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All questions of Linear Algebra for Civil Engineering (CE) Exam

If A is a non–singular matrix and the eigen values of A are 2 , 3 , -3 then the eigen values of A-1 are
  • a)
    2 , 3 , - 3
  • b)
    1/2, 1/3, -1/3
  • c)
    2|A|, 3|A|, -3|A|
  • d)
    None of these
Correct answer is option 'B'. Can you explain this answer?

Tanishq Chavan answered
If λ1 ,λ2 ,λ3 ....λare the eigen values of a non–singular matrix A, then A-1 has the eigen values  1/λ1 ,1/λ2 ,1/λ3 ....1/λn Thus eigen values of A-1are 1/2, 1/3, -1/3

The number of linearly independent eigenvectors of 
  • a)
    0    
  • b)
    1    
  • c)
    2    
  • d)
    Infinit e 
Correct answer is 'B'. Can you explain this answer?

Pranjal Sen answered
Number of linear independent vectors is equal to the sum of Geometric Multiplicity of eigen values. Here only eigen value is 2. To find Geometric multiplicity find n-r of (matrix-2I), where n is order and r is rank. Rank of obtained matrix is 1 and n=2 so n-r=1. Therefore the no of linearly independent eigen vectors is 1

The eigenvalues of
  • a)
    -19,5,37
  • b)
    19,-5,-37
  • c)
    2,-3,7
  • d)
    3,-5,37
Correct answer is option 'A'. Can you explain this answer?

Avinash Sharma answered
The eigenvalues of an upper triangular matrix are simply the diagonal entries of the matrix.
Hence 5, -19, and 37 are the eigenvalues of the matrix. Alternately, look atd
λ = 5, -19, 37

If -1 , 2 , 3 are the eigen values of a square matrix A then the eigen values of A2 are
  • a)
    -1 , 2 , 3
  • b)
    1, 4, 9
  • c)
    1, 2, 3
  • d)
    None of these
Correct answer is option 'B'. Can you explain this answer?

Aditya Patel answered
If λ1 ,λ2 ,λ3 ....λare the eigen values of  a matrix A, then A2 has the eigen values  λ12 ,λ22 ,λ32 ....λn2 So, eigen values of Aare 1, 4, 9.

Eigen values of a matrix    are 5 and 1. What are the eigen values of the matrix S2  = SS?
  • a)
    1 and 25  
  • b)
    6 and 4  
  • c)
    5 and 1  
  • d)
    2 and 10 
Correct answer is option 'A'. Can you explain this answer?

EduRev GATE answered
We know If λ be the eigen value of A ⇒λ2 is an eigen value of A2 .
For S matrix, if eigen values are λ1, λ2, λ3,... then for S² matrix, the eigen values will be λ²1 λ²2 λ²3......
For S matrix, if eigen values are 1 and 5 then for S² matrix, the eigen values are 1 and 25

If 2 and 4 are the eigen values of A then the eigenvalues of AT are
  • a)
    1/2, 1/4 
  • b)
    2, 4
  • c)
    4, 16
  • d)
    None of these
Correct answer is option 'B'. Can you explain this answer?

Sanvi Kapoor answered
The eigenvalues of a matrix A are the roots of the characteristic equation of A, which are invariant under the transpose operation. This means that the matrix A and its transpose AT have the same eigenvalues.
Given that 2 and 4 are the eigenvalues of A, the eigenvalues of AT will also be 2 and 4.

In an M × N matrix such that all non-zero entries are covered in a rows and b column. Then the maximum number of non-zero entries, such that no two are on the same row or column, is 
  • a)
    ≤ a + b    
  • b)
    ≤ max (a, b)  
  • c)
    ≤ min[ M –a, N–b]    
  • d)
    ≤ min  {a, b} 
Correct answer is option 'D'. Can you explain this answer?

Naroj Boda answered
Suppose a < b, for example let a = 3, b= 5, then we can put non-zero entries only in 3 rows and 5 columns. So suppose we put non-zero entries in any 3 rows in 3 different columns. Now we can’t put any other non-zero entry anywhere in matrix, because if we put it in some other row, then we will have 4 rows containing non-zeros, if we put it in one of those 3 rows, then we will have more than one non-zero entry in one row, which is not allowed.
So we can fill only “a” non-zero entries if a < b, similarly if b < a, we can put only “b” non-zero entries. So answer is ≤min(a,b), because whatever is less between a and b, we can put atmost that many non-zero entries.

