MATH 304
Linear Algebra
Lecture 7:
Evaluation of determinants.
The Vandermonde determinant.
Cramer’s rule.
Determinants
Determinant is a scalar assigned to each square matrix.
Notation. The determinant of a matrix
A = (aij )1≤i,j≤n is denoted det A or
a a ... a 1n 11 12
a a ... a 2n 21 22
.
..
.
.
.
.
. ... .
.
an1 an2 . . . ann Principal property: det A 6= 0 if and only if a
system of linear equations with the coefficient
matrix A has a unique solution. Equivalently,
det A 6= 0 if and only if the matrix A is invertible.
Explicit definition in low dimensions
a b
= ad − bc,
Definition. det (a) = a, c d
a11 a12 a13 a21 a22 a23 = a11 a22 a33 + a12 a23 a31 + a13 a21 a32 −
a31 a32 a33 −a13 a22 a31 − a12 a21 a33 − a11 a23 a32 .
+:
−:

*

∗
∗

∗

∗
*
∗
*
∗
∗
*
∗
 
∗
∗
 
∗ ,  ∗
*
*
 
*
∗
 
∗ ,  *
∗
∗
*
∗
∗
*
∗
∗
 
∗
∗
 
* ,  *
∗
∗
 
∗
*
 
∗ ,  ∗
*
∗
∗
∗
*
∗
∗
*

*

∗ .
∗

∗

* .
∗
Properties of determinants
Determinants and elementary row operations:
• if a row of a matrix is multiplied by a scalar r ,
the determinant is also multiplied by r ;
• if we add a row of a matrix multiplied by a scalar
to another row, the determinant remains the same;
• if we interchange two rows of a matrix, the
determinant changes its sign.
Properties of determinants
Tests for singularity:
• if a matrix A has a zero row then det A = 0;
• if a matrix A has two identical rows then
det A = 0;
• if a matrix has two proportional rows then
det A = 0.
Properties of determinants
Special matrices:
• det I = 1;
• the determinant of a diagonal matrix is equal to
the product of its diagonal entries;
• the determinant of an upper triangular matrix is
equal to the product of its diagonal entries.
Properties of determinants
Determinant of the transpose:
• If A is a square matrix then det AT = det A.
Columns vs. rows:
• if one column of a matrix is multiplied by a
scalar, the determinant is multiplied by the same
scalar;
• adding a scalar multiple of one column to
another does not change the determinant;
• interchanging two columns of a matrix changes
the sign of its determinant;
• if a matrix A has a zero column or two
proportional columns then det A = 0.
Row and column expansions
Given an n×n matrix A = (aij ), let Mij denote the
(n − 1)×(n − 1) submatrix obtained by deleting the
ith row and the jth column of A.
Theorem For any 1 ≤ k, m ≤ n we have that
n
X
det A =
(−1)k+j akj det Mkj ,
j=1
(expansion by kth row )
det A =
n
X
(−1)i+m aim det Mim .
i=1
(expansion by mth column)
Signs for row/column expansions

+
−

+


−
..
.
−
+
−
+
..
.
+
−
+
−
..
.
−
+
−
+
..
.

···
· · ·

· · ·


· · ·
...


1 2 3
Example. A = 4 5 6.
7 8 9
Expansion by


1 ∗ ∗
 ∗ 5 6
∗ 8 9
5
det A = 1 8
the 1st row:

 
∗ 2 ∗
∗ ∗
4 ∗ 6 4 5
7 8
7 ∗ 9
4 6
6 + 34
−
2
7 9
7
9

3
∗
∗
5 8
= (5 · 9 − 6 · 8) − 2(4 · 9 − 6 · 7) + 3(4 · 8 − 5 · 7) = 0.


1 2 3
Example. A = 4 5 6.
7 8 9
Expansion by the 2nd column:

 
 

1 ∗ 3
1 ∗ 3
∗ 2 ∗
 4 ∗ 6  ∗ 5 ∗  4 ∗ 6 
∗ 8 ∗
7 ∗ 9
7 ∗ 9
4 6
1 3
1 3
+ 5
det A = −2 7 9 − 84 6
7 9
= −2(4 · 9 − 6 · 7) + 5(1 · 9 − 3 · 7) − 8(1 · 6 − 3 · 4) = 0.

1 2 3
Example. A = 4 5 6.
7 8 9

Subtract the
the 3rd row:
1 2
4 5
7 8
1st row from the 2nd row and from
3 1 2 3 1 2 3 6 = 3 3 3 = 3 3 3 = 0
9 7 8 9 6 6 6
since the last matrix has two proportional rows.


