### 3.1

```3.1 Determinants
The Laplace Expansion
What’s the use?
• All square matrices have a real number determinant
• We can use the determinant to see if the matrix is
invertible.
• We can use the determinant to find the inverse.
• We can also use determinants to solve systems of
equations (using Cramer’s Rule)
Definition
• Given a (1 x 1) matrix A = [a]:
•
•
•
•
det[a] = a
We know A is invertible as long as a ≠ 0.
So for a (1x1) A is invertible iff a≠0.
A-1 = [1/a]
Definition for (2x2)
• Recall what the inverse of a (2x2) matrix was:
a b
A  

c d 
1 d  b


A 
ad  bc c a 
1
• The inverse clearly exists as long as (ad - bc) ≠ 0.
• We call det A = ad - bc
• Again, A is invertible iff det A ≠ 0
Minors and Cofactors
• Mij(A) = (i,j) minor of an n x n matrix, A:
• is the determinant of the (n-1)x(n-1) matrix
formed from A by deleting row i and column j.
• Cij(A) = (i,j) cofactor of A
• = (-1)i+jMij(A)
• C = ± M depending on row and column
…continued
• Since Cij(A) = (-1)i+jMij(A), note the simple
method we can use for finding the sign to apply to
M to get C based on the position of cofactor in A:








...
 ...

 ...

   ...

   ...
... ... ... ...





Example:
• Find the minors and cofactors of positions (2,1),(2,2)
and (2,3).
2 3 1 


A  0 1 2 


3 1 4
Determinant of an (nxn), n≥2
• Laplace Expansion:
• We can use the Laplace Expansion along any row or
column to find the determinant of (n x n) mtx, A:
– Expansion along rowi:
det(A) = ai1Ci1(A) + ai2Ci2(A) + … + ainCin(A)
– Expansion along column j:
det(A) = a1jC1j(A) + a2jC2j(A) + … + anjCnj(A)
Example
• Find the determinant of our earlier matrix:
2 3 1 


A  0 1 2 


3 1 4
Example
• Find the determinant of the following matrix:
0 3 0

2 4 1
B  
1 1 3
0 3 2

0

0

1
1

• Note that if we have a matrix with a row or column of
zeros, the determinant is 0.
• Recall also that we can apply elementary row operations
to matrix A to create a row of zeros in some matrices.
• We can create some zeros in a row by applying
elementary row operations in all matrices.
• The question is, what effect will applying the row
operations have on the determinant.
Theorem 2
• Let A be an (n x n) matrix.
1 If A has a row or column of zeros, det A = 0.
2 If two distinct rows (or columns) of A are interchanged,
the determinant of the resulting matrix is (- det A).
3 If a row (or column) of A is multiplied by a constant, u, the
determinant of the resulting matrix is u(det A).
4 If two distinct rows (or columns) of A are identical,
det A = 0.
5 If a mult of a row of A is added to a diff’t row, the
determinant of the resulting mtx is det A (cols too).
Proof
• Property 2: Proof by induction,
• Prove for n=2:
a11
A  

a21
a12 
,det A  a11 a22  a21 a12
a22 

a21
B  

a11
a22 
, det B  a21 a12  a11 a22   det A
a12 

Prop 2 (cont)
• For A = (nxn),n>2, we swap rows c and d to get B.
– In finding det A and det B, we expand on a row other than c
or d, we get the same coefficients w/ same signs, and the
minors are the same except rows c and d are swapped.
– We continue this process, always getting the same
coefficients w/ the same signs until we get down to (2x2)’s
– These (2x2)’s will be c 
 
d 
– We have shown that for (2x2)’s, det B = -det A, so for each
of our terms (which have same coeff’s so far), the det B = detA, so all of our terms will simply have opp signs.
Proof of Property 4
• If two distinct rows (or cols) of A are same, det A=0
–
–
–
–
First swap the two rows that are the same to get B
B = A so det A = det B
Also, by prop. 2, det B = - det A
So det A = - det A which can only happen if det A = 0. •
Proof of property 5
• Proof: B is obtained from A by adding k times row c
to row d.
