# Vectors, matrices, and polynomials

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5) Use cofactor expansion to find the determinant

| 4 -1 1 6 |

| 0 0 -3 3 |

| 4 1 0 14|

| 4 1 3 2 |

6)Find the characteristic equation, eigenvalues, and corresponding eigenvectors for

A = 1 2

2 1

7) Answer True or False for the following statements: (True means always true; False means sometimes false). Justify your answers. Remember, you can use counterexamples only to justify False)

a) Statement: "If T: R^n -> R^n is a one-to-one linear operator, then T is invertible".

b) Let Ta: R^n -> R^n be a linear operator with standard matrix [Ta] = A.

Statement: "If det (A) = 0 then the range of Ta = R^n. ".

c) Consider the following linear operations on R²:

T1 is a counterclockwise rotation of 60º, and T2 is an orthogonal projection on the y-axis is R².

Statement: "T1ºT2 = T2ºT1"

d) Statement: "T(x, y, z) = (2x, 0, 1) is a linear operator in R³."

8) Answer True or False for the following statements: (True means always true; False means sometimes false). Justify your answers. Remember, you can use counterexamples only to justify False)

a) The set of all invertible 3x3 matrices is a subspace in the vector space of all 3x3 matrices under standard operations of matrix addition and scalar multiplication.

b) The set of all polynomials a0 + a1x + a2x² + a3x³ for which a0= 0 is a subspace in the vector space R3[X] of all polynomials of degree ≤3 under standard operations of polynomial addition and scalar multiplication.

c) The set of all polynomials a0 + a1x + a2x^2 + a3x^3 for which a3 ≠ 0 is a subspace in the vector space R3[X] of all polynomials of degree ≤3 under standard operations of polynomial addition and scalar multiplication

9)Determine whether w is in span of the given vectors v1, v2,.. , and if it is, express w as a linear combination of v1, v2, ...

a) w = {-1, 4, 15}, v1 = ( 1,2,8) v2 = (3,0,1)

b) w= 4x+6x², v1 = 1 + x + x², v2 = 2 + 2x + 3x³, v3 = 1+5x+6x²

c) w= 0 5 v1= 1 2 v2= 3 1

-4 -1 -1 0 1 1

10) Determine whether the given vectors span R³.

a) (1,3,4), (2,0,1), (2,1,0)

b) (1,5,1), (2,6,1), (3,3,0), (4,6,2)

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##### Solution Summary

These problems deal with cofactor expansion, eigenvalues, eigenvectors, linear operators, polynomial addition, spans, and scalar multiplication.

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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...

=> ...

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