DeterminantsApplications of determinants
MTH 215: Introduction to Linear AlgebraChapter 3
Jonathan A. Chavez Casillas1
1University of Rhode IslandDepartment of Mathematics
February 13, 2019
Jonathan Chavez
DeterminantsApplications of determinants
1 DeterminantsCofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices
2 Applications of determinantsA Formula for the InverseCramer’s RulePolynomial Interpolation
Jonathan Chavez
DeterminantsApplications of determinants
Cofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices
Determinant of a 2 × 2 Matrix
Definition
Let A =[
a bc d
]. Then the determinant of A is defined as
det A = ad − bc
What about the determinant of an n × n matrix for other values of n?
Notation. For det[
a bc d
], we often write
∣∣∣∣ a bc d
∣∣∣∣, i.e., use vertical bars instead of square
brackets.
Jonathan Chavez
DeterminantsApplications of determinants
Cofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices
How do we find the determinant of an n × n matrix?The determinant of an n × n matrix is defined recursively, using determinants of (n − 1)× (n − 1)submatrices, and requires some new definitions and notation.
DefinitionLet A = [aij ] be an n × n matrix. The sign of the (i , j) position is (−1)i+j .Thus the sign is 1 if (i + j) is even, and −1 if (i + j) is odd.
Jonathan Chavez
DeterminantsApplications of determinants
Cofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices
The Minor of a MatrixDefinitionLet A = [aij ] be an n × n matrix. The ij th minor of A, denoted as minor (A)ij , is the determinant of then − 1× n − 1 matrix which results from deleting the i th row and the j th column of A.
A =
a11 a12 · · · a1j · · · a1na21 a22 · · · a2j · · · a2n
......
......
ai1 ai2 · · · aij · · · ain...
......
...
an1 an2 · · · anj · · · ann
For any matrix A, minor (A)ij is found by first removing the i th row and j th column, and taking thedeterminant of the remaining matrix.
Jonathan Chavez
DeterminantsApplications of determinants
Cofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices
ExampleFind minor(A)12 of
A =
[1 1 32 4 15 2 6
]
SolutionFirst, remove the 1st row and 2nd column from A.
A =
[1 1 32 4 15 2 6
]
The resulting matrix is A =[
2 15 6
]. Using our previous definition, we calculate the determinant of this matrix as:
(2)(6) − (5)(1) = 12 − 5 = 7
Therefore, minor(A)12 = 7.
Jonathan Chavez
DeterminantsApplications of determinants
Cofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices
The Cofactors of a MatrixDefinitionThe ij th cofactor of A is
cof(A)ij = (−1)i+j minor (A)ij
Example (continued)Find cof(A)12 if
A =
[1 1 32 4 15 2 6
]
SolutionBy the definition, we know that cof(A)12 = (−1)1+2minor (A)12. From earlier, we know that minor (A)12 = 7.
Therefore, cof(A)12 = (−1)1+2minor (A)12 = (−1)37 = −7
Jonathan Chavez
DeterminantsApplications of determinants
Cofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices
Cofactor ExpansionUsing these definitions, we can now define the determinant of the n × n matrix A:
Definitiondet A = a11cof(A)11 + a12cof(A)12 + a13cof(A)13 + · · ·+ a1ncof(A)1nThis is called the cofactor expansion of det A along row 1.
In other words,
det (A) =n∑
j=1
aij cof (A)ij =n∑
i=1
aij cof (A)ij
The first formula consists of expanding the determinant along the i th row and the second expands thedeterminant along the j th column.
Cofactor expansion is also called Laplace Expansion.
Jonathan Chavez
DeterminantsApplications of determinants
Cofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices
Cofactor ExpansionExample
Let A =
[1 1 32 4 15 2 6
]. Find det A.
