž one one times 𝑏 one one plus π‘Ž one two multiplied by 𝑏 two one. We can apply the same process to find the matrix 𝐴𝐡. We need to find the determinant of this matrix. To do this, we need to take the difference in the products of the diagonals. Doing this, distributing over our parentheses and then simplifying gives us the following expression. And this is a very complicated-looking expression.

However, we can notice something interesting. This expression is actually equal to the determinant of matrix 𝐴 multiplied by the determinant of matrix 𝐡. We can notice this by taking factors out of our expression or by using the FOIL method to distribute over these parentheses. We’ve shown for any two-by-two matrices 𝐴 and 𝐡, the determinant of 𝐴 times 𝐡 is equal to the determinant of 𝐴 multiplied by the determinant of 𝐡. And in fact, we can prove that this result holds in general. We have the third property of determinants of matrices. If 𝐴 and 𝐡 are any square matrices of the same order, then the determinant of 𝐴 times 𝐡 is equal to the determinant of 𝐴 multiplied by the determinant of 𝐡.

There’s one more property we want to prove about the determinant of matrices. This time, it’s about the determinant of a specific type of matrix. This time, we want to consider the determinant of any triangular matrix, for example, the two-by-two upper triangular matrix π‘Ž one one, π‘Ž one two, zero, π‘Ž two two. We can calculate the determinant of this matrix; it’s the difference in the products of the diagonal. This gives us π‘Ž one one times π‘Ž two two minus zero times π‘Ž one two. And the second term has a factor of zero, so it’s equal to zero. We can notice something interesting. The determinant of this matrix is the product of the entries on its main diagonal. We can show the same is true for a lower triangular matrix of order two by two. Remember, a lower triangular matrix is one where every entry above the main diagonal is equal to zero. The determinant of the two-by-two matrix 𝑏 one one, zero, 𝑏 two one, 𝑏 two two is equal to 𝑏 one one multiplied by 𝑏 two two, the product of the entries on its main diagonal.

Therefore, we’ve proven the determinant of any two-by-two triangular matrix is the product of the entries on its main diagonal. In fact, we can prove this result for upper or lower triangular matrices of order three by three, just by using the definition of a determinant. And, in turn, we can prove this for any higher-order square matrices. This gives us our fourth property of determinants. The determinant of any square triangular matrix, upper or lower, is the product of the elements on its main diagonal. Let’s now see an example of how we can use some of these properties to help us answer questions involving the determinants of different matrices.

If 𝐴 is a square matrix of order two by two and the determinant of two 𝐴 is equal to 12, then the determinant of three times 𝐴 transpose is equal to what. Is it option (A) 18, option (B) 24, option (C) 27? Or is it option (D) 36?

In this question, we’re given some information about the determinant of a two-by-two matrix 𝐴. We’re told the determinant of two 𝐴 is equal to 12, and we need to use this information to determine the determinant of three times the transpose of 𝐴. Since we’re told that 𝐴 is a two-by-two matrix, we might be tempted to start by defining 𝐴 to be a matrix of four unknowns. We could then substitute our expression for 𝐴 into our equation to find an expression for the determinant of 𝐴 and then try to use this to find an expression for the determinant of three times the transpose of 𝐴. And this would work; however, it would be very complicated. Instead, we need to notice that our equations involve determinants of matrices.

So instead, we’ll start by simplifying by using the properties of determinants. We’ll start by simplifying the expression the determinant of two times 𝐴. And to do this, we’ll start by recalling the following property. For any square matrix 𝐡 of order 𝑛 by 𝑛 and any scalar value π‘˜, the determinant of π‘˜ times 𝐡 is equal to π‘˜ to the 𝑛th power multiplied by the determinant of 𝐡. In our case, 𝐴 is a matrix of order two by two. So, our value of 𝑛 is two, giving us the determinant of two 𝐴 is equal to two squared multiplied by the determinant of 𝐴, which is, of course, four times the determinant of 𝐴. We can then substitute this expression for the determinant of two 𝐴 into the equation we’re given in the question. This then gives us that four times the determinant of 𝐴 is equal to 12. And we can solve for the determinant of 𝐴. We divide both sides of the equation by four. This gives us the determinant of 𝐴 is equal to three.

