linear algebra homework

Q1. 

S1.

A is 3x2 so B will be 2x3.

Rank(A) = 2 so it's possible to create I2 which has rank 2.

Now let B = a b c, d e f

Now when we multiply BA and equate with I2 we will have 4 equations for 6 variables.

So 2 variables will be free and hence there will be infinite such matrices.


Rank(I_3) = 3

Rank(AC) = min(Rank(A),Rank(C) = 2

So it's not possible since ranks don't match. Q2. Prove that a square row echelon matrix M is either the identity matrix I, or else its bottom row is zero. S2. If we start with the top left element as the first pivot element, then each subsequent row needs to have the pivot element to its right. So if we keep shifting the column and row by 1 we will just manage to take row echelon matrix by using the bottom left element. If at any step we take more than 1 step right, than one or more rows at bottom will have to be zero rows. Q3.


S3. We need to convert the given matrix to Identity matrix. It is possible since both columns are independent.

R2 - 2R1

E1

1 0

-2 1



R2*-1/4

E2

1 0

0 -1/4


R1 - 5R2

E3

1 -5

0 1


Pr. th.E3*E2*E1 = A inverse

prove by computing



Q4.


S4. R2 - 2R1

R3 - R1


1 2 1

0 1 1

0 1 1


R3 - R2


1 2 1

0 1 1

0 0 0 Q5.

S5.

1 2 -1 3

2 5 1 8

1 3 0 5


Get it into Row echelon form:

R2 - 2R1

R3 - R1

R3 - R2


1 2 -1 3

0 1 3 2

0 0 -2 0

z = 0

y = 2

x = -1 Q6.


S6.

R2 - 2R1 , R3 - R1 and then swap R2,R3:

1 2 3

0 0 k-3

0 0 0


  1. k = 3
  2. k != 3
  3. not possible

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