Vectors, matrices, and polynomials

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(Qu. 5)
The cofactor expansion for the determinant of a matrix, A, will give the same result no

matter which row or column of the matrix we choose - so we can choose the row or column

that will give us least work - in this case, row 2 looks the most likely because of the

two zeros.

The expansion along row i (=2) is done by summing the n cofactors. Each cofactor is the

product of three numbers:

1) the a[i,j] entry in the matrix,
2) the determinant of the smaller (n-1 by n-1) matrix obtained from A by deleting row i

and column j (we'll call the determinant of this smaller matrix: A[i,j])
3) and a +1 or -1 factor which is calculated as (-1)^(i+j)

In full:

n
det A = Sum (-1)^(i+j) a[i,j] det A[i,j]
j=1

So,

| 4 -1 1 6 |
| 0 0 -3 3 | = (-1)(0)(A[2,1]) + (1)(0)(A[2,2]) +
| 4 1 0 14 |
| 4 1 3 2 | (-1)(-3)(A[2,3]) + (1)(3)(A[2,4])

= 3 A[2,3] + 3 A[2,4]

So we need to work out A[2,3] (= B) and A[2,4] (= C):

| 4 -1 6 |
B = | 4 1 14 | = (1)(4)B[1,1] + (-1)(-1)B[1,2] + (1)(6)B[1,3]
| 4 1 2 |

to find the value of B we can again use the cofactor expansion - here we have expanded B

along the first row.

B[1,1] = | 1 14 | = (1)(1)(2) + (-1)(14)(1) = -12
| 1 2 |

We can continue carrying out the expansion until we reach 1 by 1 matrices, which have a

determinant simply equal to the value of their single entry.

Doing this repeatedly gives:

B = (4)(-12) + (-48) + (6)(0)
= -96

Similarly, you can show that C = A[2,4] = 24

So,

det A = 3(-96) + 3(24) = -216

(Qu. 6)

Recalling the definitions, we say that a nonzero vector, v, is an eigenvector with

corresponding eigenvalue, lambda, of matrix A, if and only if they satisfy the equation:

A.v = lambda.v

If you'll let me say lamdba = r (just for ease of notation, lamdba is the usual notation

for the eigenvalue though!), we can rewrite A.v = r.v as:

(A - rI).v = 0

where I is the identity matrix. We want to find the values of r for which the matrix:
(A - rI) is singular (i.e non invertible), which means the determinant of (A - rI) must

be zero:

| A - rI | = 0

This is our characteristic equation.
=>
| 1-r 2 | = 0
| 2 1-r |

=> (1-r)(1-r) - 4 = 0

=> r^2 - 2r - 3 = 0 factorising this gives...

=> ...