10.05 Cofactors and determinants

Cofactors and determinants of 3x3 matrices

When we looked at matrix equations, and solving systems of equations using matrices we learned that the inverse of a $2\times2$2×2 matrix, , was

Solving larger systems of matrices would require knowledge of the inverse of $3\times3$3×3 matrices, or $4\times4$4×4 matrices or even much larger ones.  For example in the study of bioinformatics, matrices as large as $100\times100$100×100 or larger can often be found. 

Fortunately, at this stage we only need to worry about $3\times3$3×3 (although we do look at some $4\times4$4×4's soon as well. ) 

There is no simple rule or formula we can use to write the inverse of a $3\times3$3×3 matrix straight down, but there is a process for us to be able to determine it by hand. 

First let's look at some important language that we need to be familiar with.

Minor

In a square matrix, every element has its own minor.  This is the value of the determinant of the matrix that results from crossing out the row and column of the element we are looking at.  

This interactive will help you identify the minors for elements in a $3\times3$3×3 matrix. 

 

Place Signs

Each element in a square matrix also has a place sign.  Place signs alternate from $+$+ to $-$− as you move across a row and down columns.  The place sign of element $E_{11}$E11​ is always positive. 

So for a $3\times3$3×3, the place signs are 

or for a $4\times4$4×4 the place signs would be 

or even for a $2\times2$2×2 the place signs would be 

 

Cofactors

Combining both the minors and place signs we can determine cofactors.  A cofactor is the result from multiplying the elements place sign and minor together.  

That is, the cofactor is the determinant of the remaining terms from a matrix when the rows and columns particular to that element are removed, and this determinant value is multiplied by the sign given by the place sign position of that element. 

This applet will help you determine the cofactor of an element.  

Practice question

Question 1

Determinants

We already know how to find the determinant of a $2\times2$2×2 matrix.

We learned that the value of a determinant (specifically if it is non-zero) tells us that a determinant exists. The determinant value is also used in the construction of the determinant matrix. 

Calculating the determinant of a $3\times3$3×3 requires the calculation of specific cofactors. 

Two step process

• We start by picking any row or column of the matrix and identifying the $3$3 elements in that row/column.
• We then take the sum of the product of each of these elements with their corresponding cofactor. 

 

Let's have a look at an example together. 

Worked examples

Question 2

It actually doesn't matter which row or column we pick.  But because we are going to be taking the products of each of the elements if we choose the row or column with the zero, then that product will be easy to calculate!

Looking at row $1$1.

Elements of row $1$1 are

$m_{11},m_{12}$m11​,m12​ and ,$m_{13}$m13​ which are $1,1$1,1 and $0$0.

We also need the cofactors of these elements.  

Multiplying the element by its cofactor and summing them together gives

$1\times8+1\times-1+0\times-5=8-1=7$1×8+1×−1+0×−5=8−1=7

So the determinant of this matrix is $7$7.  

Question 3

Find the determinant of the following matrix

So the determinant is found by taking each of the elements and multiplying it by the cofactor.  

$2\times-4+1\times-8+2\times7=-8-8+14=-2$2×−4+1×−8+2×7=−8−8+14=−2

Cofactors and determinants of 4x4 matrices

Recall that every square matrix $A_n$An​ possesses a determinant - a real number calculated from the elements of the matrix itself and denoted as either $det$det $A$A or $|A|$|A|. The determinant is called so because its value determines whether $A_n$An​ possesses an inverse. If $|A|=0$|A|=0, the matrix is said to be singular and does not have an inverse. If $|A|\ne0$|A|≠0 it is said to be invertible and consequently has an inverse.

As the order of the matrix increases, so does the difficulty in finding $|A|$|A|. 

 

Higher order determinants are found using a four step process. We will develop the technique using a specific order $3$3 example.

How to find the determinant of the matrix $A=$A= 

1. Choose a row to distribute along

Often row $1$1 is chosen but rows $2$2 and $3$3 can also be chosen if preferred. For this exercise, we choose row $1$1.