– The cofactors of this row do not depend on their values, only
on their position and on the rest of the matrix, so they remain
the same. So Cdj (B) = Cdj (A)
n
– If we expand along row d of B: det B  (a  ka )C (B)

dj
cj
dj
j 1
n
n
j 1
j 1
  adj Cdj (A)  k  acj Cdj (A)
 det A  k det C
Continued...
• This matrix C is just obtained from A by replacing row d by
row c since the coefficients are from row c and we expand along
d so that row c remains in the minor matrix.
• Therefore, matrix C has two identical rows, and by property 4,
det C=0
• Therefore, det B=det A•
Finding determinants is easier
• Now we can simplify the matrices of which we need to find the
determinant:
• row or column reduce to create more zeros
• swapping rows/cols changes sign
• adding multiples of a row/col to another doesn’t change
• factor out a common factor from a row or column
• pulls out as a multiplier on the determinant
Examples
2 3 1
0 1 1
2 3 4
a

if ...det d

g
b
e
h
c 

f  2

i 
2g 2h

find...det d  a e  b

 g
h
2i 

f  c

i 
Examples
• Find all x such that det A = 0 (making A not invertible).
1 x x
A x 1 x
x x 1
Examples
• show that:
1
a1
2
a1
1
a2
2
a2
1
a3  a3  a2 a3  a1 a2  a1 
2
a3
• Matrix of this type is called a Vandermonde matrix.
• Can extend formula to (n x n) case
Theorem 3
• If A is (n x n), then det(uA) = un det A for any u (try it…)
• Proof:
R1 
uR1 
 
 
R2 
uR2 
A   ,,,,uA   
... 
... 
R 
uR 
 n 
 n 
 R1 
 R1 
R1 
 
 
 
R
uR2  2 R2 
n  2 
det (uA)  u det   u   ...  u  
 ... 
 ... 
... 
uR 
uR 
R 
 n 
 n 
 n 
Triangular matrices
• Lower Triangular: all entries above main diagonal are zero
• Upper Triangular: all entries below main diagonal are zero
• Note what happens with the determinant:
a

0

0
0

d 

g 

0 h i 
0 0 j 

b c
e f
• Theorem 4: the determinant of a triangular matrix is the
product of the entries on the main diagonal.
This makes life easy!
• Can we carry all matrices to triangular form using row
operations? Of course!
• So to find the determinant, reduce to triangular form noting
the row operations you have used along the way.
• Remember:
• when you swap rows, switch sign of determinant
• when you mult a row times a value, the det of what
remains must be multiplied by recipricol
• when you add a mult of one row to another, nothing
changes.
Continued..
• Note that if when you reduce, you end up with a row of zeros,
the determinant will be zero (as it should be since the matrix
will not be invertible.
Theorem 5
• Given the following block matrices w/ A and B square:
A
det 
0
X 
 det A det B
B 
A 0 
 det A det B
det 
Y B 
Proof of Theorem 5(ind)
• show true for A (1 x 1)
A
det 
0
a11
X 
 det 
B 
0
X 
 a11 M11  a11 det B  det A det B
B 
Proof of Theorem 5(ind)
• assume true for A (k x k)
A
det 
0
X 
 det A det B
B 
For A (k x k)
• Prove true for A ((k+1) x (k+1))
A X 
T  

0 B 
det T  a11 M11  a21 M21  ... ak 1 Mk 1
Proof of Theorem 5(ind)
• Mi1 is the det of submatrix (Si1(T)) formed by deleting the
ith row and 1st col of T
Si1 (A) Xi  Since only getting rid of row of A

Si1 (T )  
B  and same row of X, and col 1 ofA
 0
and col 1 of 0.
• Si1 (A) will be (k x k).
• Mi1(T) =det(Si1(T)) = det(Si1(A))detB (by induction)
=Mi1(A)detB
Proof of Theorem 5(ind)
det T  a11 M11  a21 M21  ... ak 1 Mk 1
 a11 M11 (A) det B  a21 M21 (A)det B  ... ak 1 Mk 1 (A)det B
 det B(a11 M11 (A)  a21 M21 (A)  ... ak 1 Mk 1 ( A))
 det B(det A) •
```