MTH501 Midterm Past Paper Fall 2004

MTH501 Linear Algebra · Mid Term Examination - November 2004 · Time Allowed: 90 Minutes · Instructions · Total Questions: 11

Q1. Marks : 1

Consider the system of Linear equations x x x x x x x x x Then determinant of the Matrix of Coefficients of the above system is (a) 91 (b) 123 (c) (d) 81

Q2. Marks : 2

x y x A and B y

x y x such that 2020

2512 AB

Then the values of x and y are

(a) x = - 2 and y = 4 (b) x = 2 and y = - 6 (c) x = 0 and y = 0

(d) x = 2 and y = -4

Q3. Marks : 1

Which statement about the set S is false where S = {(1, 1, 3), (2, 3, 7),(2, 2, 6)}

(a) The set S contain an element which is solution of the equation 5x – yz = (b) The Set S is linearly independent. (c) The set S contain two elements which are multiple of each other. (d) The Set S is linearly dependent.

Q4. Marks : 2

Consider a linear transformation T such that T ⎡⎤ ⎡⎤

Then T ⎡⎤

(a)

(b)

(c)

(d)

Q5. Marks : 1

How many subspaces R 2 have?

1. only two: {0} and R 2
2. Only four: {0} x- axis and y -axis and R 2
3. Infinitely many.
4. None of the above.

Correct answer: Infinitely many.

Q6. Marks : 8

An oil refinery produces low-sulfur and high-sulfur fuel. Each ton of low-sulfur requires 5 minutes in the blending plant and 4 minutes in the refining plant; each ton of high sulfur fuel requires 4 minutes in the blending plant and 2 minutes in the refining plant. If the blending plant is available for 3 hours and the refining plant is available for 2 hours, how many tons of each type of fuel should be manufactured so that the plants are fully utilized?

Solution:

The data given in the question can be formed into system of linear equations as Low-sulfur High-sulfur Blending plant Refining Plant As we are given in the question that we have blending plant and refining plant available for 3 and 2 hours respectively. Let x tons of low-sulfur and y tons of high sulfur be the amount should be manufactured so that plants are fully utilized. Then from the above data we must have the system, 5x + 4y = 180 4x + 2y = 120 Augmented matrix for the above system

is by R R

So we have x + 2y = 60 and -6y = - 120, thus we have y= 20 tons and x = 20tons are the required manufactured tons of each low-sulfur and high-sulfur so that we can utilize both plants for the given time.