 

2. Find the three minors 

A minor $M_{i,j}$Mi,j​ is the determinant of the matrix formed by crossing out the elements horizontally along row $i$i and vertically up and down column $j$j. 

So to find the minor $M_{1,1}$M1,1​ we first cross out all the elements of row $1$1 and column $1$1, and then find the determinant of the matrix that is left. Hence:

For the minor $M_{1,2}$M1,2​, cross out row $1$1 and column $2$2, and again find the determinant of the matrix that remains:

Finally for $M_{1,3}$M1,3​, cross out the row $1$1 and column $3$3, and find the determinant of the matrix that remains:

Note that if, in step $1$1, we had chosen to move along row $2$2, then we would evaluate the minors given by $M_{2,1},M_{2,2}$M2,1​,M2,2​ and $M_{2,3}$M2,3​. The same applies if we had chosen row $3$3 - we would find the values of $M_{3,1},M_{3,2}$M3,1​,M3,2​ and $M_{3,3}$M3,3​.

 

3. Find the cofactors

The next step is to change the three minors into what are known as cofactors using the rule of signs. This is the simple part of the exercise.

A cofactor is a simply a minor with a positive sign or a negative sign attached to the front of it. It's sometimes referred to as a signed minor. 

The sign that we need to attach (a "$+$+" or a "$-$−")  depends on whether $i+j$i+j is even or odd (of course attaching a plus sign makes no difference to the minor at all). 

Specifically, if we denote $C_{i,j}$Ci,j​ as the cofactor corresponding to $M_{i,j}$Mi,j​, then the rule of signs says that if $i+j$i+j is even, then $C_{i,j}=M_{i,j}$Ci,j​=Mi,j​, and if $i+j$i+j is odd then $C_{i,j}=-M_{i,j}$Ci,j​=−Mi,j​. 

Mathematically we can write that $C_{i,j}=\left(-1\right)^{i+j}\times M_{i,j}$Ci,j​=(−1)i+j×Mi,j​. Note that whenever $i+j$i+j is even the negative disappears. If the power is odd, the negative survives.

Hence, because we distributed through row $1$1, we find:

• $C_{1,1}=+M_{1,1}=+\left(2\right)=2$C1,1​=+M1,1​=+(2)=2
• $C_{1,2}=-M_{1,1}=-\left(0\right)=0$C1,2​=−M1,1​=−(0)=0
• $C_{1,3}=+M_{1,1}=+\left(-4\right)=-4$C1,3​=+M1,1​=+(−4)=−4

 

4. Form a linear combination of the row elements with the cofactors

This linear combination will be the determinant of the matrix.

Because we used row $1$1, we take the elements of that row, namely $\left(1,5,3\right)$(1,5,3) and the cofactors $\left(2,0,-4\right)$(2,0,−4) and write:

$\left|A\right|=1\times\left(2\right)+5\times\left(0\right)+3\times\left(-4\right)=2+0-12=-10$|A|=1×(2)+5×(0)+3×(−4)=2+0−12=−10

 

Distributing along another row

Just to show that any other row would deliver the same result, we will find the determinant for the same matrix, but distribute along the second row. 

In the second row, the minors become:

$M_{2,1}$M2,1​	$=$=	$\left(5\times2-6\times3\right)=-8$(5×2−6×3)=−8
$M_{2,2}$M2,2​	$=$=	$\left(1\times2-4\times3\right)=-10$(1×2−4×3)=−10
$M_{2,3}$M2,3​	$=$=	$\left(1\times6-4\times5\right)=-14$(1×6−4×5)=−14

Then, applying the rule of signs:

$C_{2,1}=8$C2,1​=8 (note that $2+1$2+1 is odd, so sign change is required).

$C_{2,2}=-10$C2,2​=−10 ($2+2$2+2 is even so no sign change necessary).

