3.6-6 F2010
=====
The challenge of cofactor expansion of a 5x5 determinant is to begin each
expansion opportunity using a row or column that contains as many zeros
as possible. The best one for the 5x5 is row 3, which contains 4 zeros.
The expansion then gives (+1)(5)(a 4x4 determinant).
The checkerboard sign (+1) is found by marking from th eupper left corner
of the 5x5 determinant, saying "plus - minus - plus - minus -plus" in
going from the (1,1)-position to the (3,3)-poition [where 5 is located].
The progress marked, to obtain a 4x4 determinant to work on, then the next
step chooses a cofactor expansion of the 4x4 on the last row, which contains
3 zeros. This time the answer is (-1)(2)(3x3 determinant).
The process continues, with a row one expansion of the 3x3 [it has 2 zeros]
to get a 2x2 determinant, which is evaluated using the 2x2 Sarrus' rule.
3.6-20 F2010
=====
The method of elimination is a synonym for the Four Rules
(1) The triangular rule, or 1-arrow Sarrus' rule;
(2) Swap [swap(s,t) changes the sign of the determinant];
(3) Mult [mult(t,m) divides the determinant by m];
(4) Combo [combo(s,t,c) leaves the determinant unchanged].
The last 3 rules are equivalent to one rule, a special case of the determinant
product theorem, det(A) = det(EA)/det(E).
The objective is to apply combo to the matrix to get a triangular form, then
apply the one-arrow Sarrus' rule to evaluate the determinant.
3.6-32 F2010
=====
Cramer's rule for this problem is
x1 = det(A1)/det(A), x2 = det(A2)/det(A), x3 = det(A3)/det(A).
The symbols are:
A = matrix whose rows are [3,4,-3], [3,-2,4], [3,2,-1].
b = column vector with entries 5,7,3
A1 = matrix A with col(A,1)=b
A2 = matrix A with col(A,2)=b
A3 = matrix A with col(A,3)=b
You are expected to evaluate the four determinants det(A), det(A1),
det(A2), det(A3), simplify the fractions and report the answer
x1, x2, x3. Maple can be used to check the answer:
with(linalg):
A:=matrix([[3,4,-3], [3,-2,4], [3,2,-1]]);
b:=vector([5,7,3]);
linsolve(A,b);
3.6-40 F2010
=====
This problem will help you to understand the adjugate formulas
inverse(A) = adj(A)/det(A),
adj(A) = adjugate or adjoint of A
= transpose of the matrix of cofactors of A
The following maple code can be used as an answer check, but not as a solution to the problem.
The problem solution is expected to be hand-written, with all 2x2 determinant details documented.
The expected presentation is a formula for adjugate(A) as the transpose of a
big matrix whose 9 entries are written as 2x2 determinants multiplied by
checkerboard signs [cofactor = sign x minor].
For presentation details, follow Example 11 Section 3.6 in E&P.
with(linalg):
A:=matrix([[2,4,-3], [2,-3,-1], [-5,0,-3]]);
adjoint(A);
det(A);
inverse(A);
3.6-60 F2010
=====
Part (a).
Intuition to solve the problem can be obtained by writing out explicitly
the determinants B1, B2, B3, B4. Subject B3, B4 to cofactor expansion to
get the recursion identity of 3.6-60a. The work sumbitted must be for
dimension n, for full credit. Examples are not a proof of part (a). Use
"..." notation to write out B_{n}, B_{n-1} and B_{n-2}. Cite steps where
cofactor expansion by row and by column are used.
Part (b).
Induction is explained in the appendix to the Varberg calculus text
currently used for Calc I,II,III. Part 3.6-60b is expected to be solved
using the theory of mathematical induction. Graders will award two
possible grades: 55, 100. For 100 percent, the work must use induction
methods. For 55 percent, some drastic unforgivable error has to occur,
for example, submitting a work which does examples but no induction, or
choosing a sequence of details that bears little or no relationship to
the expected details.
Exposition details (a)
=========
Write out determinant B_n on paper using "..." notation. At least six (6) columns
should be written, to give enough detail. Then use cofactor expansion along row 1
to produce 2 determinants of order n-1. Finally, use column 1 cofactor expansion
on the second determinant to produce a determinant of order n-2. All of this can
fit one one sheet of paper.
Proof details (b)
=========
While many possible definitions exist for statement P_n in math
induction, all of them have to have two equations in order apply the
recursion formula of (a):
(a) B_k = 2 B_{k-1} - B_{k-2} for k > 3
The problem is that to use (a), two equations are needed for expansion
of the RHS. Bear in mind that P_n = { B_n = n+1} does not work for two
reasons: first, P_1 = { B_1=2} and B_1 is not defined, only B_2 and
higher; second, not enough equations appear in P_n to expand the RHS of
(a). Something like
P_n = { B_{n+1} = n+2, B_{n+2} = n+3 }
does work. The details do not appear here, but are expected on submitted
work.