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Mathematics 1229A/B

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Math 1229A/B
Unit 10:
Determinants
(text reference: Section 4.1)
▯V. Olds 2010 128 Unit 10
10 Determinants
Square matrices are a special class of matrices. We have already seen one instance of a concept
which is deﬁned only for square matrices — the inverse matrix. That is, only a square matrix may
have an inverse. In this unit we will (begin to) learn about another concept which is deﬁned only
for square matrices — the determinant of a matrix.
Deﬁnition of Determinant: Part One
Every square matrix has associated with it a number, called the determinant of the
matrix. The determinant of the n × n matrix A is denoted by det A.
The number which is the determinant of a square matrix measures a certain characteristic of the
matrix. In a more advanced study of matrix algebra, this characteristic is used for various purposes.
In this course, the only way in which we will use this number is in its connection to the existence of
the inverse of the matrix, and through that it’s application to SLE’s in which the coeﬃcient matrix
is a square matrix. For these purposes, what will matter to us is whether or not this number, the
determinant of the matrix, is 0. But of course, in order to determine whether or not the determinant
of a particular matrix is 0, we need to know how to calculate that number.
Calculating the determinant of a square matrix is somewhat complicated. The deﬁnition is re-
cursive, meaning that the calculation is deﬁned in a straightforward way for small matrices, and
then for larger matrices, the determinant is deﬁned as being a calculation involving the determinants
of smaller matrices, which are certain submatrices of the matrix. We could express this recursive
deﬁnition of the determinant of a square matrix of order n as applying for all n ≥ 2, speciﬁcally
deﬁning only the determinant of a square matrix of order 1, i.e. a (square) matrix containing only
a single number. However, the calculation for a 2 × 2 matrix is very straightforward — easier to
think of as a special deﬁnition all on its own — so instead we use speciﬁc deﬁnitions for n = 1 and
n = 2, and then deﬁne the determinant of a square matrix of order n > 2 in terms of determinants
of submatrices of order n − 1, which are found by expressing them in terms of determinants of
successively smaller submatrices until we get down to submatrices of order 2. The calculation of
detA as deﬁned in this way, when A is a square matrix of order n > 2, is not really as complicated
as it will look. It’s just a matter of applying a certain formula carefully, as many times as necessary
until we have expressed detA in terms of the determinants of 2 × 2 matrices. Those determinants
are easy to ﬁnd.
So we start by deﬁning detA for square matrices of order 1 and of order 2. When A is a 1 × 1
matrix, i.e. a matrix containing only one number, ﬁnding the particular number detA which is
associated with that matrix is trivial. That number is the only number around — the single number
that’s in the matrix. For a square matrix of order 2, i.e. a matrix containing 4 numbers arranged
in a square, we have to do a little more work. But it’s a simple calculation. In fact, we can think
of the calculation as “down products minus up products”, which is something we have seen before.
But this time there’s only one down product, and only one up product, so it’s actually just “down
product minus up product”.
Deﬁnition of Determinant: Part Two
If A is a square matrix of order 1, so that A = [a] for some number a, then detA = a.
That is, if A is a 1 × 1 matrix, then detA = 11.
▯ ▯
If A is a square matrix of order 2, so that A = a b for some numbers a, b, c and
c d Unit 10 129
d, then detA = ad − cb. That is, if A is a 2 × 2 matrix, then detA 11 22− a 21 12
Example 10.1. Find the determina▯ts of t▯e following matrices:▯ ▯
1 2 2 0
(a) A = [5] (b) B = (c) C =
3 4 1 3
Solution:
(a) Here, A is 1 × 1, so detA =11 = 5. That is, the determinant of this matrix is just the number
that’s in the matrix.
(b) For a 2×2 matrix, we use the formula detB = 11 22b 21 12hat 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 up to the right. So we take the down product minus the
up product. Here, the down product is b11 22= 1(4) = 4 and the up product is 21 12= 3(2) = 6.
Therefore detB = 1(4) − 3(2) = 4 − 6 = −2.
(c) Again, we have a 2 × 2 matrix, so we do the same calculation as in (b). We get
detC = c 11 22− c21 12= (2)(3) − (1)(0) = 6 − 0 = 6
Before we can deﬁne how the determinant of a larger matrix is deﬁned in terms of determinants of
certain submatrices, we need to deﬁne what those submatrices are, and some notation for indicating
what submatrix we’re referring to. We will also deﬁne terminology which means the determinant
of a particular submatrix, and for another number obtained from that determinant, which is that
same number, but sometimes with the sign changed.
