# Lesson Video: Three-by-Three Determinants Mathematics • 10th Grade

In this video, we will learn how to evaluate 3 × 3 determinants using the cofactor expansion (the Laplace expansion) or the rule of Sarrus.

15:14

### Video Transcript

In this video, we will learn how to evaluate three-by-three determinants using the cofactors, Laplace expansion, or the SARAS method. Calculating the determinant of a square matrix is really useful to gather information about the matrix, such as being able to tell whether or not the matrix is invertible.

Let’s start with a quick reminder on how to calculate the determinant of two-by-two matrix. Then we’ll go on to see how we can calculate the determinant of a general square matrix with order 𝑛 by 𝑛 using what we know about finding the determinant of a two-by-two matrix. Let’s consider this two-by-two matrix. The determinant of 𝐴 is denoted with straight lines either side of 𝐴. It’s the same symbol that we use for absolute value. For a two-by-two matrix, it’s given by the formula 𝑎𝑑 minus 𝑏𝑐. Remember, we’ve got to be really careful when handling negative entries in a matrix, especially when we’re trying to find the determinant.

As an example, find the determinant of the matrix five, one, negative five, five. Let’s call this matrix 𝐵. Then we calculate the determined of 𝐵 to be five times five minus one times negative five. We calculate this to be 25 minus negative five. And this is where we must be really careful with the negatives as this is 25 add five, which is 30. But remember, the entries of the matrix are not always numbers. For instance, find the determinant of the matrix 𝐶 equals 𝑥, 𝑦, 𝑧, 𝑥. Using the same rule, this matrix has determinant 𝑥 squared minus 𝑦𝑧.

Moving on to think about three-by-three matrices, there are specific methods for finding the determinant of a three-by-three matrix. But here we’re going to develop a more general approach so that this method can be applied to larger matrices. The first thing we’re going to look at in order to be able to calculate three-by-three determinants are matrix minors. Consider a matrix 𝐴 with order 𝑚 by 𝑛, then the matrix minor, 𝐴 𝑖𝑗, is the initial matrix 𝐴 after having removed the 𝑖th row and the 𝑗th column. This means that 𝐴 𝑖𝑗 is a matrix with order 𝑚 minus one by 𝑛 minus one. Recall that the order of a matrix means the size of the matrix, so an 𝑚-by-𝑛 matrix is a matrix with 𝑚 rows and 𝑛 columns. The easiest way to understand matrix minors is through an example.

Suppose we have this matrix 𝐴. Let’s calculate the matrix minor 𝐴 two three. This means removing the second row and the third column from the initial matrix. We can see that the entries that remain are the entries negative two, three, zero, negative three. So this is the two-by-two matrix minor 𝐴 two three. Let’s also calculate the matrix minor 𝐴 three one. This means removing the third row and the first column. And we can see that we’re left with the two-by-two matrix three, negative three, three, six.

We’re going to use this concept with the formal definition of the general form of the determinant of a matrix. Consider a square matrix 𝐴 with order 𝑛 by 𝑛, then the determinant of 𝐴 is calculated in one of two ways, both of which involve the determinant of particular matrix minors of 𝐴. We can choose to calculate the determinant by using a particular row. Alternatively, we can choose one particular column. This is quite an overwhelming definition of the determinant. And it makes a lot more sense when we see it applied to an example.

Find the determinant of the matrix one, two, three, three, two, two, zero, nine, eight.

Let’s label this matrix as 𝐴. We can calculate the determinant of 𝐴 by first selecting either one row or one column. A good strategy is to choose the row or column which contains the most number of zeros. For this matrix, this would either be the third row or the first column, which both contain one zero entry. Let’s choose the first column. And here is the formula for finding the determinant of a matrix when we use a particular column 𝑗.

For this question, as we’re using the first column, 𝑗 is equal to one. And because this is a three-by-three matrix, 𝑖 runs from one to three. Here is the same formula but adapted so that 𝑖 runs from one to three. And as we’re using the first column, 𝑗 is equal to one. Now 𝑎 one one, 𝑎 two one, and 𝑎 three one are the entries in the column that we’re using. These are the entries one, three, and zero. And these are the determinants of the matrix minors.

