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Examine the consistency of the system of equations.
x + 2y = 2
2x + 3y = 3
The given system of equations is:
x + 2y = 2
2x + 3y = 3
The given system of equations can be written in the form of AX = B, where
∴ A is non-singular.
Therefore, A−1 exists.
Hence, the given system of equations is consistent.
Examine the consistency of the system of equations.
3x − y − 2z = 2
2y − z = −1
3x − 5y = 3
This system of equations can be written in the form of AX = B, where
∴ A is a singular matrix.
Matrix A =variable coefficient matrix, rank of A=2
Matrix C = augmented matrix, rank of C =3
System of equation has an inconsistency because rank of matrix A < matrix C. The Solution doesn't exist.
Examine the consistency of the system of equations.
2x − y = 5
x + y = 4
The given system of equations is:
2x − y = 5
x + y = 4
The given system of equations can be written in the form of AX = B, where
∴ A is non-singular.
Therefore, A−1 exists.
Hence, the given system of equations is consistent.
Examine the consistency of the system of equations.
x + 3y = 5
2x + 6y = 8
The given system of equations is:
x + 3y = 5
2x + 6y = 8
The given system of equations can be written in the form of AX = B, where
∴ A is a singular matrix.
Thus, the solution of the given system of equations does not exist. Hence, the system of equations is inconsistent.
Examine the consistency of the system of equations.
x + y + z = 1
2x + 3y + 2z = 2
ax + ay + 2az = 4
The given system of equations is:
x + y + z = 1
2x + 3y + 2z = 2
ax + ay + 2az = 4
This system of equations can be written in the form AX = B, where
∴ A is non-singular.
Therefore, A−1 exists.
Hence, the given system of equations is consistent.
Examine the consistency of the system of equations.
5x − y + 4z = 5
2x + 3y + 5z = 2
5x − 2y + 6z = −1
The given system of equations is:
5x − y + 4z = 5
2x + 3y + 5z = 2
5x − 2y + 6z = −1
This system of equations can be written in the form of AX = B, where
∴ A is non-singular.
Therefore, A−1 exists.
Hence, the given system of equations is consistent.
Solve system of linear equations, using matrix method.
The given system of equations can be written in the form of AX = B, where
Thus, A is non-singular. Therefore, its inverse exists.
Solve system of linear equations, using matrix method.
The given system of equations can be written in the form of AX = B, where
Thus, A is non-singular. Therefore, its inverse exists.
Solve system of linear equations, using matrix method.
The given system of equations can be written in the form of AX = B, where
Thus, A is non-singular. Therefore, its inverse exists.
Solve system of linear equations, using matrix method.
5x + 2y = 3
3x + 2y = 5
The given system of equations can be written in the form of AX = B, where
Thus, A is non-singular. Therefore, its inverse exists.
Solve system of linear equations, using matrix method.
The given system of equations can be written in the form of AX = B, where
Thus, A is non-singular. Therefore, its inverse exists.
Solve system of linear equations, using matrix method.
x − y + z = 4
2x + y − 3z = 0
x + y + z = 2
The given system of equations can be written in the form of AX = B, where
Thus, A is non-singular. Therefore, its inverse exists.
Solve system of linear equations, using matrix method.
2x + 3y + 3z = 5
x − 2y + z = −4
3x − y − 2z = 3
The given system of equations can be written in the form AX = B, where
Thus, A is non-singular. Therefore, its inverse exists.
Solve system of linear equations, using matrix method.
x − y + 2z = 7
3x + 4y − 5z = −5
2x − y + 3z = 12
The given system of equations can be written in the form of AX = B, where
Thus, A is non-singular. Therefore, its inverse exists.
Solve system of linear equations, using matrix method.
x − y + 2z = 7
3x + 4y − 5z = −5
2x − y + 3z = 12
The given system of equations can be written in the form of AX = B, where
Thus, A is non-singular. Therefore, its inverse exists.
The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs 60. The cost of 2 kg onion, 4 kg
wheat and 6 kg rice is Rs 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is Rs 70.
Find cost of each item per kg by matrix method.
Let the cost of onions, wheat, and rice per kg be Rs x, Rs y,and Rs z respectively.
Then, the given situation can be represented by a system of equations as:
This system of equations can be written in the form of AX = B, where
Now,
X = A−1B
Hence, the cost of onions is Rs 5 per kg, the cost of wheat is Rs 8 per kg, and the cost of rice is Rs 8 per kg.
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