Notice that if we have a square matrix of order n, we can obtain various submatrices of order
n − 1 by deleting both one row and one column of the larger matrix. In fact, the matrix doesn’t
have to be square. For any m × n matrix with m > 1 and n > 1 we can obtain an (m − 1) × (n − 1)
submatrix by deleting one row and one column of the larger matrix. So we will deﬁne these sub-
matrices for any matrix that has more than one row and more than one column, but we’ll only be
using them in the context of the original matrix being a square matrix. We simply need to indicate
which row and which column are to be deleted.
Deﬁnition: For any m × n matrix A with m > 1 and n > 1, the submatrix A ij is the
(m − 1) × (n − 1) submatrix of A obtained by deleting row i and column j.
1 2 ▯ ▯
Example 10.2. For A = 3 4 and B = 1 2 ﬁnd A 21and B 11
3 4
5 6
Solution:
1 2
We ﬁnd A 21by deleting row 2 and column 1 of A = 3 4 . Since A is a 3 × 2 matrix,21 will
5 6
be a 2 × 1 matrix, consisting of the parts of rows 1 and 3 of A which are not in column 1. That is,
when we’ve deleted row 2 and column 1, all that’s left is the column 2 entries for rows 1 and 3. We
get
▯ ▯
▯ 1 2 2
▯ 3 ▯ 4 gives A21=
▯ 5 6 6 130 Unit 10
Similarly, we get 11by deleting the ﬁrst row and the ﬁrst column from B. This will give the 1 × 1
submatrix which contains only the number that’s in row 2, column 2. That is,
▯ ▯ ▯ ▯
1 2 ▯ 1 ▯ 2
For B = 3 4 ▯ 3 4 gives B11= [4]
1 2 1 3
Example 10.3. For A = 2 1 3 1 , ﬁnd A and A .
2 5 3 4 11 23
3 7 8 9
Solution:
To ﬁnd A 11we delete both the ﬁrst row and the ﬁrst column. We get
▯ 1 ▯ 2 ▯ 1 ▯ 3 1 3 1
▯ 2 1 3 1
▯ 2 5 3 4 gives A11= 5 3 4
7 8 9
▯ 2 7 8 9
For A , we delete row 2 and column 3 from the matrix A. We see that
23
1 2 ▯ 1 3
1 2 3
▯ 2 ▯ 1 ▯ 3 ▯ 1 gives A23= 2 5 4
2 5 ▯ 3 4
3 7 ▯ 8 9 3 7 9
Notice that A11 is a submatrix that we could see, intact, within matrix A, whereas the submatrix
A contains non-contiguous parts of A, because things have been deleted from the midst of rows
23
and columns. But the numbers from A that are in A 23still have the same relative positions to one
another.
We have a special name for the determinant of submatrix A ijof a matrix A. Also, in our cal-
culation of the determinant of A we will use detij, but we will need the negative of this number
whenever i and j are not both odd or both even. Notice that the sum of two odd numbers is even, as
is the sum of two even numbers. But the sum of an odd number and an even number is odd. And of
course, if we raise −1 to an odd power, the value is −1, whereas if we raise −1 to an even number, the
value is 1. So we can accomplish “use detA if ii+jd j are both odd or both even, but otherwise use the
negative of detAij by multiplying detAijby (−1) . We have a special name for this product, too.
Deﬁnition: Let A be any square matrix of order n > 1 and let A ijbe the submatrix
obtained by deleting row i and column j.
• The i,j-minor of A, denoted M ,ijs given by M ij= detA ij
• The i,j-cofactor of A, denoted C ijis given by Cij= (−1) i+jdetA ij Unit 10 131
−3 2 1
Example 10.4. For A = 4 5 6 , ﬁnd M11 M 13C 23and C31
2 −3 1
Solution:
To ﬁnd the speciﬁed minors we simply identify the corresponding submatrix and calculate its deter-
minant. We have
▯ ▯
▯ ▯ 3 ▯ 2 ▯ 1 5 6
▯ 4 5 6 ⇒ A 11=
▯ 2 −3 1 −3 1
and so M11= detA 11= 5(1)−(−3)(6) = 5−(−18) = 23. Similarly, deleting the ﬁrst row and third
column to ﬁnd A13we get
▯ ▯
M = detA = det 4 5 = 4(−3) − 2(5) = −12 − 10 = −22
13 13 2 −3
To ﬁnd a cofactor of A, we ﬁnd the corresponding minor and multiply it by −1 raised to the power
i + j (so that we multiply by −1 only if one of the row number and column number is odd and the
other is even). We get:
C 23 = (−1)2+3M 23
5
= (−1) detA 23
▯ ▯
= (−1)det −3 2
2 −3
= −[(−3)(−3) − 2(2)]
= −(9 − 4)
= −5
3+1
C 31 = (−1) M 31
= (−1) detA
31
▯ ▯
2 1
= (1)det
5 6
= (2)(6) − 5(1)
= 12 − 5
= 7
Notice that we don’t, yet, know how to ﬁnd minors or cofactors of a square matrix of order
4, or of any order larger than 3, because we haven’t yet deﬁned how to ﬁnd the determinant of a
square matrix of order larger than 2. For instance, the 1,1-minor of a 4 × 4 matrix A is simply the
determinant of 11, but since 11is a 3 × 3 matrix, we don’t know how to calculate that. But
we do, now, know everything we need to in order to deﬁne this. That is, we have assembled all
the pieces to allow us to recursively deﬁne how to ﬁnd the determinant of a square matrix of order
bigger than 2. We deﬁne it in terms of the cofactors of certain submatrices. The deﬁnition we will 132 Unit 10
now state will give a particular way of calculating detA. But then afterwards, we’ll have a theorem
that shows other, similar, ways of calculating it, using a diﬀerent series of cofactors. There are, in
fact, 2n diﬀerent ways that we could calculate the determinant of an n × n matrix, using cofactors
determined by any one particular row or column. In our deﬁnition, we’ll use the cofactors of row 1.