If the rank of a (5×6) matrix Q is 4, then which one of the following statements is correct?  
  • a)
    Q will have four linearly independent rows and four linearly independent columns  
  • b)
    Q will have four lineally independent rows and five lineally independent columns
  • c)
    QQT will be invertible
  • d)
    QTQ will be invertible 
Correct answer is option 'A'. Can you explain this answer?

Given that rank Q is 4, so it will have 4 linearly independent columns as well as 4 linearly independent rows (∵ row rank of a matrix = column rank of a matrix)

So, A is correct and accordingly B is false.

C is false as order of QQ^T will be 5 x 5 and given that rank of Q is 4 i.e. < 5. So the matrix is QQ^T will be singular and hence not invertible.

Similarly D is false as order of Q^T Q will be 6 x 6 and the matrix is Q^T Q will be singular and hence not invertible.

If 1 and 3 are the eigenvalues of a square matrix A then A3 is equal to
  • a)
    13 (A - I2 )
  • b)
    13A - 12 I2
  • c)
    12 (A - I2)
  • d)
    None of these
Correct answer is option 'B'. Can you explain this answer?

Pankaj Rane answered
Since 1 and 3 are the eigenvalues of A so the characteristic equation of A is
Also, by Cayley–Hamilton theorem, every square matrix satisfies its own characteristic equation so

The sum of the eigenvalues of    is equal to 
  • a)
    18
  • b)
    15
  • c)
    10 
  • d)
    none of these
Correct answer is option 'A'. Can you explain this answer?

Zoya Sharma answered
  • Since the sum of the eigenvalues of an n–square matrix is equal to the trace of the matrix (i.e. sum of the diagonal elements)
  • So, required sum = 8 + 5 + 5  = 18

The number of linearly independent eigenvectors of 
  • a)
    0    
  • b)
    1    
  • c)
    2    
  • d)
    Infinit e 
Correct answer is option 'B'. Can you explain this answer?

Ravi Singh answered
Number of linear independent vectors is equal to the sum of Geometric Multiplicity of eigen values. Here only eigen value is 2. To find Geometric multiplicity find n-r of (matrix-2I), where n is order and r is rank. Rank of obtained matrix is 1 and n=2 so n-r=1. Therefore the no of linearly independent eigen vectors is 1

Find the Eigenvalues of matrix
  • a)
    8, 2
  • b)
    3, -3 
  • c)
    -3, -5 
  • d)
    5, 0
Correct answer is option 'A'. Can you explain this answer?

Sanya Agarwal answered
A - λI = 0
Now, After taking the determinant:
(4 - λ)2 - 1 = 0
16 + λ2 - 8λ - 1 = 0
λ2 - 8λ + 15 = 0
 - 3) (
λ
 - 5) = 0
λ
 = 3, 5

 Find the rank of the matrix 
  • a)
    2
  • b)
  • c)
    0
  • d)
    3
Correct answer is option 'D'. Can you explain this answer?

Engineers Adda answered
To find out the rank of the matrix first find the |A| 
If the value of the |A| = 0 then the matrix is said to be reduced 
But, as the determinant of A has some finite value, then the rank of the matrix is 3.

Select a suitable figure from the four alternatives that would complete the figure matrix.

  • a)
     
    1
  • b)
     
    2
  • c)
     
    3
  • d)
    4
Correct answer is option 'D'. Can you explain this answer?

Kiran Datta answered
In each row (as well as each column), the third figure is a combination of all the, elements of the first and the second figures

The eigenvalues of
are
  • a)
    3, -5, 37
  • b)
    2, -3, 7
  • c)
    19, -5, -37
  • d)
    -19, 5,  37
Correct answer is option 'D'. Can you explain this answer?

Avinash Sharma answered
The eigenvalues of an upper triangular matrix are simply the diagonal entries of the matrix. Hence 5, -19, and 37 are the eigenvalues of the matrix. Alternately, look at 
Then  = 5,-19,37 are the roots of the equation; and hence, the eigenvalues of [A].