Recall that we find the matrix minor 𝐴 one one by removing the first row and the first column of the original matrix. We’re left with the two-by-two matrix two, two, nine, eight. We then calculate the determinant of this matrix using the usual formula for finding the determinant of a two-by-two matrix. This is two times eight minus two times nine. This is 16 minus 18, which gives us negative two. We can then calculate the matrix minor 𝐴 two one by removing the second row and the first column. This gives us the matrix two, three, nine, eight. We can then calculate its determinant to be two times eight minus three times nine. And we find that this is negative 11. We do the same, calculating the matrix minor 𝐴 three one by removing the third row and the first column of the original matrix. And we find its determinant to be two times two minus three times two, which gives us negative two.

The final step is to calculate the values negative one to the power of one add one, negative one to the power of two add one, and negative one to the power of three add one. We find these to be one, negative one, and one. Remember, raising negative one to an even power gives us one and raising negative one to an odd power gives us negative one. Let’s now substitute into this formula the values that we found. Substituting in these values becomes clear why it’s useful to choose a row or column that has the most number of zeros. Because this last term is being multiplied by zero, the whole term will be zero, which just means it’s one less thing that we actually need to calculate.

Tidying things up a bit, we find that one times one times negative two gives us negative two and negative one times three times negative 11 gives us 33. Adding these together, we get the final answer of 31.

So the main things to think about are which row or column you’re going to be using and carefully calculating each little bit of the formula. And we’ve got to make sure that we’re always really careful around negatives. Let’s now see another example, although, this time, we’ll expand along a row rather than a column.

Calculate the determinant of 𝐴 when 𝐴 equals three, zero, negative one, zero, one, zero, two, two, four.

To make things easier for ourselves, we’ll identify the row or column that contains the most number of entries which are zero. For this matrix, that’s going to be the second row of 𝐴. Let’s remind ourselves of the general formula for finding the determinant of an 𝑛-by-𝑛 matrix. Here is the formula. Remember that 𝑖 represents the row number and 𝑗 represents the column number. So as we have three columns, 𝑗 runs from one to three. And we’re expanding along the second row, so 𝑖 equals two. So this is the formula that we’re going to be using.

The entries 𝑎 two one, 𝑎 two two, and 𝑎 two three are zero, one, and zero, respectively. Because two of these entries are zero, we find that both of these terms are going to be zero because they’re both being multiplied by zero. So there’s actually only one term that we need to calculate here. Let’s begin by calculating the matrix minor 𝐴 two two. This is going to be the two-by-two matrix we get by removing the second row and the second column. So we see that the matrix minor 𝐴 two two is three, negative one, two, four. But what we actually need for our formula is its determinant. So we calculate this in the usual way for finding determinants of a two-by-two matrix. This gives us three times four minus negative one times two. This gives us 12 minus negative two, which gives us 14.

The final thing we can do is calculate the value of negative one to the power of two add two. As this is negative one to the fourth power, which is an even power, this is going to give us one. So we calculate the determinant of 𝐴 to be one times one times 14. And this, of course, just gives us 14. So carefully selecting the row or column that you choose when you’re finding the determinant of a matrix can really reduce the amount of calculations necessary.

Remember that not all matrices contain numeric entries. Let’s have a look at this in an example.

Consider the determinant 𝑥, 𝑧, 𝑦, 𝑦, 𝑥, 𝑧, 𝑧, 𝑦, 𝑥. Given that 𝑥 cubed add 𝑦 cubed add 𝑧 cubed equals negative 73 and 𝑥𝑦𝑧 equals negative eight, determine the determinant’s numerical value.

Remember that we can choose to calculate this determinant by picking a row or column. We often choose the row or column with the most amount of zeros. But as we have three unknowns, 𝑥, 𝑦, and 𝑧, let’s just choose the top row. And here is the general formula to find the determinant of an 𝑛-by-𝑛 matrix, using a particular row, 𝑖. As we’re using a three-by-three matrix, 𝑗 runs from one to three. And as we’re using the top row, 𝑖 is equal to one. So here is the particular version of this general formula applied for our question. Let’s clear some space before we continue.

I’ve kept the formula that we’re using on the screen. Let’s call the matrix that we’re using 𝐴. The straight lines either side of the matrix denote the determinant. Let’s begin by finding the different components in the formula that we’re using. The lowercase 𝑎 one one, 𝑎 one two, and 𝑎 one three are the entries in the row that we’re using. So 𝑎 one one is 𝑥, 𝑎 one two is 𝑧, and 𝑎 one three is 𝑦.