We multiply each entry of this row by the cofactor with the same index. (That is, 1jlbyply a
C1j) And then we add them all up.
Deﬁnition of Determinant: Part Three
For any square matrix A of order n ≥ 3, the determinant of A is given by
detA = a 11 11+ a12 12+ ... +1nC 1n
That is, we multiply each entry of row 1 by the corresponding cofactor of A, which is to
say that we multiply that entry by the determinant of the submatrix obtained by deleting
the row and column in which that entry occurs, or the negative of that determinant if
▯
the row and column are not either both odd or both even. Usingo denote “do this
j=1
calculation for every value of j from 1 to n, and add them all up”, we have
▯n
detA = a1j−1)1+j detA1j
j=1
1 2 3
Example 10.5. Find detA, where A = 2 1 3 .
3 2 1
Solution:
We calculate detA as it says in the deﬁnition:
▯3 ▯3
detA = a C = a (−1) 1+jdetA
1j 1j 1j 1j
j=1 j=1
1+1 1+2 1+3
= a11−1) detA 11+ a12−1) detA 12+ a13−1) detA 13
▯ 1 3 ▯ ▯ 2 3 ▯ ▯ 2 1 ▯
= (1)(−1) det + (2)(−1) det + (3)(−1) det
2 1 3 1 3 2
= [(1)(1) − (2)(3)] − 2[(2)(1) − (3)(3)] + 3[(2)(2) − (3)(1)]
= (1 − 6) − 2(2 − 9) + 3(4 − 3)
= −5 − 2(−7) + 3(1) = −5 + 14 + 3 = 12
Notice that the (−1)jmultipliers made the sign alternate across the row. That is11ftermhe a
it is +1, then for th12term it is −1 and for the13term it is +1 again. So we could express
detA as
detA = a 11detA11− a12detA 12+ a13detA 13
1 −2 −4 5
0 3 0 0
Example 10.6. If A = , ﬁnd detA.
0 −1 3 2
0 4 −5 2 Unit 10 133
Solution: Again, we use the deﬁnition to express detA in terms of the determinants of certain
submatrices of A:
1+1 1+2 1+3 1+4
detA = a11−1) detA 11+ a12−1) detA 12+ a13−1) detA 13+ a14−1) detA14
3 0 0 0 0 0
= 1(−1) det −1 3 2 + (−2)(−1) det 0 3 2
4 −5 2 0 −5 2
0 3 0 0 3 0
4 5
+(−4)(−1) det 0 −1 2 + 5(−1) det 0 −1 3
0 4 2 0 4 −5
3 0 0 0 0 0
−1 3 2 0 3 2
= 1det − (−2)det
4 −5 2 0 −5 2
0 3 0 0 3 0
+(−4)det 0 −1 2 − 5det 0 −1 3
0 4 2 0 4 −5
Notice that once again the +’s and −’s resulting from the’s are alternating. Now, to ﬁnd
the determinant of each of those 3 × 3 submatrices of A, we need to use the deﬁnition again, to
express each in terms of determinants of 2 × 2 submatrices. Of course, this time the 1 and j indices
in the (−1)+jterm (as well as the1jand A1j terms) are the row and column numbers in the
submatrix whose determinant we are currently calculating, not the row and column numbers from
the original matrix.
