• 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)

• 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]

• Recall what the inverse of a (2x2) matrix was:
A 
 a b c d 
A
 1 
1 ad  bc   d  b c a 
• The inverse clearly exists as long as (ad - bc) ≠ 0.
• We call det A = ad - bc
• Again, A is invertible iff det A ≠ 0

• M ij
(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.
• C ij
(A) = (i,j) cofactor of A
• = (-1) i+j M ij
(A)
• C = ± M depending on row and column

• Since C ij
(A) = (-1) i+j M ij
(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:



   
   
   
   
...
...
...
...
 ... ... ... ... ...






• Find the minors and cofactors of positions (2,1),(2,2) and (2,3).
A 


2 3 1
0 1 2
 3  1 4




• 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) = a i1
C i1
(A) + a i2
C i2
(A) + … + a in
C in
(A)
– Expansion along column j: det(A) = a
1j
C
1j
(A) + a
2j
C
2j
(A) + … + a nj
C nj
(A)

• Find the determinant of our earlier matrix:
A 


2 3 1
0 1 2
 3  1 4




• Find the determinant of the following matrix:
B 


0
2

4
3 0
1
0
0
1  1 3 1

0 3  2 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.

• 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).

• Property 2: Proof by induction,
A
B
• Prove for n=2:

 a a a a
11
21
21
11 a
12 a
22


,det A  a
11 a
22
 a
21 a
12 a
22 a
12


, det B  a
21 a
12
 a
11 a
22
  det A

• 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.

• 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: B is obtained from A by adding k times row c to row d.
– So row d of B is now: [a d1
+ ka c1
, a d2
+ ka c2
,…,a dn
+ ka cn
)
– 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 C dj
(B) = C dj
(A)
– If we expand along row d of B: det B  n
 j  1
( a dj
 ka cj
) C dj
( B )
 n
 j  1 a dj
C dj
( A )  k n
 j  1 a cj
C dj
( A )
 det A  k det C

• 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 •

• 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

2 3 1
0 1  1
2 3 4
if ...det


 a b c d e g h i f




 2 find ...det


 2 g  2 h  2 i d  a e  b f  c

 g h i

• Find all x such that det A = 0 (making A not invertible).
A 
1 x x x 1 x x x 1

• show that: a
1 a
1
2
1 1 1 a a
2
2
2 a
3 a
3
2

 a
3
 a
2
  a
3
 a
1
  a
2
 a
1

• Matrix of this type is called a Vandermonde matrix.
• Can extend formula to (n x n) case

• If A is (n x n), then det(uA) = u n det A for any u (try it…)
• Proof:
A 





R
1
R
...
2
R n





,,,, uA 




 uR
1 uR
2
...
uR n




 det( uA )  u det





R
1 uR
2
...





 u 2





R
R
2
...
1 uR n uR n





 ...
 u n





R
1
R
...
2





R n

• 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 b c d
0 e f
0 0 h i g

 0 0 0 j
• Theorem 4: the determinant of a triangular matrix is the product of the entries on the main diagonal.

• 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.

• 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.

• Given the following block matrices w/ A and B square: det

A X
0 B

 det A det B det

A 0
Y B

 det A det B

• show true for A (1 x 1) det 
A X
 0 B

 det a
11
X
 0 B

 a
11
M
11
 a
11 det B  det A det B

• assume true for A (k x k) det 
A X
 0 B

 det A det B For A (k x k)
• Prove true for A ((k+1) x (k+1))
T 



A X
0 B


 det T  a
11
M
11
 a
21
M
21
 ...
 a k  1
M k  1

• M i1 is the det of submatrix (S i1
(T)) formed by deleting the ith row and 1st col of T
S i 1
( T ) 



S i 1
( A ) X
0 B i



Since only getting rid of row of A and same row of X, and col 1 ofA and col 1 of 0.
• S i1
(A) will be (k x k).
• M i1
(T) =det(S i1
(T)) = det(S i1
(A))detB (by induction)
=M i1
(A)detB

det T  a
11
M
11
 a
21
M
21
 ...
 a k  1
M k  1
 a
11
M
11
( A ) det B  a
21
M
21
( A )det B  ...
 a k  1
M k  1
( A )det B
 det B ( a
11
M
11
( A )  a
21
M
21
( A )  ...
 a k  1
M k  1
( A ))
 det B (det A )
•