For what value of λ, do the simultaneous equation 2x + 3y = 1, 4x + 6y = λ have infinite solutions?
  • a)
    λ = 0 
  • b)
    λ = 1
  • c)
    λ ≠ 2
  • d)
    λ = 2
Correct answer is option 'D'. Can you explain this answer?

Understanding Infinite Solutions
To determine the value of λ for which the simultaneous equations have infinite solutions, we need to analyze the equations:
1. Original Equations:
- 2x + 3y = 1 (Equation 1)
- 4x + 6y = λ (Equation 2)
2. Condition for Infinite Solutions:
- For two linear equations to have infinite solutions, they must be equivalent. This means that one equation can be expressed as a multiple of the other.
Transforming the Equations
- We can manipulate Equation 1 to see if it can transform into Equation 2:
- Multiply Equation 1 by 2:
- 2 * (2x + 3y) = 2 * 1
- 4x + 6y = 2
Setting Equations Equal
- Now we compare this with Equation 2:
- From our manipulation, we get:
- 4x + 6y = 2
- For the two equations to be equivalent:
- λ must equal 2.
Conclusion
- Therefore, the value of λ that allows the equations to have infinite solutions is:
- λ = 2.
Thus, the correct option is D) λ = 2.

There are two containers, with one containing 4 Red and 3 Green balls and the other containing 3 Blue and 4 Green balls. One bal is drawn at random form each container.The probability that one of the ball is Red and the other is Blue will be       
  • a)
    1/7
  • b)
    9/49
  • c)
    12/49
  • d)
    3/7
Correct answer is option 'C'. Can you explain this answer?

Snehal Nair answered
Problem Analysis

To solve this problem, we need to find the probability of drawing one Red ball and one Blue ball from two different containers. Let's analyze the given information:

- Container 1: 4 Red balls and 3 Green balls
- Container 2: 3 Blue balls and 4 Green balls

Solution

Step 1: Find the probability of drawing a Red ball from Container 1

The probability of drawing a Red ball from Container 1 can be calculated as the ratio of the number of Red balls to the total number of balls in Container 1.
Probability of drawing a Red ball from Container 1 = 4/7

Step 2: Find the probability of drawing a Blue ball from Container 2

Similarly, the probability of drawing a Blue ball from Container 2 can be calculated as the ratio of the number of Blue balls to the total number of balls in Container 2.
Probability of drawing a Blue ball from Container 2 = 3/7

Step 3: Find the probability of drawing one Red and one Blue ball

Now, we need to find the probability of drawing one Red ball from Container 1 and one Blue ball from Container 2. Since the two events are independent, we can multiply the probabilities of each event.
Probability of drawing one Red and one Blue ball = Probability of drawing a Red ball from Container 1 × Probability of drawing a Blue ball from Container 2

Probability of drawing one Red and one Blue ball = (4/7) × (3/7) = 12/49

Therefore, the probability that one of the balls is Red and the other is Blue is 12/49.

Conclusion

The correct answer is option 'C', 12/49.

The system of equations
x + y + z = 6;
x + 4y + 6z = 20;
x + 4y + λz = μ
has NO solution for values of λ and μ given by
  • a)
    λ = 6, μ ≠ 20
  • b)
    λ = 6, μ = 20
  • c)
    λ ≠ 6, μ = 20
  • d)
    λ ≠ 6, μ ≠ 20
Correct answer is option 'A'. Can you explain this answer?

Meghana Kaur answered
Concept:
The number of solutions can be determined by finding out the rank of the Augmented matrix and the rank of the Coefficient matrix.
  • If rank(Augmented matrix) = rank(Coefficient matrix) = no. of variables then no of solutions = 1.
  • If rank(Augmented matrix)  ≠ rank(Coefficient matrix) then no of solutions = 0.
  • If rank(Augmented matrix) = rank(Coefficient matrix) < no. of variables, no of solutions = infinite.
Calculation:
The augmented matrix for the system of equations is
Performing: R3 → R3 – R2
      …
If λ = 6 and μ ≠ 20 then
Rank (A | B) = 3 and Rank (A) = 2
∵ Rank (A | B) ≠ Rank (A)
∴ Given the system of equations has no solution for λ = 6 and μ ≠ 20

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