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please, just be neat and clear. write well so I can see clearly. use theorems with short explanations. thank you very much. ' let me know if you have any questions.

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Exam 2S [Chapter 2 material] All work to be done on your own. Be complete and justify each step.
1. For which values of the constant b and c is the following matrix invertible? Do this without using
determinants.
0 1
1 0
0
c
c
2. Suppose
2 4
2 5
 
=  
 
A .
(a) Use ROW REDUCTION to find its inverse -1A but write down the elementary row matrices
1 2 pE ,E ,...E that co
esponds to each step in the row reduction. No credit without the
elementary matrices.
(b) Explain what each step means geometrically, using the elementary row matrices to help you.
(c) Prove that every invertible matrix is a product of elementary matrices.
(d) Write the above matrix
2 4
2 5
 
=  
 
A as a product of elementary matrices (find them and
then write A as their product.)
3. Consider the matrix
2 2 2
4 7 7
6 18 22
A .
(a) Find lower triangular elementary matrices 1,..., pE E such that the product 2 1pE E E A is
an upper triangular matrix U . (Hint: Use elementary row operations to obtain an echelon
matrix (doesn’t have to be reduced echelon) and do not try get 1’s in the pivot position).
(b) Now find lower triangular elementary matrices 1,..., pF F and an upper triangular matrix U
such that 1 2 pA F F F U . Hint: Get the iF from the above iE .
(c) Find a lower triangular matrix L and an upper triangular matrix U such that A LU . This
exercise should help you understand the theory behind finding an LU factorization.
(d) Now just find the LU factorization in the way that we did it in class – how you would do it in
practice.
4. Let A and B be two n x n matrices such that their product AB is invertible. Prove that both A and
B are invertible, and give their respective inverses. DO NOT ASSUME what you are proving! Do this
without using determinants.
5. Suppose 4 3:T → is given by
1
3 4
2
2 3
3
3 1
4
2
2
x
x x
x
T x x
x
x x
x
  
−   
    = +
   
 −     
  
.
(a) Give its standard matrix representation.
(b) Is it 1-1? Is it onto?
(c) Find a basis for its null space.
(d) Find a basis for its column space.
(e) Are its columns linearly independent? If not give a dependence relation.
6. Find the formulas for X,Y,Z in terms of A,B,C and justify your calculations. These are all block
matrices, not numerical entries.

     
=     
     
A B I 0 0 Ι
C 0 X Y Ζ 0
7. Suppose A is a 7 x 4 matrix and B is a 4 x 7 matrix with columns 1 2 7b ,b , ...,b with
6 2 54 9b b b . Show why matrix AB is not invertible.
8. (a) Prove directly from the definition that if square matrix A has a right inverseC and also a left
inverse D , then they must be the same. In other words, show that if AC = I and DA = I , then
C = D. We actually showed this in class.
(b) Show that if
2
A is invertible, then A is invertible.
(c) Let square n x n matrices A, B be such that AB = B . Then nA = I . Show whether this statement
is true or false and prove if true, and justify completely if false.
9. Let
1 1 0
1 0 1
6 2 3
− 
 
= −
 
 − 
A . Let
1 2 0
3 1 2
2 3 2
 
 
= −
 
 − − 
B .
(a) Find -1A using row reduction.
(b) Show -1B does not exist using row reduction.
(c) Prove AB does not have an inverse, without using determinants. Hint: Use Problem 4.
Answered 1 days After Oct 01, 2022

Solution

Baljit answered on Oct 03 2022
55 Votes
1. Given matrix
A=
Using row reduction method
A=IA
A
· R1 R2 we get
A
· R1(-1)R1 we get
A
· R3R3+b*R1 we get
A
· R3R3+c*R2 we get
A
R3 of LHS matrix becomes completely zero.So inverse of Matrix A does not exist.
So inverse of given matrix is does not exist i
espective of values of b and c.
*************************************************************
2. Given matrix
A=
a. We know that
A=IA
=A
· Apply R2 R2-R1 we get
=A
Here E1=
· Apply R1 R1-4*R1 we get
=A
Here E2=
· Apply R1 (¼)R1 we get
=E2E1A
Here E3=
I=E3E2E1A …….(1)
So inverse of matrix A is
A-1=
.
c. Suppose matrix B having elementary row matrices E1 ,E2 ,...Ep that co
esponds to each step in the row reduction.
So that means
EpEp-1…..E3E2E1B=I
· Ep-1…..E3E2E1B=(Ep)-1
· Ep-2…..E3E2E1B=(Ep-1)-1(Ep)-1
Similarly
· B=(E1)-1(E2)-1……..(Ep-1)-1(Ep)-1
So hence we proved every invertible matrix is product of Elementry matrices.
d. Now from equation (1)
I=E3E2E1A
· A=(E1)-1(E2)-1(E3)-1
· E1=
E1=IE1
= E1
Apply R2 R2+R1
= E1
so (E1)-1=
Similarly for
· E2= so (E2)-1=
· E3= so (E3)-1=
So
A=
*************************************************************
3.
A=
a. To find U matrix
Ep…..E2E1A=U
R2 R2-2R1 we get
R3 R3-3R1 we get
R3 R3-4R2 we get
So
U=
So elementary matrices are
...
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