$C_{2,3}=14$C2,3​=14 ($2+3$2+3 is odd so a sign change is required).

Finally, forming the linear combination: 

$\left|A\right|=2\times\left(8\right)+4\times\left(-10\right)+1\times\left(14\right)=16-40+14=-10$|A|=2×(8)+4×(−10)+1×(14)=16−40+14=−10

 

Developing a formula

The four steps above suggests a formula for the determinant of an order $3$3 matrix.

The procedure began with deciding a row to distribute along.

Suppose we chose row $i$i.

• Denote the row elements  $e_{i,1},e_{i,2},e_{i,3}$ei,1​,ei,2​,ei,3​.
• Denote the cofactors $\left(-1\right)^{i+1}M_{i,1},\left(-1\right)^{i+2}M_{i,2},\left(-1\right)^{i+3}M_{i,3}$(−1)i+1Mi,1​,(−1)i+2Mi,2​,(−1)i+3Mi,3​  

So putting these together the determinant is given by:

$\left|A\right|$|A|	$=$=	$e_{i,1}\times C_{i,1}+e_{i,1}\times C_{i,2}+e_{i,1}\times C_{i,3}$ei,1​×Ci,1​+ei,1​×Ci,2​+ei,1​×Ci,3​
	$=$=	$e_{i,1}\times\left(-1\right)^{i+1}M_{i,1}+e_{i,1}\times\left(-1\right)^{i+2}M_{i,2}+e_{i,1}\times\left(-1\right)^{i+3}M_{i,3}$ei,1​×(−1)i+1Mi,1​+ei,1​×(−1)i+2Mi,2​+ei,1​×(−1)i+3Mi,3​

 

Distributing to 4x4 matrices

The method to find an order $4$4 determinant proceeds in exactly the same way, although we are faced with evaluating four order $3$3 determinants. 

Let's find the determinant for the matrix given by $A=$A=

We will show the procedure using a row $1$1 distribution. 

$|A|$|A|	$=$=	$2\times\left(C_{1,1}\right)-1\times\left(C_{1,1}\right)+0\times\left(C_{1,1}\right)+3\times\left(C_{1,1}\right)$2×(C1,1​)−1×(C1,1​)+0×(C1,1​)+3×(C1,1​)
$|A|$|A|	$=$=	$2\times\left(M_{1,1}\right)+1\times\left(M_{1,1}\right)+0\times\left(M_{1,1}\right)-3\times\left(M_{1,1}\right)$2×(M1,1​)+1×(M1,1​)+0×(M1,1​)−3×(M1,1​)
$|A|$|A|	$=$=	$2\times$2×  $+1\times$+1×  $+0\times$+0×  $-3\times$−3×
$|A|$|A|	$=$=	$2\left(-60\right)-1\left(-62\right)+0\left(16\right)-3\left(42\right)$2(−60)−1(−62)+0(16)−3(42)
$\therefore$∴ $|A|$|A|	$=$=	$-308$−308

Of course the four determinants showing were evaluated separately using the same strategy. Without a suitable software program, this does take some time to complete!

 

Shortcuts

We can appreciate the work that might be involved in finding even higher order determinants. There are however a couple of short cuts to discuss.

Look for rows that contain zeros, because distributing along these rows greatly cuts the workload down. Hopefully you can see why.

Also understand that the rule of signs can be thought of as a patchwork of alternating pluses and minuses beginning from the top left corner of the matrix to the bottom right. For example, here is what the rule of signs becomes for an order $4$4 matrix:

$+$+	$-$−	$+$+	$-$−
$-$−	$+$+	$-$−	$+$+
$+$+	$-$−	$+$+	$-$−
$-$−	$+$+	$-$−	$+$+

Finally, someone in mathematical history noticed that an order $3$3 determinant could be worked out cleverly.