However, we’re not asked to find the determinant of 𝐴; we’re asked to find the determinant of three times the transpose of 𝐴. To do this, let’s try simplifying this expression by using the properties of determinants. First, we recall when we take the transpose of a matrix, we switch the rows with the columns. So, the transpose of matrix 𝐴 is also a matrix of order two by two. This means we can once again apply the same property. 𝐴 transpose is a two-by-two matrix. Therefore, the determinant of three multiplied by the transpose of 𝐴 is equal to three squared multiplied by the determinant of 𝐴 transpose. And we can simplify this to get nine multiplied by the transpose of 𝐴, but we can simplify this expression even further by using another one of our properties of determinants.

We recall for any square matrix 𝐡, the determinant of 𝐡 transpose is just equal to the determinant of 𝐡. And we know 𝐴 transpose is a square matrix, so we can replace this with the determinant of 𝐴 to get nine multiplied by the determinant of 𝐴. And we know what the determinant of matrix 𝐴 is. It’s equal to three. Therefore, we can just substitute three for the determinants of 𝐴 to get nine times three, which is equal to 27, which we can see is given as option (C).

Therefore, we’ve shown if 𝐴 is a square matrix of order two by two and the determinant of two 𝐴 is equal to 12, then the determinant of three times the transpose of 𝐴 is equal to 27.

Let’s now go for another example where we use the properties of the determinant of the product of two matrices.

If the determinant of 𝐴 times 𝐡 is equal to 18 and the determinant of 𝐴 is equal to two, find the determinant of 𝐡.

In this question, we’re given some information about two matrices, 𝐴 and 𝐡. We’re told the determinant of matrix 𝐴 multiplied by matrix 𝐡 is 18 and the determinant of matrix 𝐴 is equal to two. We need to use this information to determine the determinant of 𝐡. To answer this question, let’s start by recalling the property of determinants, which links the products of matrices with the determinant of each individual matrix.

We recall if 𝐴 and 𝐡 are square matrices of the same order, then the determinant of 𝐴 times 𝐡 is equal to the determinant of 𝐴 multiplied by the determinant of 𝐡. We might be tempted to apply this property directly to our question. However, there’s a problem; we’re not told the orders of matrices 𝐴 and 𝐡. And to apply this property, we do need the matrices 𝐴 and 𝐡 are square matrices of the same order. However, we can show that this must be the case from the information given. First, we’re told the determinant of matrix 𝐴 is equal to two. And we recall we can only find the determinant of square matrices, so 𝐴 is a square matrix. Similarly, the determinant of 𝐴𝐡 is equal to 18, so 𝐴 times 𝐡 is also a square matrix.

We can use this information to find an expression for the order of matrix 𝐡. First, since 𝐴 is a square matrix, let’s start by saying its order is of the form 𝑛 by 𝑛. Next, we don’t know the order of matrix 𝐡. Let’s say its order is π‘š by 𝑙. Then, there are two different ways of finding an expression for the order of 𝐴 times 𝐡. First, remember, whenever we multiply two matrices together, its order will be the number of rows of our first matrix by the number of columns of our second matrix. So, 𝐴𝐡 must have order 𝑛 by 𝑙. However, we’ve already shown that 𝐴𝐡 is a square matrix. So, the number of rows must be equal to the number of columns. In other words, 𝑛 is equal to 𝑙. So, we can replace 𝑙 with 𝑛.

Finally, to determine the value of π‘š, we notice we’re allowed to multiply 𝐴 on the right by matrix 𝐡. And recall for matrix multiplication to be well defined, the number of columns of our first matrix must be equal to the number of rows of our second matrix. Therefore, 𝑛 must be equal to π‘š. Therefore, we’ve shown 𝐴 is a square matrix, 𝐡 is a square matrix, and the orders of matrices 𝐴 and 𝐡 are equal. So, we can just apply our property to the question. We have the determinant of 𝐴𝐡 is equal to the determinant of 𝐴 multiplied by the determinant of 𝐡. Next, we’re told in the question the determinant of 𝐴𝐡 is 18 and the determinant of 𝐴 is two. So, 18 is equal to two times the determinant of 𝐡. Finally, we can divide both sides of the equation through by two to get the determinant of 𝐡 is nine.

Therefore, we were able to show if the determinant of 𝐴𝐡 is 18 and the determinant of 𝐴 is two, then the determinant of 𝐡 must be equal to nine.