The uppercase, 𝐴 one one, 𝐴 one two, and 𝐴 one three denote the matrix minors. Remember that we get these from removing a particular row and a particular column from the original matrix. For example, we get 𝐴 one one by removing the first row and the first column, like so. And we’re left with the two-by-two matrix 𝑥, 𝑧, 𝑦, 𝑥. We then do the same for 𝐴 one two. We get the matrix minor by removing the first row and the second column. This leaves us with the two-by-two matrix 𝑦, 𝑧, 𝑧, 𝑥. And finally, we get the matrix minor 𝐴 one three by removing the first row and the third column. This leaves us with the two-by-two matrix 𝑦, 𝑥, 𝑧, 𝑦.

But what we actually need for our formula is the determinant of each of these matrices. Let’s recall a method to find the determinant of a two-by-two matrix. So the determinant of the matrix minor 𝐴 one one is 𝑥 times 𝑥 minus 𝑧 times 𝑦. We can equivalently write this as 𝑥 squared minus 𝑦𝑧. In the same way, we find the determinant of the matrix minor 𝐴 one two to be 𝑦 times 𝑥 minus 𝑧 times 𝑧, or equivalently 𝑥𝑦 minus 𝑧 squared. And we find the determinant of the matrix minor 𝐴 one three to be 𝑦 times 𝑦 minus 𝑥 times 𝑧, which is equivalently 𝑦 squared minus 𝑥𝑧.

The final thing we need to calculate for our formula are the values of negative one to the power of one add one, negative one to the power of one add two, and negative one to the power of one add three. Remember that raising negative one to an even power gives us one and raising negative one to an odd power gives us negative one. So negative one to the power of one add one is negative one squared, which is one. Negative one to the power of one add two is negative one cubed, which is negative one. And finally, negative one to the power of one add three is negative one to the fourth power, which gives us one.

So let’s not write out the determinants of this matrix with the components that we found. From here, let’s just try and simplify a little bit. Let’s begin by multiplying together the brackets with algebraic terms. For the first one, we have 𝑥 multiplied by 𝑥 squared minus 𝑦𝑧. So this term becomes one times 𝑥 cubed minus 𝑥𝑦𝑧. We then do the same with the next term. This gives us negative one times 𝑥𝑦𝑧 minus 𝑧 cubed. And repeating the same with the last term, we end up with one times 𝑦 cubed minus 𝑥𝑦𝑧.

So now, we’ve multiplied together the brackets containing algebraic terms. Let’s multiply each term by the value at the front. The first term is one times 𝑥 cubed minus 𝑥𝑦𝑧, which is of course going to give us 𝑥 cubed minus 𝑥𝑦𝑧. The next term is negative one times 𝑥𝑦𝑧 minus 𝑧 cubed. So we can multiply the second bracket through by negative one. But we’ve got to be careful because the 𝑧 cubed already has a minus at the front of it, so that’s going to become a positive. So we’re gonna end up with minus 𝑥𝑦𝑧 add 𝑧 cubed. Finally, the last term is just one times 𝑦 cubed minus 𝑥𝑦𝑧. So this gives us 𝑦 cubed minus 𝑥𝑦𝑧.

We can now simplify this a little bit. We can gather together the 𝑥𝑦𝑧 terms. Bringing these together gives us minus three 𝑥𝑦𝑧. From here, we notice that we have 𝑥 cubed add 𝑧 cubed add 𝑦 cubed. We’re told in the question that 𝑥 cubed add 𝑦 cubed add 𝑧 cubed gives us negative 73. Additionally, we have negative three 𝑥𝑦𝑧, and we’re told in the question that 𝑥𝑦𝑧 is equal to negative eight. So substituting in those values gives us negative 73 minus three times negative eight, which is negative 73 add 24. And this gives us negative 49. So sometimes, we encounter matrices which don’t have numerical values, but we can still find the determinant in exactly the same way.

Let’s now summarize the main points from this lesson. For a matrix 𝐴, a matrix minor 𝐴 𝑖𝑗 is the initial matrix 𝐴 after having removed the 𝑖th row and the 𝑗th column. And here is the general formula for the determinant of an 𝑛-by-𝑛 matrix 𝐴. We can either choose a particular row or a particular column. But it often helps to choose a row or column that contains the most number of zeros.