The method is to write down the elements as they occur in the matrix, but then repeat the first two columns as shown here using the same example we dealt with above:

Then identify the $6$6 diagonals as shown in the diagram (shown in three red rectangles leaning one way and three green rectangles leaning the other way).

Find the products of the numbers in the three red rectangles and write them down as shown, and do the same for the green rectangles.

The determinant is the sum of the "red" products minus the sum of the "green" products. In other words:

$\left|A\right|=\left(8+20+36\right)-\left(48+6+20\right)=64-74=-10$|A|=(8+20+36)−(48+6+20)=64−74=−10

Applying determinants to triangle area

The area of a triangle with vertices given by the coordinates $\left(a,b\right),\left(c,d\right)$(a,b),(c,d) and $\left(e,f\right)$(e,f) is given by the absolute value of $A$A where:

There are a number of ways to prove this result. One way is shown here:

The diagram shows the triangle $PQR$PQR and three perpendiculars dropped from each vertex to $S$S, $T$T and $U$U.

The area of the triangle $PQR$PQR is given by:

$A_{PQR}=A_{SPQT}+A_{TQRU}-A_{SPRU}$APQR​=ASPQT​+ATQRU​−ASPRU​

where the expressions on the right hand side represent the areas of three trapezoids.

Thus we have:

$A_{PQR}$APQR​	$=$=	$A_{SPQT}+A_{TQRU}-A_{SPRU}$ASPQT​+ATQRU​−ASPRU​
	$=$=	$\frac{1}{2}\left(b+d\right)\left(c-a\right)+\frac{1}{2}\left(d+f\right)\left(e-c\right)-\frac{1}{2}\left(b+f\right)\left(e-a\right)$12​(b+d)(c−a)+12​(d+f)(e−c)−12​(b+f)(e−a)
	$=$=	$\frac{1}{2}\left[bc-ab+cd-ad+ed-cd+ef-cf-be+ab-ef+af\right]$12​[bc−ab+cd−ad+ed−cd+ef−cf−be+ab−ef+af]
	$=$=	$\frac{1}{2}\left[bc-ad+ed-cf-be+af\right]$12​[bc−ad+ed−cf−be+af]

If we now use cofactors to evaluate $A$A given above, we find, distributing along row $1$1, that:

$A$A	$=$=	$\frac{1}{2}\left[a\left(d-f\right)-b\left(c-e\right)+1\left(cf-de\right)\right]$12​[a(d−f)−b(c−e)+1(cf−de)]
	$=$=	$\frac{1}{2}\left[ad-af-bc+be+cf-de\right]$12​[ad−af−bc+be+cf−de]
	$=$=	$-\frac{1}{2}\left[bc-ad+ed-cf-be+af\right]$−12​[bc−ad+ed−cf−be+af]

The absolute value of $A$A is thus exactly the same quantity as $A_{PQR}$APQR​ shown above. 

 

Polygonal regions

Surveyors can estimate the area of any planar polygonal region using this formula by simply dividing the region up into triangles and overlaying a suitable set coordinate axes so that coordinates can be established for the vertices. 

Here is a simple example showing the concept. The position of the axes is arbitrary, however it is desirable to put at least one of the axes along one of the sides of the polygon. After the coordinates are determined, the areas of the four triangles are found by repeated application of the determinant rule.

 

Worked example

Question 4

Find the area of the triangle with vertices $P\left(-2,5\right),Q\left(3,1\right)$P(−2,5),Q(3,1) and $R\left(0,-4\right)$R(0,−4).

Applying the formula, we have  and distributing along row $3$3, we find:

$A$A	$=$=	$\frac{1}{2}\left[0\left(5-1\right)-\left(-4\right)\left(-2-3\right)+1\left(-2-15\right)\right]$12​[0(5−1)−(−4)(−2−3)+1(−2−15)]
	$=$=	$\frac{1}{2}\left(-20-17\right)$12​(−20−17)
	$=$=	$-18\frac{1}{2}$−1812​

Hence the area of the triangle is $18.5$18.5  $u^2$u2.

