Test: Engineering Mathematics- 1 - Computer Science Engineering (CSE) MCQ

In the given system of equations, the ratio of the coefficients of x equals the ratio of coefficients of y.

Therefore, they would be consistent only if this ratio equals the constant terms.

That is, If 10/k = 2/6 = -5/-15

Hence if k = 30, then the given system of equations is consistent.

The given system of equations are of the form AX = Bwhich represents the system of  non homogeneous equations.

Augmented matrix is given by

by applying row transformation R₂ → R₂ − (R₁+R₃), we get

so, rank of matrix = rank of augmented matrix = 2 < number of variables

∴ System has infinite solutions

Alternatively we can simply observe that second equation is redundant as it can be obtained from rest two. Hence we have 2 primary equations and 3 variables hence infinite solution.

An invertible matrix A has an LU decomposition provided that all it's leading submatrices have non-zero determinants.

All the matrices in the options are invertible because 

In option 4

 in this leading sub matrix A = [0]
Hence LU decomposition is not possible

• The statements now correctly describe the conditions for consistency and the number of solutions for a system of linear equations based on the rank of the coefficient matrix (C) and the augmented matrix (A):

  • If c ≠ a, the equations are inconsistent, meaning no solution is possible.
  • If c = a < n, the equations are consistent with an infinite number of solutions.
  • If c = a = n, the equations are consistent and there is a unique solution.

  Therefore, none of the statements are incorrect, leading to the answer being 0 incorrect statements.

For given system of equations to have no solution, the determinant of the coefficient matrix should be equal to 0

1 × (4x−24) − 1 × (2x−12) + 1(8−8) = 0

2x − 12 = 0

∴ x = 6
Augmented matrix is given by

For given system of equations to have no solution, rank of augmented matrix should be greater than coefficient matrix
∴ y − 12 ≠ 0
∴ y ≠ 12

We get x = 1 and z = 2.

Hence tr(L) = 3.

Augmented matrix will be

R₃ → R₃ – R₁

R₂ → 2R₂ – 3R₁

R₃ → R₃ – 2R₂

 

Rank of A ≠ Rank of Augmented matrix

Hence given system of equations has no solution.

Set S1 contains homogeneous equations, homogeneous equations are always consistent because it always satisfied a trivial solution i.e. x = y = z = 0

For S2

R2 → 3R2

 

Rank Augmented matrix is 3 while rank of coefficient matrix has rank is 2

Hence the S2 is inconsistent

∴ only S1 is consistent

(k + 1)x + 8y = 4k

kx + (k + 3)y = 3k – 1
The given system of linear equations has infinitely many solutions if ρ(A) = ρ(A|B) < n = 2

To get the rank less than 2, one row should be dependent on another.

nk = (k + 1), n(k + 3) = 8, n(3k – 1) = 4k

by solving the above equations, we get

n = 2, k = 1

Hence for k = 1 only the system has infinitely many solutions.

For given system of equations to have unique solution, the determinant of the coefficient matrix should be not equal to 0

 

2 (9−8) − a(3−4) + (2−3) ≠ 0

2 + a − 1 ≠ 0

a ≠ −1
 

Since Δ of coefficient matrix is not equal to 0 then b and c can take any value

Note:

If coefficient matrix’s Δ ≠ 0 then rank of coefficient matrix is always equal rank of augmented matrix so right-hand side of given equation can take any value