# Unit 18 Determinants

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### Unit 18 Determinants

Every square matrix has a number associated with it, called its determi-nant. In this section, we determine how to calculate this number, and also look at some of the properties of the determinant of a (square) matrix.

Calculating the determinant of a small matrix is easy. But as the order of the square matrix increases, the calculation of the determinant becomes more complicated.

We define the determinant recursively. That is, for a square matrix of order 1 or 2, we will define a simple formula for calculating the determinant. However, for a square matrix of order 3 or higher, we define the determinant of the matrix in terms of the determinants of certain submatrices. Thus, we calculate the determinant of a large matrix by calculating the determinants of various smaller matrices.

Definition 18.1. The Definition of Determinant, Part 1 (n 2)

Let A= (aij) be an n×n matrix. Then A has associated with it a number, called the determinant of A and denoted by detA. For n= 1 or 2, the value of detA is calculated as follows:

Ifn = 1, then detA=a11.

Ifn = 2, then detA=a11a22a12a21.

Example 1. Find the determinants of the following matrices: (a) A= [4] (b) B =

a b c d

(c) C =

1 2 3 5

Solution:

(a) Here, A is 1×1, so detA=a11=4.

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(b) For a 2×2 matrix, we use the formuladetB =b11b22b12b21. That is, we take the product of the numbers going diagonally down to the right (i.e., on the main diagonal) and then subtract from that the product of the numbers going diagonally down to the left (i.e., on the other diagonal).

So assuming thata, b,candd in matrixB are any scalars, the numberdetB is given by detB =adbc.

(c) Again, we have a 2×2 matrix, so we do the same calculation as in (b). We get detC =c11c22c12c21 = (1)(5)(2)(3) = 5(6) = 11.

Now we are ready to define the notation and terminology needed to build up to the definition of detA for A a square matrix of order n 3. We will need to express certainsubmatricesof a matrix, as well as the determinant of such a submatrix and also a particular scalar multiple of this number. These are the focus of the next 2 definitions, after which we will be able to complete our definition of determinant.

Definition 18.2. Consider any square matrix A of order n 2. We define the submatrixAij to be the (n1)×(n1) matrix obtained by deleting the

ith row and the jth column from the matrix A.

Example 2. ForA=

   

1 2 1 3 2 1 3 1 2 5 3 4 3 7 8 9

  

, findA11 and A23.

Solution: To find A11, we delete both the first row and the first column.

We get A11=

 1 3 15 3 4 7 8 9

 

For A23, we must delete row 2 and also column 3 from the matrix A.

We see that A23 =

 1 2 32 5 4 3 7 9

 

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Definition 18.3. Consider anyn×nmatrixA= (aij) with submatricesAij. (1) The (i, j)-minor of A, denoted Mij, is defined to be

Mij =detAij

(2) The (i, j)-cofactor of A, denoted Cij, is defined to be

Cij = (1)i+jMij

That is,Cij =Mij if i+j is even, butCij =Mij if i+j is odd.

Example 3. Find the minors and cofactors of:

A =

43 25 16 2 3 1

 

We get:

M11=det

5 6

3 1

= 23 and C11= (1)2M11= 23

M12 =det

4 6 2 1

=8 and C12= (1)3M12= 8

M13=det

4 5

2 3

=22 and C13= (1)4M13=22

M21 =det

2 1

3 1

= 5 and C21= (1)3M21=5 Similarly, we can calculate:

C22 =M22 =5,

M23= 5 andC23=5,

C31 =M31 = 7,

M32=22 andC32= 22 and finally C33 =M33 =23.

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Definition 18.4. The Definition of Determinant, Part 2 (n 3)

Let A= (aij) be ann×n matrix withn 3. LetCij be the (ij)-cofactor of

A. Then the value of the determinant of A can be calculated using:

detA=a11C11+a12C12+· · ·+a1nC1n. That is, detA= Pn

j=1a1jC1j

.

Note: This particular calculation for detA is called expanding the determi-nant along the first row of A.

Example 4. FinddetA, where A =

 1 2 32 1 3 3 2 1

 .

Solution:

Recall that Cij = (1)i+jMij, where Mij =detAij. We calculate detA as:

detA = P3 j=1a1jC1j

=a11(1)1+1detA11+a12(1)1+2detA12+a13(1)1+3detA13 = (1)(1)2det

1 3 2 1

+ (2)(1)3det

2 3 3 1

+ (3)(1)4det

2 1 3 2

= [(1)(1)(3)(2)]2[(2)(1)(3)(3)] + 3[(2)(2)(1)(3)] =52(7) + 3(1) =5 + 14 + 3 = 12

Notice: The (1)i+j multipliers on the minors alternate between + and as we expand along the row.

When we defined the determinant of an n × n matrix, for n 3, we expanded along the first row. The following Theorem tells us that we may, in fact, expand the determinant along any row or any column of a matrix and still get the same answer.

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Theorem 18.5. Suppose A is a square (n×n) matrix with n > 1. Then the value of detA can be found by expanding along any row or column of A. That is:

(1) for any fixed value of i, with 1in,

detA = ai1(1)i+1(detAi1) +ai2(1)i+2(detAi2) +· · ·+aij(1)i+j(detAij) + · · ·+ain(1)i+n(detAin)

= n

X

j=1

aij(1)i+j(detAij) = n

X

j=1 aijCij

(this is expansion along row i) and

(2) for any fixed value of j, with 1j n,

detA = a1j(1)1+j(detA1j) +a2j(1)2+j(detA2j) +· · ·+aij(1)i+j(detAij) + · · ·+anj(1)n+j(detAnj)

= n

X

i=1

aij(1)i+j(detAij) = n

X

i=1 aijCij (this is expansion along column j)

Notice: As before, the sign of (1)i+j always alternates between + andas we expand along any row or column, although whether the pattern begins with + or depends on which row or column we expand along (i.e., whether the fixed i orj is odd or even).