Practice questions

Question 5

Question 6

Question 7

Properties of determinants

The following properties of determinants can be used to greatly simplify the work involved in evaluating them.

While not a complete proof, we will demonstrate the validity of each of the six properties using the method of distribution into cofactors of the general order $3$3 determinant.

 

Using an order $3$3 matrix, two possibilities for the property are:

A row of zeros:  $=0$=0

A column of zeros:  $=0$=0

To show the first of these is true, we might distribute along row $1$1, so that:

$|A|$|A|	$=$=	$a\left[0\times i-h\times0\right]-b\left[0\times i-g\times0\right]+c\left[0\times h-g\times0\right]$a[0×i−h×0]−b[0×i−g×0]+c[0×h−g×0]
	$=$=	$a\left[0\right]-b\left[0\right]+c\left[0\right]$a[0]−b[0]+c[0]
	$=$=	$0$0

To show the second of these is true, we might distribute along row $1$1 again, so that:

$|A|$|A|	$=$=	$0\left[ei-fh\right]-b\left[0\times i-0\times f\right]+c\left[0\times h-0\times e\right]$0[ei−fh]−b[0×i−0×f]+c[0×h−0×e]
	$=$=	$0$0

 

 

Two possibilities for the property are:

Rows $1$1 and $2$2 interchanged: 

Columns $1$1 and $2$2 interchanged:   

Verifying the first of these:

In the first instance, distributing separately the left had side along row $1$1, and the right hand side along row $2$2:

$LHS$LHS	$=$=	$a\left(ei-fh\right)-b\left(di-fg\right)+c\left(dh-eg\right)$a(ei−fh)−b(di−fg)+c(dh−eg)
$RHS$RHS	$=$=	$-\left[-a\left(ei-fh\right)+b\left(di-fg\right)-c\left(dh-eg\right)\right]$−[−a(ei−fh)+b(di−fg)−c(dh−eg)]
	$=$=	$a\left(ei-fh\right)-b\left(di-fg\right)+c\left(dh-eg\right)$a(ei−fh)−b(di−fg)+c(dh−eg)

In the second instance, distributing along row $1$1 on both the left and right hand sides:

$LHS$LHS	$=$=	$a\left(ei-fh\right)-b\left(di-fg\right)+c\left(dh-eg\right)$a(ei−fh)−b(di−fg)+c(dh−eg)
$RHS$RHS	$=$=	$-\left[b\left(di-fg\right)-a\left(ei-fh\right)+c\left(eg-dh\right)\right]$−[b(di−fg)−a(ei−fh)+c(eg−dh)]
	$=$=	$a\left(ei-fh\right)-b\left(di-fg\right)+c\left(dh-eg\right)$a(ei−fh)−b(di−fg)+c(dh−eg)

 

Two possibilities for the property are:

Row scalar:   

Column scalar:      

We will just show the proof of one of these. Essentially, the scalar $z$z comes out as a common factor in both cases. For the row scalar:

$LHS$LHS	$=$=	$za\left(ei-fh\right)-zb\left(di-fg\right)+zc\left(dh-eg\right)$za(ei−fh)−zb(di−fg)+zc(dh−eg)
	$=$=	$z\left[a\left(ei-fh\right)-b\left(di-fg\right)+c\left(dh-eg\right)\right]$z[a(ei−fh)−b(di−fg)+c(dh−eg)]
	$=$=	$RHS$RHS

 

Two possibilities for the property are:

Equal rows:   

Equal columns:  

Again we will prove only of the these. The proof becomes obvious when distributing along row $2$2:

$LHS$LHS	$=$=	$-d\left(bc-bc\right)+e\left(ac-ac\right)-f\left(ab-ab\right)$−d(bc−bc)+e(ac−ac)−f(ab−ab)
	$=$=	$0$0

 

Two possibilities for the property are:

Row multiples:   

Column multiples:  

Notice that, in the first instance, a multiple of row 1 has been added to row 2. In the second instance, a multiple of column 1 has been added to row 3.