1 2 3
Another example. B = 4 5 6 .
7 8 13
Let’s do some row reduction.
Add −4 times the 1st row to the 2nd row:
1 2 3 1 2 3
4 5 6 = 0 −3 −6 7 8 13 7 8 13 Add −7 times the 1st row to the 3rd row:
1 2 3 1 2 3
0 −3 −6 = 0 −3 −6 7 8 13 0 −6 −8 Expand the determinant by the 1st column:
1 2 3
−3 −6 0 −3 −6 = 1 −6 −8 0 −6 −8 Thus
1 2
−3 −6 = (−3) det B = −6 −8 −6 −8 1 2
= (−3)(−2)(−2) = −12.
= (−3)(−2) 3 4

2 −2 0 3
−5 3 2 1

Example. C = 
 1 −1 0 −3, det C =?
2 0 0 −1

Expand the determinant by the 3rd column:
2 −2 0 3 2 −2 3 −5 3 2 1 = −2 1 −1 −3 1 −1 0 −3 2 0 −1 2 0 0 −1 Add −2 times the 2nd row to the 1st row:
0 0 9
2 −2 3 det C = −2 1 −1 −3 = −2 1 −1 −3 2 0 −1 2 0 −1 Expand the determinant by the 1st row:
0 0 9
1 −1 det C = −2 1 −1 −3 = −2 · 9 2
0
2 0 −1 Thus
1 −1 = −18 · 2 = −36.
det C = −18 2 0
Problem. For what values of a will the following
system have a unique solution?

 x + 2y + z = 1
−x + 4y + 2z = 2

2x − 2y + az = 3
The system has a unique solution if and only if the
coefficient matrix is invertible.


1 2 1
A = −1 4 2, det A =?
2 −2 a
Add −2 times the
1 2
−1 4
2 −2
3rd column to the 2nd column:
1 1
0
1 2 = −1
0
2 a 2 −2 − 2a a Expand the determinant
1
0
0
det A = −1
2 −2 − 2a
by the 2nd column:
1 1 1
2 = −(−2 − 2a) −1
2
a
Hence det A = −(−2 − 2a) · 3 = 6(1 + a).
Thus A is invertible if and only if a 6= −1.
More properties of determinants
Determinants and matrix multiplication:
• if A and B are n×n matrices then
det(AB) = det A · det B;
• if A and B are n×n matrices then
det(AB) = det(BA);
• if A is an invertible matrix then
det(A−1 ) = (det A)−1 .
Determinants and scalar multiplication:
• if A is an n×n matrix and r ∈ R then
det(rA) = r n det A.
Examples




−1 2 1
1 0 0
X =  0 2 −2, Y = −1 3 0.
0 0 −3
2 −2 1
det X = (−1) · 2 · (−3) = 6, det Y = det Y T = 3,
det(XY ) = 6 · 3 = 18, det(YX ) = 3 · 6 = 18,
det(Y −1 ) = 1/3, det(XY −1 ) = 6/3 = 2,
det(XYX −1 ) = det Y = 3, det(X −1 Y −1 XY ) = 1,
det(2X ) = 23 det X = 23 · 6 = 48,
det(−3X T XY −4 ) = (−3)3 · 6 · 6 · 3−4 = −12.
The Vandermonde determinant
Definition. The Vandermonde determinant is
the determinant of the following matrix


1 x1 x12 · · · x1n−1


1 x2 x22 · · · x2n−1 


n−1 
2
V =
1
x
x
·
·
·
x
3
3
3

,
..
.
 ... ...
.
. . .. 
.


1 xn xn2 · · · xnn−1
where x1 , x2 , . . . , xn ∈ R. Equivalently,
V = (aij )1≤i,j≤n , where aij = xij−1 .
Examples.
1 x1 = x2 − x1 .
• 1 x2 1 x1 x12 1 x1
0
2
2
• 1 x2 x2 = 1 x2 x2 − x1 x2 1 x x2 1 x x2 − x x 3
3
1 3
3
3
1
0
0
x2 − x1 x22 − x1 x2 2
= 1 x2 − x1 x2 − x1 x2 = 2
x
−
x
x
−
x
x
3
1
1
3
3
1 x − x x2 − x x 3
1
1
3
3
1
1 x2 x2
= (x2 − x1 ) = (x2 − x1 )(x3 − x1 ) x3 − x1 x32 − x1 x3 1 x3 = (x2 − x1 )(x3 − x1 )(x3 − x2 ).
Theorem
1 x
1
1 x2
1 x3
... ...
1 xn
x12
x22
x32
···
···
···
...
..
.
xn2 · · ·
Y
=
(xj − xi ).
..
.
1≤i<j≤n
xnn−1 x1n−1
x2n−1
x3n−1
Corollary The Vandermonde determinant is not
equal to 0 if and only if the numbers x1 , x2 , . . . , xn
are distinct.
Let x1 , x2 , . . . , xn be distinct real numbers.
Theorem For any b1 , b2 , . . . , bn ∈ R there exists a
unique polynomial p(x) = a0 +a1 x+ · · · +an−1 x n−1
of degree less than n such that p(xi ) = bi ,
1 ≤ i ≤ n.

