Lesson Explainer: Three-by-Three Determinants Mathematics • 10th Grade

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

Let us first recall the definition of a 2Γ—2 determinant.

Definition: Determinant of a 2 Γ— 2 Matrix

Let 𝐴 be a 2Γ—2 matrix given by 𝐴=ο€Όπ‘Žπ‘π‘π‘‘οˆ.

Then, the determinant of 𝐴 (denoted by (det𝐴) or |𝐴|) is given by det(𝐴)=|||π‘Žπ‘π‘π‘‘|||=π‘Žπ‘‘βˆ’π‘π‘.

Compared to the 2Γ—2 case, the calculation of determinants of 3Γ—3 matrices is much more complicated. In particular, it is not exactly clear how one would naturally extend the above definition to a bigger matrix. As it turns out, the two cases are actually intrinsically linked to each other. As we will see in the first of the two methods we will explore, a 3Γ—3 determinant can essentially be reduced to an equation involving multiple 2Γ—2 determinants, which we already know how to calculate. In order to properly explain how this works, we first have to introduce a new concept known as minors.

Definition: Minors

Let 𝐴=ο€Ήπ‘Žο…οƒο… be a matrix of order π‘šΓ—π‘š. Then, the minor of element π‘Žοƒο… (denoted by 𝐴) is the determinant of the (π‘šβˆ’1)Γ—(π‘šβˆ’1) matrix obtained after removing row 𝑖 and column 𝑗 from 𝐴.

We note that although this definition applies to square matrices of any size, we will only be considering matrices of order 3Γ—3 in this explainer, meaning that the minors we will deal with will always be 2Γ—2 determinants.

It is easiest to illustrate this concept with an example. Suppose that we have the 3Γ—3 matrix 𝐴=ο€βˆ’23βˆ’3βˆ’4360βˆ’39.

Suppose we wanted to find the minor of element π‘ŽοŠ§οŠ¨, that is, 𝐴. Calculating 𝐴 means removing row 1 and column 2 from 𝐴 and finding the determinant of the resulting matrix. We can start this process by identifying entry π‘ŽοŠ§οŠ¨ and removing the entries that are horizontally and vertically aligned with it. Let us highlight this as follows:

Then, by taking the determinant of the 2Γ—2 matrix, we get 𝐴=||βˆ’4609||=(βˆ’4)Γ—9βˆ’6Γ—0=βˆ’36βˆ’0=βˆ’36.

This process can be repeated for any element of 𝐴. For instance, let us consider the minor of element π‘ŽοŠ©οŠ§. As shown above, we can begin by removing π‘ŽοŠ©οŠ§ and the row and column that it belongs to:

Then, we can take the determinant of the resulting matrix to get 𝐴=||3βˆ’336||=3Γ—6βˆ’(βˆ’3)Γ—3=18+9=27.

Before we proceed further with the calculation of 3Γ—3 determinants, let us consider an example where we can practice finding the minor of a matrix.

Example 1: Identifying the Minor of a Matrix

Consider 𝐴=ο€βˆ’65βˆ’326βˆ’899βˆ’7.

Write the determinant whose value is equal to the minor of the element π‘ŽοŠ¨οŠ©.

Answer

Recall that if 𝐴 is a matrix of order π‘šΓ—π‘›, then the minor of element π‘Žοƒο… is the determinant of the (π‘šβˆ’1)Γ—(π‘›βˆ’1) matrix obtained after removing row 𝑖 and column 𝑗 from 𝐴.

Here, since 𝐴 is a 3Γ—3 matrix, the minor of element π‘ŽοŠ¨οŠ© (in other words 𝐴) is the determinant of the 2Γ—2 matrix obtained by removing row 2 and column 3 from 𝐴. Let us first calculate this reduced matrix by highlighting entry π‘ŽοŠ¨οŠ© and the corresponding rows and columns that it is in and then removing them from the matrix:

Now that we have the 2Γ—2 matrix, we can therefore find the minor of π‘ŽοŠ¨οŠ© by taking its determinant. Note that we do not need to simplify this since we have only been asked to write the determinant.

Thus, the minor of π‘ŽοŠ¨οŠ© is ||βˆ’6599||.

