### 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}*= (

*) be an*

_{a}_{ij}*matrix. Then A has associated with it a number, called the determinant of*

_{n}×_{n}*and denoted by*

_{A}*. For*

_{detA}*= 1 or 2, the value of*

_{n}*is calculated as follows:*

_{detA}If* _{n}* = 1, then

*=*

_{detA}*.*

_{a11}If* _{n}* = 2, then

*=*

_{detA}*.*

_{a11a22}−_{a12a21}*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.

(b) For a 2*×*2 matrix, we use the formula* _{detB}* =

*. 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).*

_{b11b22}−_{b12b21}So assuming that* _{a}*,

*,*

_{b}*and*

_{c}*in matrix*

_{d}*are any scalars, the number*

_{B}*is given by*

_{detB}*=*

_{detB}*.*

_{ad}−_{bc}(c) Again, we have a 2*×*2 matrix, so we do the same calculation as in (b).
We get * _{detC}* =

*= (1)(5)*

_{c11c22}−_{c12c21}*−*(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 square matrix of order*

_{A}

_{n}*≥*3. We will need to express certain

*submatrices*of 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

*submatrix*to be the (

_{A}_{ij}*1)*

_{n}−*×*(

*1) matrix obtained by deleting the*

_{n}−*ith* row and the * _{j}th* column from the matrix

*.*

_{A}*Example 2.* For* _{A}*=

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

, find* _{A11}* 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

**Definition 18.3.** Consider any* _{n}×_{n}*matrix

*= (*

_{A}*) with submatrices*

_{a}_{ij}*. (1) The (*

_{A}_{ij}*)-*

_{i, j}*of*

_{minor}*, denoted*

_{A}*, is defined to be*

_{M}_{ij}*Mij* =*detAij*

(2) The (* _{i, j}*)-cofactor of

*, denoted*

_{A}*, is defined to be*

_{C}_{ij}*Cij* = (*−*1)*i*+*jMij*

That is,* _{C}_{ij}* =

*if*

_{M}_{ij}*+*

_{i}*is even, but*

_{j}*=*

_{C}_{ij}*−*if

_{M}_{ij}*+*

_{i}*is odd.*

_{j}*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)2

*= 23*

_{M11}*M12* =_{det}

4 6 2 1

=*−*8 and * _{C12}*= (

*−*1)3

*= 8*

_{M12}*M13*=_{det}

4 5

2 *−*3

=*−*22 and * _{C13}*= (

*−*1)4

*=*

_{M13}*−*22

*M21* =_{det}

2 1

*−*3 1

= 5 and * _{C21}*= (

*−*1)3

*=*

_{M21}*−*5 Similarly, we can calculate:

*C22* =* _{M22}* =

*−*5,

*M23*= 5 and* _{C23}*=

*−*5,

*C31* =* _{M31}* = 7,

*M32*=*−*22 and* _{C32}*= 22
and finally

*=*

_{C33}*=*

_{M33}*−*23.

**Definition 18.4.** The Definition of Determinant, Part 2 (_{n}*≥*3)

Let * _{A}*= (

*) be an*

_{a}_{ij}*matrix with*

_{n}×_{n}

_{n}*≥*3. Let

*be the (*

_{C}_{ij}*)-cofactor of*

_{ij}*A*. Then the value of the determinant of * _{A}* can be calculated using:

*detA*=* _{a11C11}*+

*+*

_{a12C12}*· · ·*+

*. That is,*

_{a1}_{n}_{C1}_{n}*= P*

_{detA}*n*

*j*=1*a1jC1j*

.

Note: This particular calculation for * _{detA}* is called

*expanding the*

*determi-nant along the first row of A.*

*Example 4.* Find* _{detA}*, where

*=*

_{A}

1 2 32 1 3 3 2 1

.