a0 + a1 x1 + a2 x12 + · · · + an−1 x1n−1 = b1



a0 + a1 x2 + a2 x22 + · · · + an−1 x2n−1 = b2
············



a0 + a1 xn + a2 xn2 + · · · + an−1 xnn−1 = bn
a0 , a1 , . . . , an−1 are unknowns. The coefficient
matrix is the Vandermonde matrix.
A general system of n linear equations in n variables:

a11 x1 + a12 x2 + · · · + a1n xn = b1



a21 x1 + a22 x2 + · · · + a2n xn = b2
⇐⇒ Ax = b,
·········



an1 x1 + an2 x2 + · · · + ann xn = bn
where

a11 a12
a a
 21 22
A =  ..
..
 .
.
an1 an2
 
 

b1
x1
. . . a1n





. . . a2n 
 b2 
x2 
, x =  .. , b =  .. 
. . . ... 
.
.

bn
xn
. . . ann
Cramer’s rule

a11 x1 + a12 x2 + · · · + a1n xn = b1



a21 x1 + a22 x2 + · · · + a2n xn = b2
⇐⇒ Ax = b
·
·
·
·
·
·
·
·
·



an1 x1 + an2 x2 + · · · + ann xn = bn
Theorem Assume that the matrix A is invertible.
Then the only solution of the system is given by
det Ai
xi =
, i = 1, 2, . . . , n,
det A
where the matrix Ai is obtained by substituting the
vector b for the ith column of A.
Example.

 x + 2y + 3z = 0
4x + 5y + 6z = 0

7x + 8y + 13z = 1
Augmented matrix of the system:


1 2 3 0
(A | b) = 4 5 6 0
7 8 13 1
As obtained earlier in this lecture, det A = −12.
Since det A 6= 0, there exists a unique solution of
the system.
Example.

 x + 2y + 3z = 0
4x + 5y + 6z = 0

7x + 8y + 13z = 1


1 2 3 0
4 5 6 0
7 8 13 1
By Cramer’s rule,
0 2 3 0 5 6 2 3
1 8 13 5 6 −3
1
=
x=
=
=
1 2 3 −12
−12 4
4 5 6 7 8 13 Example.

 x + 2y + 3z = 0
4x + 5y + 6z = 0

7x + 8y + 13z = 1
By Cramer’s
1
4
7
y=
1
4
7
rule,


1 2 3 0
4 5 6 0
7 8 13 1
0 3 1 3
0 6 − 4 6
1 13 6
1
=
=
=−
−12
−12
2
2 3 5 6 8 13 Example.

 x + 2y + 3z = 0
4x + 5y + 6z = 0

7x + 8y + 13z = 1


1 2 3 0
4 5 6 0
7 8 13 1
By Cramer’s rule,
1 2 0
4 5 0
1 2
7 8 1
4 5
−3
1
=
z=
=
=
1 2 3 −12
−12 4
4 5 6 7 8 13 System of linear equations:

 x + 2y + 3z = 0
4x + 5y + 6z = 0

7x + 8y + 13z = 1
Solution: (x, y , z) =
1 1
1
.
,− ,
4
2 4
Determinants and the inverse matrix
Given an n×n matrix A = (aij ), let Mij denote the
(n − 1)×(n − 1) submatrix obtained by deleting the ith row
and the jth column of A.
e = (αij )
The cofactor matrix of A is an n×n matrix A
i+j
defined by αij = (−1) det Mij .
e T A = AA
e T = (det A)I .
Theorem A
Corollary If det A 6= 0 then the matrix A is invertible and
eT .
A−1 = (det A)−1 A
e T = (det A)I means that
AA
Pn
k+j
akj det Mkj = det A for all k,
j=1 (−1)
Pn
k+j
amj det Mkj = 0 for m =
6 k.
j=1 (−1)