The next step in calculating 3Γ—3 determinants using cofactor expansion is to find the cofactors of the matrix. Let us first define what cofactors are.

Definition: Cofactors

Let 𝐴=ο€Ήπ‘Žο…οƒο… be a matrix of order π‘šΓ—π‘š. Then, the cofactor of element π‘Žοƒο… (denoted by 𝐢) is 𝐢=(βˆ’1)𝐴,οƒο…οƒοŠ°ο…οƒο… where 𝐴 is the minor of element π‘Žοƒο….

As we can see from the above definition, cofactors can be obtained from minors by multiplying by a factor of (βˆ’1)οƒοŠ°ο…, or in other words by 1 or βˆ’1, depending on the index of π‘Žοƒο…. To demonstrate this, let us recall once more the matrix we previously used to demonstrate how to find minors: 𝐴=ο€βˆ’23βˆ’3βˆ’4360βˆ’39.

Recall that we found 𝐴=βˆ’36 and 𝐴=27. Using the above definition, we can calculate cofactors 𝐢 and 𝐢 using these minors. We have 𝐢=(βˆ’1)𝐴=(βˆ’1)Γ—(βˆ’36)=(βˆ’1)Γ—(βˆ’36)=36.

So, in the case of 𝐢, since (βˆ’1)=βˆ’1, the sign is reversed. For 𝐢, we have 𝐢=(βˆ’1)𝐴=(βˆ’1)Γ—27=1Γ—27=27.οŠͺ

Here, since (βˆ’1)=1, we have 𝐢=𝐴.

As we can see, the only complication that we might face while calculating the cofactors once we have the minors is to make sure we get the correct sign. One way to do this is to verify whether 𝑖+𝑗 is odd or even. This is because ifisoddifiseven𝑖+𝑗,(βˆ’1)=βˆ’1,𝑖+𝑗,(βˆ’1)=1.οƒοŠ°ο…οƒοŠ°ο…

Aside from calculating (βˆ’1)οƒοŠ°ο… directly, we can also refer to the following matrix, which shows how the signs of the cofactors alternate in a chessboard-like pattern: +βˆ’+βˆ’+βˆ’+βˆ’+.

For instance, the entry in position (3,2) of this matrix shows us that the cofactor of entry π‘ŽοŠ©οŠ¨ would have a factor of βˆ’1 compared to its minor.

Having defined minors and cofactors, we are now in a position to define a formula for calculating the determinant of a 3Γ—3 matrix.

Definition: Determinant of a 3 Γ— 3 Matrix (Cofactor Expansion)

Let 𝐴=ο€Ήπ‘Žο…οƒο… be a 3Γ—3 matrix. Then, for any fixed 𝑖=1,2, or 3, the determinant of 𝐴 is det(𝐴)=π‘ŽπΆ+π‘ŽπΆ+π‘ŽπΆ,οƒοŠ§οƒοŠ§οƒοŠ¨οƒοŠ¨οƒοŠ©οƒοŠ© where each 𝐢 is the cofactor of entry π‘Žοƒο…. This is known as the cofactor expansion (or Laplace expansion) along row 𝑖. Alternatively, for any fixed 𝑗=1,2, or 3, we have det(𝐴)=π‘ŽπΆ+π‘ŽπΆ+π‘ŽπΆ.οŠ§ο…οŠ§ο…οŠ¨ο…οŠ¨ο…οŠ©ο…οŠ©ο… This is the cofactor expansion along column 𝑗.

We note that it can be useful to write the above formulas in terms of minors rather than cofactors. For instance, the cofactor expansion along row 1 is equal to det(𝐴)=π‘ŽπΆ+π‘ŽπΆ+π‘ŽπΆ=π‘Žπ΄βˆ’π‘Žπ΄+π‘Žπ΄, where we have substituted 𝐢=𝐴, 𝐢=βˆ’π΄οŠ§οŠ¨οŠ§οŠ¨, and 𝐢=𝐴. We can also write this explicitly as det(𝐴)=π‘Ž|||π‘Žπ‘Žπ‘Žπ‘Ž|||βˆ’π‘Ž|||π‘Žπ‘Žπ‘Žπ‘Ž|||+π‘Ž|||π‘Žπ‘Žπ‘Žπ‘Ž|||.

