Q1: Examine the consistency of the system of equations.
x + 2y = 2
2x + 3y = 3
Ans: 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.
Q2: Examine the consistency of the system of equations.
2x − y = 5 x
x + y = 4
Ans: 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.
Q3: Examine the consistency of the system of equations.
x + 3y = 5
2x + 6y = 8
Ans: 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
Thus, the solution of the given system of equations does not exist. Hence, the system of equations is inconsistent.
Q4: Examine the consistency of the system of equations.
x + y + z = 1
2x + 3y + 2z = 2
ax + ay + 2az = 4
Ans: 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.
Q5: Examine the consistency of the system of equations.
3x − y − 2z = 2
2y − z = −1
3x − 5y = 3
Ans: The given system of equations is:
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.
Thus, the solution of the given system of equations does not exist. Hence, the system of equations is inconsistent.
Q6: Examine the consistency of the system of equations.
5x − y + 4z = 5
2x + 3y + 5z = 2
5x − 2y + 6z = −1
Ans: 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
Therefore, A^{−1} exists.
Hence, the given system of equations is consistent.
Q7: Solve system of linear equations, using matrix method.
Ans: The given system of equations can be written in the form of AX = B, where
Thus, A is non-singular. Therefore, its inverse exists.
Q8: Solve system of linear equations, using matrix method.
Ans: The given system of equations can be written in the form of AX = B, where
Thus, A is non-singular. Therefore, its inverse exists.
Q9: Solve system of linear equations, using matrix method.
Ans: The given system of equations can be written in the form of AX = B, where
Thus, A is non-singular. Therefore, its inverse exists.
Q10: Solve system of linear equations, using matrix method.
5x + 2y = 3
3x + 2y = 5
Ans: The given system of equations can be written in the form of AX = B, where
Thus, A is non-singular. Therefore, its inverse exists.
Q11: Solve system of linear equations, using matrix method.
Ans: The given system of equations can be written in the form of AX = B, where
Thus, A is non-singular. Therefore, its inverse exists.
Q12: Solve system of linear equations, using matrix method.
x − y + z = 4
2x + y − 3z = 0
x + y + z = 2
Ans: The given system of equations can be written in the form of AX = B, where
Thus, A is non-singular. Therefore, its inverse exists.
Q13: Solve system of linear equations, using matrix method.
2x + 3y + 3z = 5
x − 2y + z = −4
3x − y − 2z = 3
Ans: The given system of equations can be written in the form AX = B, where
Thus, A is non-singular. Therefore, its inverse exists.
Q14: Solve system of linear equations, using matrix method.
x − y + 2z = 7
3x + 4y − 5z = −5
2x − y + 3z = 12
Ans: The given system of equations can be written in the form of AX = B, where
Thus, A is non-singular. Therefore, its inverse exists.
Q15: If
find A^{−1}. Using A^{−1} solve the system of equations
Ans:
Now, the given system of equations can be written in the form of AX = B, where
Q16: 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.
Ans: 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^{−1} B
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.