SolutionUsing cofactor expansion along row 1,
det A = 1cof11(A) + 1cof12(A) + 3cof13(A)
= 1(−1)2∣∣∣ 4 1
2 6
∣∣∣ + 1(−1)3∣∣∣ 2 1
5 6
∣∣∣ + 3(−1)4∣∣∣ 2 4
5 2
∣∣∣= 1(24 − 2) − 1(12 − 5) + 3(4 − 20)= 22 − 7 + 3(−16)= 22 − 7 − 48= −33
Jonathan Chavez
DeterminantsApplications of determinants
Cofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices
Solution (continued)
A =
[1 1 32 4 15 2 6
]
Now try cofactor expansion along column 2.
We get the same answer!
Jonathan Chavez
DeterminantsApplications of determinants
Cofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices
The Determinant is Well Defined
TheoremThe determinant of an n × n matrix A can be computed using cofactor expansion along any row or columnof A.
What is the significance of this theorem?This theorem allows us to choose any row or column for computing cofactor expansion, which gives us theopportunity to save ourselves some work!
Jonathan Chavez
DeterminantsApplications of determinants
Cofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices
Problem
Let A =
0 1 −2 15 0 0 70 1 −1 03 0 0 2
. Find det A.
SolutionCofactor expansion along row 1 yields
whereas cofactor expansion along, row 3 yields
so, in the first case we must compute three cofactors, but just two cofactors in the second case.
Jonathan Chavez
DeterminantsApplications of determinants
Cofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices
Solution (continued)Therefore, we can save ourselves some work by using cofactor expansion along row 3 rather than row 1.
A =
0 1 −2 15 0 0 70 1 −1 03 0 0 2
Each of the two determinants above can easily be evaluated using cofactor expansion along column 2.Jonathan Chavez
DeterminantsApplications of determinants
Cofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices
Solution (continued)
Jonathan Chavez
DeterminantsApplications of determinants
Cofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices
A Row or Column of Zeros
ExampleLet
A =
−8 1 0 −45 7 0 −7
12 −3 0 8−3 11 0 2
.
By choosing column 3 for cofactor expansion, we get det A = 0, i.e.,
det A = 0× cof(A)13 + 0× cof(A)23 + 0× cof(A)33 + 0× cof(A)43 = 0.
Important FactIf A is an n × n matrix with a row or column of zeros, then det A = 0.
Jonathan Chavez
DeterminantsApplications of determinants
Cofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices
Determinants of Inverses
TheoremAn n × n matrix A is invertible if and only if det A , 0. In this case,
det(A−1) = 1det A .
Jonathan Chavez
DeterminantsApplications of determinants
Cofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices
Example (Illustration of the above Theorem.)
Let A =[
2 −3−6 4
].
Then det A = 2(4)− (−3)(−6) = −10.
Since det A , 0, the above Theorem tell us that A is invertible, and that det(A−1) should be equal to − 110 .
We can verify this directly.
Using the formula for the inverse of a 2× 2 matrix
A−1 = 1−10
[4 36 2
].
Therefore, det A−1 =(− 1
10
)2∣∣∣∣ 4 3
6 2
∣∣∣∣ =(− 1
10
)2(8− 18) = 1
100 × (−10) = − 110 .
Jonathan Chavez
DeterminantsApplications of determinants
Cofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices
Problem
Find all values of c for which A =
[c 1 00 2 c
−1 c 5
]is invertible.
Solution
Jonathan Chavez
DeterminantsApplications of determinants
Cofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices
Determinants of Products and Transposes
TheoremLet A and B be n × n matrices. Then
det (AB) = det A× det B.
TheoremIf A is an n × n matrix, then the determinant of its transpose is given by
det(AT ) = det A.
Jonathan Chavez
DeterminantsApplications of determinants
Cofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices
ProblemSuppose A, B and C are 4 × 4 matrices with
det A = −1, det B = 2, and det C = 1.
Find det(2A2(B−1)(CT )3B(A−1)).
Solution
Jonathan Chavez
DeterminantsApplications of determinants
Cofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices
ProblemSuppose A is a 3 × 3 matrix. Find det A and det B if
det(2A−1) = −4 = det(A3(B−1)T ).