Rather than going so far as to calculate the cofactors in each situation, it is usually quicker to just find the minors and alternate the signs accordingly.

The best way to demonstrate this process is with an example.

Example 2: Evaluating a 3 Γ— 3 Determinant Using Cofactor Expansion

Find the value of ||||226βˆ’31βˆ’2βˆ’5βˆ’1βˆ’4||||.

Answer

Let the given matrix be 𝐴=ο€Ήπ‘Žο…οƒο…. To calculate the determinant of a 3Γ—3 matrix, we can use the method of cofactor expansion by choosing a specific row or column of the matrix, calculating the minors for each entry of that row or column, and alternating the signs according to their corresponding cofactors.

In some situations, it can be useful to choose a specific row or column such that, upon finding the cofactor expansion of that row, the calculations will be simplified. However, since we have no way of identifying such cases yet, we will simply choose the first row of the matrix for simplicity. Now, we can find a formula for the determinant in terms of the minors by multiplying them by 1 or βˆ’1, depending on the corresponding signs in the following matrix: +βˆ’+βˆ’+βˆ’+βˆ’+.

Since we are concerned with the first row, we take note that the signs are +, βˆ’, and +. We can then use these to form the following formula for the determinant: det(𝐴)=π‘ŽπΆ+π‘ŽπΆ+π‘ŽπΆ=π‘Žπ΄βˆ’π‘Žπ΄+π‘Žπ΄, where we have used the fact that 𝐢=𝐴, 𝐢=βˆ’π΄οŠ§οŠ¨οŠ§οŠ¨, and 𝐢=𝐴.

To evaluate this, let us begin by finding 𝐴, 𝐴, and 𝐴. We can find each minor by removing the row and column containing that entry and taking the determinant of the result. As we are considering the first row, let us highlight it below:

We demonstrate this process for each entry as follows:

Using these matrices, we can now explicitly write a formula for the determinant: det(𝐴)=π‘Žπ΄βˆ’π‘Žπ΄+π‘Žπ΄=2||1βˆ’2βˆ’1βˆ’4||βˆ’2||βˆ’3βˆ’2βˆ’5βˆ’4||+6||βˆ’31βˆ’5βˆ’1||.

Let us find the determinants of each of these matrices in turn. The first one is 𝐴=||1βˆ’2βˆ’1βˆ’4||=1Γ—(βˆ’4)βˆ’(βˆ’1)Γ—(βˆ’2)=βˆ’4βˆ’2=βˆ’6.

The second one is 𝐴=||βˆ’3βˆ’2βˆ’5βˆ’4||=(βˆ’3)Γ—(βˆ’4)βˆ’(βˆ’5)Γ—(βˆ’2)=12βˆ’10=2.

Lastly, the third one is 𝐴=||βˆ’31βˆ’5βˆ’1||=(βˆ’3)Γ—(βˆ’1)βˆ’(βˆ’5)Γ—(1)=3+5=8.

Altogether, we have det(𝐴)=2||1βˆ’2βˆ’1βˆ’4||βˆ’2||βˆ’3βˆ’2βˆ’5βˆ’4||+6||βˆ’31βˆ’5βˆ’1||=2Γ—(βˆ’6)βˆ’2Γ—2+6Γ—8=βˆ’12βˆ’4+48=32.

Having seen the process for calculating a 3Γ—3 determinant using cofactor expansion, it is important to realize that this method works because the cofactors of a 3Γ—3 matrix are 2Γ—2 determinants, which we know how to solve. We also note that although we will not attempt this in this explainer, cofactor expansion can be applied to matrices of higher order than 3Γ—3. For instance, the cofactors of a 4Γ—4 matrix are 3Γ—3 determinants, which we can then find using cofactor expansion in a recursive manner. Similarly, we can reduce 5Γ—5 determinants to 4Γ—4 determinants, which we can find using the above method, and so on.

One of the main problems with cofactor expansion is that it is quite complicated, although it does become easier to use with practice. An alternative way to obtain the same result is to use the Sarrus rule, which we will now explain.

How To: Finding 3 Γ— 3 Determinants Using the Sarrus Rule

Let 𝐴 be a 3Γ—3 matrix given by 𝐴=οπ‘Žπ‘Žπ‘Žπ‘Žπ‘Žπ‘Žπ‘Žπ‘Žπ‘Žο.

