MTH501 Midterm Solved Paper 2013

Mth501 subjective for mid term Latest spring 2013 By ~“Librainsmine”~

Q1. Applying certain elementary row operation to the elementary matrix

to produce an identity matrix · . [2 marks] · Solution:

Multiply R2 by 1/5

E

We get the identity matrix

Q2. Why is it NOT possible to solve the following system of linear equations applying the Cramer’rule? [2 marks]

Solution:

E

I 

I 

x x x x A=

det

det since it is an singular matric and its determinant is zero. we need determinant of A f.or applying crammers rule x let x b x

A

A

Q3. Determine whether or not the inverse of the matrix

exists? Justify your answer with appropriate reason. [3 marks]

Solution:

1(0 0) 2(0 6) 3(0 4) 1(0) 2( 6) 3( 4) ( 12) 12 A

A A A A

As the determinant of the given matrix is zero its mean it is a singular matrix the inverse of the singular matrix does not exist.

Q4. by using a cofactor expansion across the third row, where [3 marks]

Solution:

Using cofactor expansion along the first column: A

det( A A

Show that · is a subspace of · . [5 marks] · Solution:

To check if W is the subspace of ,we 1st check that axiom 1 and 6 of a vector space holds.

Let and v= be vectors in W then u+v= is in W and also if k ia any scalar and u= in any vector W,then ku=k Hence W is a subspace. u a b c a b c a b c a b c a a b b c c a b c a b c s

Spring 2013

Q1. If A = B, then determine the values of x and y; where

y A x and

B

Solution:

As we know that matrix A=B its mean that the every entry in A = to the coresponding entry in B so its clearly seen that value of x+2=4 .....x=4/-2=x=-2 y+2=2......y=-1

Q2. Determine which of the following condition(s) hold(s) for a vector space V over R. Justify your answer with appropriate reason.

W a b c a b c

W R

R