Solution
Jonathan Chavez
DeterminantsApplications of determinants
Cofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices
ProblemLet A be an n × n matrix. Find all conditions that ensure det(−A) = det A.
Solution
Jonathan Chavez
DeterminantsApplications of determinants
Cofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices
Elementary Row Operations and DeterminantsExampleLet
A =
[2 0 −30 4 01 0 −2
].
Computing det A by cofactor expansion along row (or column) 2 yields
det A = 4(−1)4∣∣∣ 2 −3
1 −2
∣∣∣ = 4(−1) = −4.
Let B1, B2, and B3 be obtained from A by interchanging rows 1 and 2, multiplying row 3 by −3, and adding twice row 1 torow 3, respectively, i.e.,
B1 =
[2 0 −31 0 −20 4 0
], B2 =
[2 0 −30 4 0
−3 0 6
], B3 =
[2 0 −30 4 05 0 −8
].
We are interested in how elementary operations affect the determinant. Computing det B1, det B2, and det B3:
Jonathan Chavez
DeterminantsApplications of determinants
Cofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices
Example (continued)
det B1 = 4(−1)5∣∣∣∣ 2 −3
1 −2
∣∣∣∣ = (−4)(−1) = 4 = (−1) det A.
det B2 = 4(−1)4∣∣∣∣ 2 −3−3 6
∣∣∣∣ = 4(12− 9) = 4× 3 = 12 = −3 det A.
det B3 = 4(−1)4∣∣∣∣ 2 −3
5 −8
∣∣∣∣ = 4(−16 + 15) = 4(−1) = −4 = det A.
The general effects of elementary row operations on the determinant are summarized in the next theorem.
Jonathan Chavez
DeterminantsApplications of determinants
Cofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices
TheoremLet A be an n × n matrix and B be an n × n matrix as defined below.
1 Let B be a matrix which results from switching two rows of A. Then det (B) = − det (A) .
2 Let B be a matrix which results from multiplying some row of A by a scalar k. Thendet (B) = k det (A) .
3 Let B be a matrix which results from adding a multiple of a row to another row. Thendet (A) = det (B).
4 If A contains a row which is a multiple of another row of A, then det (A) = 0
An analogous theorem holds for elementary column operation. If A is a matrix, then an elementary columnoperation on A is simply the corresponding elementary row operation performed on the transpose of A, AT .
Jonathan Chavez
DeterminantsApplications of determinants
Cofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices
Determinants and Scalar MultiplicationProblemSuppose A is a 3 × 3 matrix with det A = 7. What is det(−2A)?
SolutionWrite A =
[a11 a12 a13a21 a22 a23a31 a32 a33
]. Then
Jonathan Chavez
DeterminantsApplications of determinants
Cofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices
Solution (continued)Think about the matrix −2A as the matrix obtained from A be multiplying each of its rows by −2. Thisinvolves three elementary row operations, each of which contributes a factor of −2 to the determinant, andthus det A = (−2)× (−2)× (−2)× 7 = (−2)3 × 7.
TheoremIf A is an n × n matrix and k is any scalar, then
det(kA) = kn det A.
Jonathan Chavez
DeterminantsApplications of determinants
Cofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices
Computing the Determinant
Example
det
[ −3 5 −61 −1 32 −4 1
]=
∣∣∣∣∣ 0 2 31 −1 32 −4 1
∣∣∣∣∣ row 1 + 3×(row 2)
=
∣∣∣∣∣ 0 2 31 −1 30 −2 −5
∣∣∣∣∣ row 3 − 2×(row 2)
= (1)(−1)2+1∣∣∣∣ 2 3−2 −5
∣∣∣∣ cofactor expansion: column 1
= −(−10 + 6)= 4.
Jonathan Chavez
DeterminantsApplications of determinants
Cofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices
Example
det
3 1 2 4−1 −3 8 0
1 −1 5 51 1 2 −1
=
Jonathan Chavez
DeterminantsApplications of determinants
Cofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices
Problem
If det
[a1 a2 a3b1 b2 b3c1 c2 c3
]= 4, find det
[−b1 −b2 −b3
a1 + 2b1 a2 + 2b2 a3 + 2b33c1 3c2 3c3
].