The above theorem is very important in simplifying our calculations for finding the determinant of a matrix. Suppose we are asked to find the deter-minant of a matrix such as:

A =

   

1 2 3 4 2 1 3 1 2 0 0 0 3 7 8 9

   

Then expanding along the first row, as in our original definition of detA, we have:

detA = (1)C11+ (2)C12+ (3)C13+ (4)C14

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Each of the matrices A11, A12, A13, and A14 is a 3×3 matrix, and so each of the 4 determinants we need to calculate, in order to calculate detA, requires as much calculation as the determinant in the earlier example.

We can reduce the amount of work required to calculate the determinant of a relatively large matrix by choosing wisely which row or column to ex-pand along. Generally, we want to choose a row or column containing as many zeroes as possible, so that we only need to calculate determinants of the fewest submatrices possible.

Example 5. FinddetA, where A =

 2 40 3 61 0 0 5

 

Solution: Looking carefully at A, we see that it will save some work if we expand along column 1. (Alternatively, we could choose row 3.)

detA =a11C11+a21C21+a31C31

= (2)C11+ (0)C21+ (0)C31

= (2)(1)1+1M11+ 0 + 0 = (2)(1)2det

3 1 0 5

= (2)(15) = 30

Example 6. FinddetA, where A =

   

1 2 3 4 2 1 3 1 2 0 0 0 3 7 8 9

   

Solution: For this matrix, as we saw previously, expanding along row 1 re-quires us to calculate determinants of four 3×3 submatrices. However, we see that row 3 contains 3 zeroes. This means that if we calculate detA by expanding along row 3, then 3 of the 4 cofactors are going to be multiplied by 0, so there is only one 3×3 submatrix whose determinant we actually need to calculate. Choosing row 3 to expand along, we get:

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detA =a31C31+a32C32+a33C33+a34C34

= 2C31+ 0C32+ 0C33+ 0C34

= (2)(1)3+1M31+ 0 + 0 + 0 = (2)(1)4detA31

= (2)det

 2 3 41 3 1 7 8 9

 

= (2)

2det

3 1 8 9

3det

1 1 7 9

+ 4det

1 3 7 8

= 2{2[3(9)1(8)]3[1(9)1(7)] + 4[1(8)3(7)]} = 2[(2)(19)(3)(2) + (4)(13)] = 2(20) =40 This is much faster than if we had expanded along row 1.

Notice: Here, detA31was calculated by expanding along row 1. However, if submatrixA31 had contained some zeroes, we could have reduced the work even more by selecting an appropriate row or column to expand along for this calculation as well.

There are several theorems which can be used to find the determinants of certain kinds of matrices with little or no work.

Theorem 18.6. If a square matrixAhas any row or column containing only zeroes, then detA= 0.

Proof: If matrix A has a row (column) of zero’s we simply choose to ex-pand the determinant along that row (column), giving us the resultdetA= 0.

Example 7. FinddetA, where

A =

 2 4 21 3 1 0 0 0

 

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Solution: We see, without any calculation, that detA= 0 since the last row contains only zeroes.

Theorem 18.7. If a square matrix A has two equal rows, or two equal columns, then detA= 0.

Proof: We leave this proof until later. In the next section we will encounter another theorem which makes this proof very easy. (We will simply subtract one of the equal rows (columns) from the other, yielding a zero row (column), and then use the previous theorem).

Example 8. FinddetA, where

A =

 21 43 21 5 0 5

 

Looking at A, we see that the first and last columns are equal, so detA= 0.

Definition 18.8. A square matrix A is called upper triangular if all entries below the main diagonal are zero, and is called lower triangular if all entries above the main diagonal are zero. If all entries above the main diagonal and all entries below the main diagonal are zero, then A is called a diagonal matrix.

For instance, among the matrices shown below, Ais upper triangular, B is lower triangular and C is a diagonal matrix.

A=

 10 43 12

0 0 5

B =

 21 02 00 5 0 1

C =

 2 00 7 00 0 0 3

 

Theorem 18.9. If a square matrix A= (aij) is either upper or lower trian-gular, or is a diagonal matrix, then the determinant of A is the product of the elements lying on the main diagonal.

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Sketch of proof: Repeatedly choosing column 1 (for an upper triangular or diagonal matrix) or row 1 (for a lower triangular matrix) to expand along yields the stated result.

Corollary 18.10. For any n 1, the determinant of the identity matrix of order n is detI = 1.

Proof: By definition, the identity matrix is a diagonal matrix with each of the main diagonal entries 1, so by the above theorem we have:

detI = (1)(1)...(1) = 1

Example 9. Find the determinants of

(a)A=

 10 43 12

0 0 5

 (b)B =

 21 02 00 5 0 1

 and (c)C =

 2 00 7 00 0 0 3

 

Solution:

(a) Here, A is a 3×3 upper triangular matrix, so we get

detA= (a11)(a22)(a33) = (1)(3)(5) =15

(b) We see that B is lower triangular, so detB = (2)(2)(1) = 4. (c) For this diagonal matrix, we get detC = (2)(7)(3) = 42.

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