To find det(𝐴), we can do the following.

  1. First, write the first and second columns of 𝐴 again on the right-hand side of 𝐴.
  2. Next, highlight the diagonal entries as shown. For each diagonal, take the products of the entries and add them up:
  3. Now, highlight the diagonals in the opposite direction as shown. For each diagonal, take the products of the entries and subtract them all:
  4. Lastly, combine the expressions from steps 2 and 3 to obtain in the following expression for the determinant det(𝐴)=π‘Žπ‘Žπ‘Ž+π‘Žπ‘Žπ‘Ž+π‘Žπ‘Žπ‘Žβˆ’π‘Žπ‘Žπ‘Žβˆ’π‘Žπ‘Žπ‘Žβˆ’π‘Žπ‘Žπ‘Ž.

The Sarrus rule provides us with a way to calculate the determinant without having to find the minors or cofactors first. We should note, however, that the number of calculations we have to do is still roughly the same, except that it may be easier to remember this method. To demonstrate this, let us briefly show how cofactor expansion along row 1 applied to the same general matrix works. We have det(𝐴)=π‘Žπ΄βˆ’π‘Žπ΄+π‘Žπ΄=π‘Ž|||π‘Žπ‘Žπ‘Žπ‘Ž|||βˆ’π‘Ž|||π‘Žπ‘Žπ‘Žπ‘Ž|||+π‘Ž|||π‘Žπ‘Žπ‘Žπ‘Ž|||=π‘Ž(π‘Žπ‘Žβˆ’π‘Žπ‘Ž)βˆ’π‘Ž(π‘Žπ‘Žβˆ’π‘Žπ‘Ž)+π‘Ž(π‘Žπ‘Žβˆ’π‘Žπ‘Ž)=π‘Žπ‘Žπ‘Žβˆ’π‘Žπ‘Žπ‘Žβˆ’π‘Žπ‘Žπ‘Ž+π‘Žπ‘Žπ‘Ž+π‘Žπ‘Žπ‘Žβˆ’π‘Žπ‘Žπ‘Ž, which is the same result after a slight rearrangement to the terms.

Let us consider an example where we calculate the determinant using the Sarrus rule.

Example 3: Evaluating a 3 Γ— 3 Determinant Using the Sarrus Rule

Evaluate ||||1βˆ’9βˆ’6βˆ’8412βˆ’19||||.

Answer

In this question, we are being asked to evaluate the determinant of a 3Γ—3 matrix. One way to do this is to use the Sarrus rule. To do this, we begin by rewriting the matrix with the first two columns added to the right.

Now, we highlight the three diagonals going from left to right, take their products, and sum them up:

Next, we highlight the three diagonals going from right to left, take their products, and subtract them:

Therefore, it can be seen that the determinant is equal to the sum of these two parts: βˆ’30+βˆ’599=βˆ’629.

We can also use cofactor expansion to verify whether this value for the determinant is correct. Recall that if we expand along the first row of the matrix, the cofactor expansion formula gives us det(𝐴)=π‘Žπ΄βˆ’π‘Žπ΄+π‘Žπ΄, where 𝐴, 𝐴, and 𝐴 are the minors of π‘ŽοŠ§οŠ§, π‘ŽοŠ§οŠ¨, and π‘ŽοŠ§οŠ© respectively. We can calculate the minor in each case by removing the corresponding row and column for that entry and calculating the determinant of the resulting matrix. We can demonstrate this process as follows:

Thus, using these explicit forms of the minors, the determinant is equal to det(𝐴)=1||41βˆ’19||βˆ’(βˆ’9)||βˆ’8129||+(βˆ’6)||βˆ’842βˆ’1||=4Γ—9βˆ’(βˆ’1)Γ—1+9((βˆ’8)Γ—9βˆ’1Γ—2)βˆ’6((βˆ’8)Γ—(βˆ’1)βˆ’4Γ—2)=37+9(βˆ’74)βˆ’6(0)=βˆ’629.

Having seen two methods for calculating the determinant, we note that both of them will help us arrive at the same answer, so we should use whichever we are most comfortable with. For the remaining two examples in this explainer, let us consider algebraic problems involving determinants, where we alternate between the two methods.