A proof of the first instance becomes:

$LHS$LHS	$=$=	$a\left(ei-fh\right)-b\left(di-fg\right)+c\left(dh-eg\right)$a(ei−fh)−b(di−fg)+c(dh−eg)
$RHS$RHS	$=$=	$\left[azbi+aei-azch-afh\right]-\left[bzai+bdi-zbcg-bfg\right]+\left[czah+cdh-czbg-ceg\right]$[azbi+aei−azch−afh]−[bzai+bdi−zbcg−bfg]+[czah+cdh−czbg−ceg]
	$=$=	$\left[aei-afh\right]-\left[bdi-bfg\right]+\left[cdh-ceg\right]$[aei−afh]−[bdi−bfg]+[cdh−ceg]
	$=$=	$a\left(ei-fh\right)-b\left(di-fg\right)+c\left(dh-eg\right)$a(ei−fh)−b(di−fg)+c(dh−eg)
	$=$=	$LHS$LHS

The second instance is proved the same way. Note that all terms that contain four factors disappear.

 

Written symbolically, this says: 

To prove this property, we will distribute the determinant on the left along the first row, and also distribute the determinant on the right along the first column:

$LHS$LHS	$=$=	$a\left(ei-fh\right)-b\left(di-fg\right)+c\left(dh-eg\right)$a(ei−fh)−b(di−fg)+c(dh−eg)
$RHS$RHS	$=$=	$a\left(ei-hf\right)-b\left(di-gf\right)+c\left(dh-ge\right)$a(ei−hf)−b(di−gf)+c(dh−ge)
	$=$=	$a\left(ei-fh\right)-b\left(di-fg\right)+c\left(dh-eg\right)$a(ei−fh)−b(di−fg)+c(dh−eg)
	$=$=	$LHS$LHS

We can distribute along any row or column of one matrix and still get the same answer by choosing the matching column or row in the other.

Property $6$6 is really pointing to the fact that the determinant of any matrix $A$A is also the determinant of its transpose $A^T$AT. 

 

How to use the properties

Regarding property $1$1 and property $4$4 their usefulness is fairly obvious. If you spot either two rows or two columns that are identical, then the determinant is zero. Now working is required beyond that observation. If you spot a row of zeros or a column of zeros, then again the determinant is zero.

With respect to property $2$2, we can eliminate a negative sign in front of the determinant by swapping either two rows or two columns.

For property $3$3, if you spot a common factor in either a row or a column, you can take it out to the front of the determinant. This common factor extraction might alert you to something hidden in the determinant elements left behind.

For example, observe how this order $2$2 determinant is simplified:

Property $5$5 is perhaps the most powerful. Adding to any row of a determinant a non-zero multiple of any other row, without changing the determinant's value, can greatly simplify working. The same applies to columns. We'll show one example with an order $3$3 matrix: 

Here, row $2$2 has been altered. The new row $2$2 has become the old row two plus twice row $1$1. The size of the numbers have shrunk and a zero has appeared in the first column. Also a common factor of $3$3 was taken out of row $3$3 and consequently the determinant looks much easier to deal with.

By distributing along row $2$2, we find that: 

$3\left[-0\left(21+16\right)+1\left(-18+8\right)+1\left(-12-7\right)\right]=-87$3[−0(21+16)+1(−18+8)+1(−12−7)]=−87

Applying property $6$6 is useful whenever you notice that two matrices have the same entries, but that the positions of the entries have been changed by reflecting them across the main diagonal. If you calculate the determinant of one of them, the other one is guaranteed to have the same determinant. So without doing any calculation, we know that the determinant 

is equal to $-87$−87, since we calculated the determinant of the matrix obtained by exchanging the rows and columns in the previous paragraph, and the property tells us our answers will be the same.

Practice questions

Question 8

Question 9

Question 10