Solution∣∣∣∣∣ −b1 −b2 −b3a1 + 2b1 a2 + 2b2 a3 + 2b3
3c1 3c2 3c3
∣∣∣∣∣ =
Jonathan Chavez
DeterminantsApplications of determinants
Cofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices
Problem
Let A =
[2 3 53 5 91 2 4
]. Find det A.
SolutionNotice:
row1 + row3 − row2 =[
0 0 0]
Hence
Jonathan Chavez
DeterminantsApplications of determinants
Cofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices
ProblemLet
A =
[ a b cp q rx y z
]and B =
[ 2a + p 2b + q 2c + r2p + x 2q + y 2r + z2x + a 2y + b 2z + c
].
Show that det B = 9 det A.
Jonathan Chavez
DeterminantsApplications of determinants
Cofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices
Solution
Jonathan Chavez
DeterminantsApplications of determinants
Cofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices
Using Row OperationsProblemFind det(A) if:
A =
[5 1 21 3 22 6 0
]
Solution
Jonathan Chavez
DeterminantsApplications of determinants
Cofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices
Solution (continued)
Jonathan Chavez
DeterminantsApplications of determinants
Cofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices
Determinants of a Triangular Matrices
Definitions1 An n × n matrix A is called upper triangular if all entries below the main diagonal are zero.2 An n × n matrix A is called lower triangular if all entries above the main diagonal are zero.3 An n × n matrix A is called triangular if it is upper triangular or lower triangular.
TheoremIf A = [aij ] is an n × n triangular matrix, then
det A = a11 × a22 × a33 × · · · × ann,
i.e., det A is the product of the entries of the main diagonal of A.
Jonathan Chavez
DeterminantsApplications of determinants
Cofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices
Example
det
[ 1 2 30 5 60 0 9
]= 1× det
[5 60 9
]= 1× 5× det
[9]
= 1× 5× 9= 45.
Notice that 45 is the product of the entries on the main diagonal.[ 1 2 30 5 60 0 9
]
Jonathan Chavez
DeterminantsApplications of determinants
A Formula for the InverseCramer’s RulePolynomial Interpolation
Table of Contents
1 DeterminantsCofactors and n × n DeterminantsProperties of the DeterminantObtaining determinants using Row OperationsTriangular Matrices
2 Applications of determinantsA Formula for the InverseCramer’s RulePolynomial Interpolation
Jonathan Chavez
DeterminantsApplications of determinants
A Formula for the InverseCramer’s RulePolynomial Interpolation
The Cofactor Matrix
DefinitionLet A = [aij ] be an n × n matrix. The cofactor matrix of A, is the matrix
[cof(A)ij ] ,
i.e., the matrix whose (i , j)-entry is the (i , j)-cofactor of A.
Reminder: the (i , j)-cofactor
cof(A)ij = (−1)i+jminor (A)ij ,
where minor (A)ij is the determinant of the matrix obtained from A by deleting row i and column j.
Jonathan Chavez
DeterminantsApplications of determinants
A Formula for the InverseCramer’s RulePolynomial Interpolation
ProblemFind the cofactor matrix [cof(A)ij ] of the matrix
A =
[4 0 31 9 70 6 4
].
SolutionFor each i and j, 1 ≤ i, j ≤ 3, we need to compute cof(A)ij , so there are 9 cofactors to compute.
cof(A)11 = (−1)1+1 det A11 =∣∣∣ 9 7
6 4
∣∣∣ = 9 × 4 − 6 × 7 = 36 − 42 = −6.
cof(A)12 = (−1)1+2 det A12 =∣∣∣ 1 7
0 4
∣∣∣ = −(4 − 0) = −4.
cof(A)13 = (−1)1+3 det A12 =∣∣∣ 1 9
0 6
∣∣∣ = (6 − 0) = 6.