*Solution:*

Recall that * _{C}_{ij}* = (

*−*1)

*i*+

*j*, where

_{M}_{ij}*=*

_{M}_{ij}*. We calculate*

_{detA}_{ij}*as:*

_{detA}*detA* = P3
*j*=1*a1jC1j*

=* _{a11}*(

*−*1)1+1

*+*

_{detA11}*(*

_{a12}*−*1)1+2

*+*

_{detA12}*(*

_{a13}*−*1)1+3

*= (1)(*

_{detA13}*−*1)2

_{det}

1 3 2 1

+ (2)(*−*1)3_{det}

2 3 3 1

+ (3)(*−*1)4_{det}

2 1 3 2

= [(1)(1)*−*(3)(2)]*−*2[(2)(1)*−*(3)(3)] + 3[(2)(2)*−*(1)(3)]
=*−*5*−*2(*−*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.

**Theorem 18.5.** *Suppose* _{A}*is a square* (* _{n}×_{n}*)

*matrix with*

*1. Then*

_{n >}*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* 1

*≤*

_{i}≤_{n},*detA* = _{a}_{i}_{1}(*−*1)*i*+1(_{detA}_{i}_{1}) +_{a}_{i}_{2}(*−*1)*i*+2(_{detA}_{i}_{2}) +*· · ·*+* _{a}_{ij}*(

*−*1)

*i*+

*j*(

*) +*

_{detA}_{ij}*· · ·*+

*(*

_{a}_{in}*−*1)

*i*+

*n*(

*)*

_{detA}_{in}=
*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* 1

*≤*

_{j}*≤*

_{n},*detA* = * _{a1}_{j}*(

*−*1)1+

*j*(

*) +*

_{detA1}_{j}*(*

_{a2}_{j}*−*1)2+

*j*(

*) +*

_{detA2}_{j}*· · ·*+

*(*

_{a}_{ij}*−*1)

*i*+

*j*(

*) +*

_{detA}_{ij}*· · ·*+

*(*

_{a}_{nj}*−*1)

*n*+

*j*(

*)*

_{detA}_{nj}=
*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 + and*−*as
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}* or

*is odd or even).*

_{j}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)

*+ (3)*

_{C12}*+ (4)*

_{C13}

_{C14}Each of the matrices * _{A11, A12, A13,}* and

*is a 3*

_{A14}*×*3 matrix, and so each of the 4 determinants we need to calculate, in order to calculate

*, requires as much calculation as the determinant in the earlier example.*

_{detA}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.* Find* _{detA}*, 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)

*+ (0)*

_{C21}

_{C31}= (2)(*−*1)1+1* _{M11}*+ 0 + 0
= (2)(

*−*1)2

_{det}

3 1
0 *−*5

= (2)(*−*15) = *−*30

*Example 6.* Find* _{detA}*, 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:

*detA* =* _{a31C31}*+

*+*

_{a32C32}*+*

_{a33C33}

_{a34C34}= 2* _{C31}*+ 0

*+ 0*

_{C32}*+ 0*

_{C33}

_{C34}= (2)(*−*1)3+1* _{M31}*+ 0 + 0 + 0
= (2)(

*−*1)4

_{detA31}= (2)_{det}

2 3 41 3 1 7 8 9

= (2)

2_{det}

3 1 8 9

*−*3_{det}

1 1 7 9

+ 4_{det}

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, * _{detA31}*was calculated by expanding along row 1. However,
if submatrix

*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.*

_{A31}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 matrix _{A}has any row or column containing only*

*zeroes, then*

*= 0.*

_{detA}**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 result

*= 0.*

_{detA}*Example 7.* Find* _{detA}*, where

*A* =

2 4 21 3 1 0 0 0

*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.* Find* _{detA}*, where

*A* =

21 43 21
*−*5 0 *−*5

Looking at * _{A}*, we see that the first and last columns are equal, so

*= 0.*

_{detA}**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

*is called a*

_{A}*diagonal*matrix.

For instance, among the matrices shown below, * _{A}*is upper triangular,

*is lower triangular and*

_{B}*is a diagonal matrix.*

_{C}*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}*= (

*)*

_{a}_{ij}*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.*

*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}*) = (1)(*

_{a33}*−*3)(5) =

*−*15

(b) We see that * _{B}* is lower triangular, so

*= (2)(2)(*

_{detB}*−*1) =

*−*4. (c) For this diagonal matrix, we get

*= (2)(7)(*

_{detC}*−*3) =

*−*42.