Example 4: Solving an Equation Involving a 3 Γ— 3 Determinant

Find the solution set of ||||π‘₯00βˆ’1βˆ’5π‘₯021π‘₯||||=βˆ’80π‘₯.

Answer

In this question, we have been asked to solve an equation for π‘₯ involving a determinant. The most natural thing to do is to first calculate the determinant, so let us do this using cofactor expansion.

To use cofactor expansion, we first have to choose a row or column to expand along. Let the given matrix be 𝐴=ο€Ήπ‘Žο…οƒο…. Recall the formula we have for expanding along row 𝑖: det(𝐴)=π‘ŽπΆ+π‘ŽπΆ+π‘ŽπΆ,οƒοŠ§οƒοŠ§οƒοŠ¨οƒοŠ¨οƒοŠ©οƒοŠ© where each 𝐢 is the cofactor of π‘Žοƒο…. Note that as each term in the equation has its corresponding row entry as a factor, any row where multiple entries are zero is a good choice for cofactor expansion. To see this, consider the first row of the given matrix:

Here, we have π‘Ž=π‘₯, π‘Ž=0, and π‘Ž=0. Since two of the entries are zero, the formula for row expansion when 𝑖=1 will simplify to det(𝐴)=π‘₯𝐢+0𝐢+0𝐢=π‘₯𝐢.

Recall also that 𝐢=(βˆ’1)𝐴=𝐴, so we can rewrite this in terms of the minor of π‘ŽοŠ§οŠ§ as det(𝐴)=π‘₯𝐴.

Therefore, we only need to calculate 𝐴 if we expand along this row. We can find it by removing row 1 and column 1 from the matrix and taking the determinant of the result:

Thus, the determinant is det(𝐴)=π‘₯𝐴=π‘₯||βˆ’5π‘₯01π‘₯||=π‘₯((βˆ’5π‘₯)Γ—π‘₯βˆ’1Γ—0)=π‘₯ο€Ήβˆ’5π‘₯βˆ’0=βˆ’5π‘₯.

Recall that we have been given that det(𝐴)=βˆ’80π‘₯, so we have βˆ’80π‘₯=βˆ’5π‘₯.

To find the solution set of this equation, we will take both terms to one side and factor them: 5π‘₯βˆ’80π‘₯=05π‘₯ο€Ήπ‘₯βˆ’16=05π‘₯(π‘₯βˆ’4)(π‘₯+4)=0.

We can see that this equation is satisfied if π‘₯=0, π‘₯=4, or π‘₯=βˆ’4. Thus, the solution set is {0,4,βˆ’4}.

Note that, in the above example, we found that the calculations were simplified to a large extent using the fact that the matrix had a row that had only one nonzero entry. While this is outside the scope of this explainer, we do note that the calculations of determinants can be simplified if we take advantage of such cases.

For our final example, let us approach a determinant problem with unknown variables by using the Sarrus rule.

Example 5: Evaluating a Determinant Involving Variables Using Given Equations

Consider the determinant ||||π‘₯𝑧𝑦𝑦π‘₯𝑧𝑧𝑦π‘₯||||.

Given that π‘₯+𝑦+𝑧=βˆ’73 and π‘₯𝑦𝑧=βˆ’8, determine the determinant's numerical value.

Answer

As we have been given a determinant with variable values and the given equations do not immediately appear to simplify the entries of the matrix, let us proceed by calculating the determinant using the Sarrus rule and see whether we get an expression that can be simplified.

To use the Sarrus rule to find a determinant, we begin by writing the first and second columns on the right-hand side of the matrix.

Now, we take the products of the diagonal entries (as highlighted below) and add them together:

Since we have been given that π‘₯+𝑦+𝑧=βˆ’73, this is already looking promising. Next, we take the products of the diagonal entries in the opposite direction (as highlighted below) and subtract them all:

As we have an expression for π‘₯𝑦𝑧, this will also simplify. First, adding these two components together gives us an expression for the determinant: π‘₯+𝑦+π‘§βˆ’3π‘₯𝑦𝑧.