3 0 0 0 0 0
detA = 1det −1 3 2 − (−2)det 0 3 2
4 −5 2 0 −5 2
0 3 0 0 3 0
+(−4)det 0 −1 2 − 5det 0 −1 3
0 4 2 0 4 −5
▯ ▯ ▯ ▯ ▯ ▯ ▯▯
= 1 3(−1)1+1det 3 2 + 0(−1)1+2det −1 2 + 0(−1)1+3det −1 3
−5 2 4 2 4 −5
▯ ▯ 3 2▯ ▯ 0 2 ▯ ▯ 0 3 ▯▯
−(−2) 0(−1) 1+1det + 0(−1)1+2det + 0(−1)1+3det
−5 2 0 2 0 −5
▯ ▯ ▯ ▯ ▯ ▯ ▯▯
1+1 −1 2 1+2 0 2 1+3 0 −1
+(−4) 0(−1) det + 3(−1) det + 0(−1) det
▯ ▯ 4 ▯ ▯ 0 2 ▯ ▯ 0 4 ▯▯
−1 3 0 3 0 −1
−5 0(−1)1+1det + 3(−1)1+2det + 0(−1)1+3det
4 −5 0 −5 0 4
Now, we have detA expressed in terms of determinants of 2 × 2 matrices, so for each of them we 134 Unit 10
▯ ▯
simply use the formula: det b = ad − cb.
c d
detA = 1{3(1)[3(2) − (−5)(2)] + 0(−1)[(−1)(2) − 4(2)] + 0(1)[(−1)(−5) − 4(3)]}
+2{0(1)[3(2) − (−5)(2)] + 0(−1)[0(2) − 0(2)] + 0(1)[0(−5) − 0(3)]}
−4{0(1)[(−1)(2) − 4(2)] + 3(−1)[0(2) − 0(2)] + 0(1)[0(4) − 0(−1)]}
−5{0(1)[(−1)(−5) − 4(3)] + 3(−1)[0(−5) − 0(3)] + 0(1)[0(4) − 0(−1)]}
= 1{3[6 − (−10)] − 0[−2 − 6] + 0[5 − 12]}
+2{0[6 − (−10)] − 0[0 − 0] + 0[0 − 0]}
−4{0[−2 − 8] − 3[0 − 0] + 0[0 − 0]}
−5{0[5 − 12] − 3[0 − 0] + 0[0 − 0]}
= [3(16) − 0(−8) + 0(−7)] + 2[0(16) − 0(0) + 0(0)]
−4[0(−10) − 3(0) + 0(0)] − 5[0(−7) − 3(0) + 0(0)]
= (48 − 0 + 0) + 2(0 − 0 + 0) − 4(0 − 0 + 0) − 5(0 − 0 + 0)
= 48 − 0 − 0 − 0
= 48
i+j
Notice that once again, for each determinant, the eﬀect of tterms is to make the +’s and
−’s alternate. And since we’re always starting with i = j = 1, the pattern always starts with +.
Well, that was a lot of work! But as long as we take it slowly and carefully, paying attention to
all the details of what we’re doing, none of it is diﬃcult. Notice, though, that the way the calculation
was expressed above, many calculations were done unnecessarily, because we knew they were going
to be multiplied by 0. Let’s restate that calculation, exactly the same, but taking advantage of those
zero multipliers to not bother expressing the calculations that don’t matter because they won’t be
used.
Exampl 10.6. Revisite:
1 −2 −4 5
0 3 0 0
If A = , ﬁnd detA.
0 −1 3 2
0 4 −5 2
Solution:
detA = a11−1)1+1 detA11+ a12−1) 1+2detA12+ a13−1) 1+3detA 13+ a14−1)1+4detA 14
3 0 0 0 0 0
2 3
= 1(−1) det −1 3 2 + (−2)(−1) det 0 3 2
4 −5 2 0 −5 2
0 3 0 0 3 0
+(−4)(−1) det 0 −1 2 + 5(−1) det 0 −1 3
0 4 2 0 4 −5
▯ ▯ ▯ ▯
1+1 3 2
= 1(1) 3(−1) det −5 2 − 0 + 0 + (−2)(−1){0 + 0 + 0}
▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯
1+2 0 2 1+2 0 3
+(−4)(1) 0 + 3(−1) det 0 2 + 0 + 5(−1) 0 + 3(−1) det 0 −5 + 0 Unit 10 135
so detA = 1{3(1)[3(2) − (−5)(2)]} − (−2)(0) + (−4){3(−1)[0(2) − 0(2)]} − 5{3(−1)[0(−5) − 0(3)]}
= 1{3[6 − (−10)]} + 0 − 4{(−3)[0 − 0]} − 5{−3(0 − 0)}
= 3(16) − 4[(−3)(0)] − 5[(−3)(0)]
= 48 − 0 − 0
= 48
Well, that was somewhat better! But what if we could have taken advantage of more 0’s earlier
on? Consider, for instance, the determinantfor the matrix A in that example. The transpose
of the matrix has 3 zeroes in its ﬁrst row. And that would mean that we would only need to calculate
the determinant of one 3 × 3 matrix, because all of the others will simply be multiplied by 0 and
therefore don’t need to be calculated. That would certainly be convenient.
Actually, we could use that in the calculation of detA as given, too. Because the cal

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