Find the value of k for which the following system of equations will be consistent.
2x – 5y = 10 and 6x – 15y = k
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 system of equation, 5x + 2y + z = 3,7x + 10y + 2z = 7,2x + 8y + z = 4 has
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 R_{2 }→ R_{2 }− (R_{1}+R_{3}), 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.
Matrix for which LU decomposition is not possible?
An invertible matrix A has an LU decomposition provided that all it's leading submatrices have nonzero 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
If the coefficient matrix is C_{n×n} and its corresponding rank is c also the augmented matrix is A_{n×(n+1)} and its corresponding rank is a, then how many statements given below are incorrect?
I. If c ≠ a, the equations are inconsistent with an infinite number of solutions.
II. If c = a < n, the equations are inconsistent
III. If c = a =n, equation may be inconsistent.
Find the value of x and y for the below given simultaneous equations that has no solutions?
a + b + c = 18
2a + 4b + 6c = 12
2a + xc + 4b = y
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
In the LU decomposition of the matrix, , if the diagonal elements of U are both 1, then the trace of L is:
We get x = 1 and z = 2.
Hence tr(L) = 3.
Consider the following system of equations in three real variables x, y, z.
2x – 3y + 7z = 5
3x + y – 3z = 13
2x + 19y – 47z = 32
The system of the equation has
Augmented matrix will be
R_{3} → R_{3} – R_{1}
R_{2} → 2R_{2} – 3R_{1
}
R_{3} → R_{3} – 2R_{2
}
Rank of A ≠ Rank of Augmented matrix
Hence given system of equations has no solution.
Among the given two sets which is/are consistent:
S1 = {3x + ay + 4z = 0,bx + 2y + z = 0,5x + 7z + 9z = 0
S2 = {2x + 6y= −11, 6x + 20y − 6z = −3, 6y − 18z = −1}
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
The number of values of k for which the system of equations (k + 1) x + 8y = 4k, kx + (k + 3)y = 3k – 1 has infinitely many solutions, is
(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(AB) < 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.
The simultaneous equations
2x + ay + z = 20
x + 3y + 4z = b
x + 2y + 3z = c
has unique solution then what is the value of a, b and c respectively?
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 righthand side of given equation can take any value
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