- Type
- Midterm past paper
- Course
- MTH501 Linear Algebra
- Size
- PDF 96 KB
- Updated
- 23 Oct 2026
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make an appropriate assumption and use it to solve the problem.
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Questions: 11
Q1. Consider the system of Linear equations
x1 4x2� 5x3 0 2x1� x2 8x3 9 x1 x2� 8x3 2 Then determinant of the Matrix of Coefficients of the above system is (a) 91 (b) 123 (c) 141 (d) 81
Total
Marks : 1
A and B2
A B
T 9
18
0
45
9
Q2. Marks : 2
2 0 20
2 5 12 1 2 x y1� 2
2 3 x� y0 2
Then the values of x and y are x� 5 y
3x such that
(a) (b) (c)
(d)
x = - 2 and y = 4 x = 2 and y = - 6 x = 0 and y = 0
x = 2 and y = -4
Q3. Which statement about the set S is false where S = {(1, 1, 3), (2, 3, 7),(2, 2, 6)}
Marks : 1
(a)
(b)
(c) (d) The set S contain an element which is solution of the equation 5x – yz = 0 The Set S is linearly independent. The set S contain two elements which are multiple of each other. The Set S is linearly dependent.
Q4. Marks : 2
Consider a linear transformation T such that 1 6
Then 5
T25 15
9
(a)
(b) 1
5
30
(c)
12
(d) is
Q5. How many subspaces R2 have?
Marks : 1
(a) (b)
(c)
(d) only two: {0} and R2 Only four: {0} x- axis and y -axis and R2 Infinitely many. None of the above.
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 5 4 1801 4 2 120 ∼4 2 60
2 120 by R1� R2
1 2 60 0� 6�120 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.
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