Let’s now go through a question which asks us to find the value of the determinant of a triangular matrix.

Find the value of the determinant of the three-by-three matrix five, negative one, negative eight, zero, two, 60, zero, zero, zero.

In this question, we’re asked to evaluate the determinant of a given three-by-three matrix. We could do this by using the definition of a determinant. However, there’s actually an easier method if we can just notice the property of this matrix. We need to notice that this is an upper triangular matrix. This means every entry below the leading diagonal of this matrix is equal to zero. The leading diagonal of a matrix is the entries whose row number is equal to the column number. So, for this matrix, that’s five, two, and zero. So, this matrix is an upper triangular matrix.

And we recall the determinant of any square triangular matrix is the product of all of the entries on its leading diagonal. In our case, the leading diagonal has terms five, two, and zero. Therefore, the determinant of this matrix is five multiplied by two multiplied by zero, which we can evaluate is equal to zero.

Therefore, the determinants of the three-by-three matrix five, negative one, negative eight, zero, two, 60, zero, zero, zero is equal to zero.

Let’s now go through an example where we need to use the determinant of a diagonal matrix to find the value of a variable.

Consider the equation the determinant of the three-by-three matrix π‘₯ minus one, zero, zero, zero, π‘₯ squared plus π‘₯ plus one, zero, zero, zero, one is equal to two. Determine the value of π‘₯ to the sixth power.

In this question, we’re given an equation involving the determinant of a three-by-three matrix which has a variable π‘₯. We need to use this to determine the value of π‘₯ to the sixth power. To do this, we need to find an expression for the determinant of this matrix. We could do this by using the definition of a determinant, expanding over one of the rows or columns. However, we can also notice that every element not on the main diagonal of this matrix is equal to zero. In other words, this three-by-three matrix is a diagonal matrix. And we can use this to find the determinant of this matrix because every diagonal matrix is an upper and lower triangular matrix. For example, every entry below the main diagonal of our matrix is zero. So, this is an example of an upper triangular matrix.

Now, we just need to recall that the determinant of any square triangular matrix is the product of the elements on its main diagonal. And it’s worth noting since all diagonal matrices are both upper and lower triangular matrices, this property will also hold for any square diagonal matrix. Therefore, we can evaluate the determinant of this matrix by finding the product of its leading diagonal, π‘₯ minus one times π‘₯ squared plus π‘₯ plus one multiplied by one. We can then distribute over our parentheses. We get π‘₯ cubed plus π‘₯ squared plus π‘₯ minus π‘₯ squared minus π‘₯ minus one. And if we simplify this expression, we see it’s equal to π‘₯ cubed minus one. And remember, we’re told in the question this determinant is equal to two.

Therefore, our expression for the determinant is equal to two. π‘₯ cubed minus one is equal to two. We want to use this to find the value of π‘₯ to the sixth power. So, let’s start by adding one to both sides of the equation to give us that π‘₯ cubed is equal to three. And then, we can find an expression for π‘₯ to the sixth power by squaring both sides of our equation, giving us that π‘₯ to the sixth power is equal to nine, which is our final answer.

Therefore, we’ve shown if the determinant of the three-by-three matrix π‘₯ minus one, zero, zero, zero, π‘₯ squared plus π‘₯ plus one, zero, zero, zero, one is equal to two, then the value of π‘₯ to the sixth power is nine.

Let’s now go over the key points of this video. We found and proved four useful results involving the determinants of matrices. First, we showed for any square matrix 𝐴 of order 𝑛 by 𝑛 and scalar π‘˜, the determinant of π‘˜π΄ is equal to π‘˜ to the 𝑛th power multiplied by the determinant of 𝐴. Next, we showed for any square matrix 𝐴, the determinant of the transpose of 𝐴 is just equal to the determinant of 𝐴. Next, we showed for any square matrices 𝐴 and 𝐡 of the same order, the determinant of 𝐴 times 𝐡 is equal to the determinant of 𝐴 multiplied by the determinant of 𝐡. Finally, we showed the determinant of any square triangular matrix is the product of all of the terms on its principal diagonal. And this result holds for both upper and lower triangular matrices. And it even holds for square diagonal matrices because all diagonal matrices are both upper and lower triangular matrices.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.