We can now use the given equations, π‘₯+𝑦+𝑧=βˆ’73 and π‘₯𝑦𝑧=βˆ’8, to simplify the equation. Combining these together gives us π‘₯+𝑦+π‘§βˆ’3π‘₯𝑦𝑧=βˆ’73βˆ’3(βˆ’8)=βˆ’73+24=βˆ’49.

Thus, the determinant is βˆ’49.

We can also verify this using the cofactor method. Let us recall that expanding along the first row of the matrix gives us the formula det(𝐴)=π‘Žπ΄βˆ’π‘Žπ΄+π‘Žπ΄=π‘Ž|||π‘Žπ‘Žπ‘Žπ‘Ž|||βˆ’π‘Ž|||π‘Žπ‘Žπ‘Žπ‘Ž|||+π‘Ž|||π‘Žπ‘Žπ‘Žπ‘Ž|||.

Substituting in the values of the given matrix into this formula and calculating the determinants, we get det(𝐴)=π‘₯||π‘₯𝑧𝑦π‘₯||βˆ’π‘§||𝑦𝑧𝑧π‘₯||+𝑦||𝑦π‘₯𝑧𝑦||=π‘₯ο€Ήπ‘₯βˆ’π‘¦π‘§ο…βˆ’π‘§ο€Ήπ‘₯π‘¦βˆ’π‘§ο…+π‘¦ο€Ήπ‘¦βˆ’π‘₯𝑧=π‘₯βˆ’π‘₯π‘¦π‘§βˆ’π‘§π‘₯𝑦+𝑧+π‘¦βˆ’π‘¦π‘₯𝑧=π‘₯+𝑦+π‘§βˆ’3π‘₯𝑦𝑧.

This is the same expression as the one we found before. Thus, using either method, the final value for the determinant is βˆ’49.

Let us summarize the key points that we have learned in this explainer.

Key Points

  • Let 𝐴=ο€Ήπ‘Žο…οƒο… be a matrix of order π‘šΓ—π‘š. Then, the minor of element π‘Žοƒο… (denoted by 𝐴) is the determinant of the (π‘šβˆ’1)Γ—(π‘šβˆ’1) matrix obtained after removing row 𝑖 and column 𝑗 from 𝐴.
  • Using the above definition of minors, the cofactor of element π‘Žοƒο… (denoted by 𝐢) is equal to 𝐢=(βˆ’1)𝐴.οƒο…οƒοŠ°ο…οƒο…
  • From the cofactors, for any fixed 𝑖=1,2, or 3, the determinant of a 3Γ—3 matrix 𝐴 is equal to det(𝐴)=π‘ŽπΆ+π‘ŽπΆ+π‘ŽπΆ.οƒοŠ§οƒοŠ§οƒοŠ¨οƒοŠ¨οƒοŠ©οƒοŠ© This is known as the cofactor expansion (or Laplace expansion) along row 𝑖. Alternatively, for any fixed 𝑗=1,2, or 3, we have det(𝐴)=π‘ŽπΆ+π‘ŽπΆ+π‘ŽπΆ.οŠ§ο…οŠ§ο…οŠ¨ο…οŠ¨ο…οŠ©ο…οŠ©ο… This is the cofactor expansion along column 𝑗.
  • If we expand along the first row, we can write this explicitly as det(𝐴)=π‘Ž|||π‘Žπ‘Žπ‘Žπ‘Ž|||βˆ’π‘Ž|||π‘Žπ‘Žπ‘Žπ‘Ž|||+π‘Ž|||π‘Žπ‘Žπ‘Žπ‘Ž|||.
  • Alternatively, we can calculate the determinant using the Sarrus rule.
    • First, write the first and second columns of 𝐴 again on the right-hand side of 𝐴.
    • Next, highlight the diagonal entries as shown. For each diagonal, take the products of the entries and add them up:
    • Now, highlight the diagonals in the opposite direction as shown. For each diagonal, take the products of the entries and subtract them all:
    • Finally, combine the expressions from steps 2 and 3 to obtain det(𝐴)=π‘Žπ‘Žπ‘Ž+π‘Žπ‘Žπ‘Ž+π‘Žπ‘Žπ‘Žβˆ’π‘Žπ‘Žπ‘Žβˆ’π‘Žπ‘Žπ‘Žβˆ’π‘Žπ‘Žπ‘Ž.

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