Computing the six remaining cofactors results in the cofactor matrix[−6 −4 618 16 −24
−27 −25 36
].
Jonathan Chavez
DeterminantsApplications of determinants
A Formula for the InverseCramer’s RulePolynomial Interpolation
Definition (The Adjugate)If A is an n × n matrix, then the adjugate of A is defined by
adj A =[
cof(A)ij]T
,
where cof(A)ij is the (i, j)-cofactor of A, i.e., adj A is the transpose of the cofactor matrix.
Example
A =
[4 0 31 9 70 6 4
], has cofactor matrix
[−6 −4 618 16 −24
−27 −25 36
].
Therefore, the adjugate of A is
adj A =
[−6 −4 618 16 −24
−27 −25 36
]T
=
[−6 18 −27−4 16 −25
6 −24 36
].
Jonathan Chavez
DeterminantsApplications of determinants
A Formula for the InverseCramer’s RulePolynomial Interpolation
Problem
Find adj A when A =
[ 2 1 35 −7 13 0 −6
].
Solution
adj A =
[ 42 6 2233 −21 1321 3 −19
].
Jonathan Chavez
DeterminantsApplications of determinants
A Formula for the InverseCramer’s RulePolynomial Interpolation
The Adjugate of a 2 × 2 MatrixExampleLet A be a 2 × 2 matrix, say A =
[a bc d
].
Then
adj A =[
cof(A)11 cof(A)12cof(A)21 cof(A)22
]T=
[(−1)2d (−1)3c(−1)3b (−1)4a
]T
=[
d −b−c a
].
We’ve seen this matrix before: if det A , 0, then
A−1 =1
det A
[d −b
−c a
]=
1det A
adj A.
This actuallyholds true for all matrices.
Jonathan Chavez
DeterminantsApplications of determinants
A Formula for the InverseCramer’s RulePolynomial Interpolation
The Adjugate Formula
TheoremIf A is an n × n matrix, then
A(adj A) = (det A)I = (adjA)A.
Furthermore, if det A , 0, then we get a formula for A−1, i.e.,
A−1 = 1det A adj A.
Inverting a matrix using the adjugateExcept in the case of a 2× 2 matrix, the adjugate formula is a very inefficient method for computing theinverse of a matrix; the matrix inversion algorithm is much more practical. However, the adjugate formulais of theoretical significance.
Jonathan Chavez
DeterminantsApplications of determinants
A Formula for the InverseCramer’s RulePolynomial Interpolation
Using the Adjugate to Find the Inverse of a MatrixExample
Let A =
[ 4 0 31 9 70 6 4
]. As we saw earlier, det A = −6 , 0, so A is invertible, and
adj A =
[ −6 18 −27−4 16 −25
6 −24 36
].
A−1 = 1det A adj A = 1
−6
[ −6 18 −27−4 16 −25
6 −24 36
]=
1 −3 9
223 − 8
3256
−1 4 −6
.
You can check this by computing AA−1.
Jonathan Chavez
DeterminantsApplications of determinants
A Formula for the InverseCramer’s RulePolynomial Interpolation
ProblemLet A be an n × n invertible matrix. Show that det(adj A) = (det A)n−1.
Solution
Even if A is not invertible, det(adj A) = (det A)n−1, but the proof is more complicated.
Jonathan Chavez
DeterminantsApplications of determinants
A Formula for the InverseCramer’s RulePolynomial Interpolation
Cramer’s Rule
If A is an n × n invertible matrix, then the solution to AX = B can be given in terms of determinants ofmatrices.
Theorem (Cramer’s Rule)Let A be an n× n invertible matrix, and consider the system AX = B, where X =
[x1 x2 · · · xn
]T .We define Ai to be the matrix obtained from A by replacing column i with B.Then for each value of i , 1 ≤ i ≤ n,
xi = det Aidet A
Jonathan Chavez
DeterminantsApplications of determinants
A Formula for the InverseCramer’s RulePolynomial Interpolation
Example (Cramer’s Rule)Solve the following system of linear equations using Cramer’s Rule.
