1MA01: Mathematical Methods Tutorial Sheet
1
Remember, to find the eigenvalues (λ) you use the characteristic equation
of a matrix A:
det(A − λI) = 0
and then for each λ solve Av = λv for the eigenvector v.
1. applications of matrix inverse: cryptography
You receive a coded message! You know that each letter of the original
message is replaced with a number corresponding to its placement in the
alphabet: eg “E” is represented by “5”, “W” by “23” and so on. Spaces
are indicated by zeroes. You also know the message was transformed
(encoded) by multiplying the message on the left by the following


1 2 3


A =  0 1 4 .
5 6 0
(1)
Translate the coded message:


14 87 73 24 38 111 2


B =  1 84 67 17 46 94 1  .
66 117 112 55 17 198 6
(2)
First find the inverse of A eg by Gauss-Jordan. You should get

A−1

−24 18
5

20
−15
−4
=


−5
4
1
(3)
Then to decode multiply the coded message by A−1 , after some careful
and tedious multiplication you should get


12 9 14 5 1 18 0


−1
A B =  1 12 7 5 2 18 1 
0 18 15 3 11 19 0
1
(4)
Sinéad Ryan, ryan@maths.tcd.ie, see also http://www.maths.tcd.ie/˜ryan/123.html
and matching each entry in the array to a letter gives


l i n e a r


 a l g e b r a .
r o c k s
2. matrix eigenvalues and eigenvectors
Find the eigenvalues and eigenvectors of the matrices
3 6
1 4
A=
!
0
1
−2 −3
; B=
!
.
(5)
For A:
To find the eigenvalues, solve the characteristic equation det(A − λI) =
0. ie
A − λI =
=
3 6
1 4
!
λ 0
0 λ
−
3−λ
6
1
4−λ
!
!
Then
!
0 =
=
=
=
3−λ
6
det
1
4−λ
(3 − λ)(4 − λ) − 6(1)
λ2 − 7λ + 6
(λ − 6)(λ − 1)
and in the usual way λ = 6 and λ = 1, are the eigenvalues.
Calculate first the eigenvector for λ = 6. Solve
0 = (A − λI)v
!
3 6
=
−
1 4
=
−3 6
1 −2
!
6 0
0 6
a
b
!
!
a
b
!
yielding −3a+6b = 0 and a−2b = 0 and these equations mean that any
vector with components satisfying a = 2b is an eigenvector, eg (2,1).
For λ = 1, solve
0 = (A − λI)v
!
3 6
=
−
1 4
=
2 6
1 3
!
1 0
0 1
a
b
!
a
b
!
!
yielding 2a + 6b = 0 and a + 3b = 0 and these equations mean that any
vector with components satisfying a = −3b is an eigenvector, eg (-3,1).
3. triangular matrices and their determinants
A square triangular matrix has all zeros either above or below the
main diagonal. The matrices below are all examples of this. Compute
their determinants in the usual way and notice that for these special
matrices det(A) = a11 a22 . . . ann ie the determinant is just the product
of the entries on the diagonal!


5 0 0

A=
 0 −3 0  ; B =
0 0 4
det(A) =
=
det(B) =
=
det(C) =
=
6 0
2 −1
!


10 5 1

; C=
 0 0 −4  (6)
0 0 6
5[−3(4) − 0] − 0[0(4) − 0(0)] + 0[0(0) − (−3)(0)]
−60
6(−1) − 0(2)
−6
10[0(6) − (−4)(0)] − 5[0(6) − (−4)(0)] + 1[0(0) − 0(0)]
0
4. triangular matrices and eigenvalues
Consider the matrix
A=
6 0
9 −4
!
(7)
Determine its eigenvalues. What do you notice about the eigenvalues
when compared with the matrix entries?
Compute the eigenvalues by solving the characteristic equation.
0 =
=
=
=
6−λ
0
det
9
−4 − λ
(6 − λ)(−4 − λ) − 0
λ2 − 2λ − 24
(λ − 6)(λ + 4)
!
and so the eigenvalues are λ = 6 and λ = −4. Note that these are
the same numbers that appear on the main diagonal of the matrix in
question.
In general for triangular matrices - square matrices which have all zeroes
either above or below the main diagonal - the eigenvalues are simply
the entries on the main diagonal.