3x1 + x2 − x3 = −15x1 + 2x2 = 2x1 + x2 − x3 = 1
First, x1 = det A1det A , where
A =
[ 3 1 −15 2 01 1 −1
]and A1 =
[ −1 1 −12 2 01 1 −1
].
Computing the determinants of these two matrices,
det A = −4 and det A1 = 4,
and thus x1 = 4−4 = −1.
Jonathan Chavez
DeterminantsApplications of determinants
A Formula for the InverseCramer’s RulePolynomial Interpolation
Example (continued)Secondly, x2 = det A2
det A where det A = −4 and
det A2 =
∣∣∣∣∣ 3 −1 −15 2 01 1 −1
∣∣∣∣∣ = −14,
and thus x2 = −144 = 7
2 . Finally, x3 = det A3det A , where det A = −4 and
det A3 =
∣∣∣∣∣ 3 1 −15 2 21 1 1
∣∣∣∣∣ = −6,
and thus x3 = −6−4 = 3
2 . Therefore, the solution to the system is given by
X =
[ x1x2x3
]=
−17232
.
You can check this by substituting these values into the original system.Jonathan Chavez
DeterminantsApplications of determinants
A Formula for the InverseCramer’s RulePolynomial Interpolation
Polynomial InterpolationProblemGiven data points (0, 1), (1, 2), (2, 5) and (3, 10), find an interpolating polynomial p(x) of degree at most three, and thenestimate the value of y corresponding to x = 3
2 .
SolutionWe want to find the coefficients r0, r1, r2 and r3 of
p(x) = r0 + r1x + r2x2 + r3x3
so that p(0) = 1, p(1) = 2, p(2) = 5, and p(3) = 10.
p(0) = r0 = 1p(1) = r0 + r1 + r2 + r3 = 2p(2) = r0 + 2r1 + 4r2 + 8r3 = 5p(3) = r0 + 3r1 + 9r2 + 27r3 = 10
Jonathan Chavez
DeterminantsApplications of determinants
A Formula for the InverseCramer’s RulePolynomial Interpolation
Solution (continued)Solve this system of four equations in the four variables r0, r1, r2 and r3. 1 0 0 0 1
1 1 1 1 21 2 4 8 51 3 9 27 10
→ · · · → 1 0 0 0 1
0 1 0 0 00 0 1 0 10 0 0 1 0
Therefore r0 = 1, r1 = 0, r2 = 1, r3 = 0, and so
p(x) = 1 + x2.
The estimate for the value of y corresponding to x = 32 is
y = p(3
2
)= 1 +
(32
)2= 13
4 ,
resulting in the point ( 32 , 13
4 ).
Jonathan Chavez
DeterminantsApplications of determinants
A Formula for the InverseCramer’s RulePolynomial Interpolation
TheoremGiven n data points (x1, y1), (x2, y2), . . . , (xn, yn) with the xi distinct, there is a unique polynomial
p(x) = r0 + r1x + r2x2 + · · · + rn−1xn−1
such that p(xi ) = yi for i = 1, 2, . . . , n. The polynomial p(x) is called the interpolating polynomial for the data points.
To find p(x), set up a system of n linear equations in the n variables r0, r1, r2, . . . , rn−1.
r0 + r1x1 + r2x21 + · · · + rn−1xn−1
1 = y1r0 + r1x2 + r2x2
2 + · · · + rn−1xn−12 = y2
.
.
....
.
.
.
r0 + r1xn + r2x2n + · · · + rn−1xn−1
n = yn
The fact that the xi are distinct guarantees that the coefficient matrix1 x1 x2
1 · · · xn−11
1 x2 x22 · · · xn−1
2...
.
.
....
.
.
....
1 xn x2n · · · xn−1
n
has determinant not equal to zero, and so the system has a unique solution, i.e., there is a unique interpolating polynomialfor the data points.
Jonathan Chavez