Examine the consistency of the system of equations in Exercises 1 to 6.
Question 1. x + 2y = 2
2x + 3y = 3
Solution:
Matrix form of the given equations is AX = B
where, A =
, B = and, X = ∴
Now, |A| =
∵ Inverse of matrix exists, unique solution.
∴ System of equation is consistent.
Question 2. 2x – y = 5
x + y = 4
Solution:
Matrix form of the given equations is AX = B
where, A =
, B = and, X = ∴
Now, |A| =
∵ Inverse of matrix exists, unique solution.
∴ System of equation is consistent.
Question 3. x + 3y = 5
2x + 6y = 8
Solution:
Matrix form of the given equations is AX = B
where, A =
, B = and, X = ∴
Now, |A| =
And, adj. A =
∴ (adj. A) B =
∵ Have no common solution.
∴ System of equation is inconsistent.
Question 4. x + y + z = 1
2x + 3y + 2z = 2
ax + ay + 2az = 4
Solution:
Matrix form of the given equations is AX = B
where, A =
, B = and, X = ∴
Now, |A| =
∵ Inverse of matrix exists, unique solution.
∴ System of equation is consistent.
Question 5. 3x – y – 2z = 2
2y – z = -1
3x – 5y = 3
Solution:
Matrix form of the given equations is AX = B
where, A =
, B= and, X = ∴
Now, |A| =
And, adj. A =
∴ (adj. A) B =
∴ System of equation is inconsistent.
Question 6. 5x – y + 4z = 5
2x + 3y + 5z = 2
5x – 2y + 6z = –1
Solution:
Matrix form of the given equations is AX = B
where, A =
, B = and, X= ∴
Now, |A| =
∵ Inverse of matrix exists, unique solution.
∴ System of equation is consistent.
Solve system of linear equations, using matrix method, in Exercises 7 to 14.
Question 7. 5x + 2y = 4
7x + 3y = 5
Solution:
Matrix form of the given equations is AX = B
where, A=
, B= , X= ∴
Now, |A|=
∴Unique solution
Now, X = A-1B =
(adj.A)B
Therefore, x=2 and y=-3
Question 8. 2x – y = -2
3x + 4y = 3
Solution:
Matrix form of the given equations is AX = B
where, A=
, B= , X= ∴
Now, |A|=
∴Unique solution
Now, X = A-1B
(adj.A)B
Therefore, x=-5/11 and y=12/11
Question 9. 4x – 3y = 3
3x – 5y = 7
Solution:
Matrix form of the given equations is AX = B
where, A=
, B= , X= ∴
Now, |A|=
∴Unique solution
n Now, X =A-1B
A(adj.A)B
Therefore, x= -6/11 and y= -19/11
Question 10. 5x + 2y = 3
3x + 2y = 5
Solution:
Matrix form of the given equations is AX = B
where, A=
, B= , X= ∴
Now, |A|=
∴Unique solution
Now, X = A-1B
A(adj.A)B
Therefore, x= -1 and y= 4
x – 2y – z = 3/2
3y – 5z = 9
Solution:
Matrix form of the given equation is AX = B
i.e.
∴ |A| =
∴ Solution is unique.
Now, X = A-1B =
(adj.A)B
Therefore, x=1, y=1/2, z=3/2
Question 12. x – y + z = 4
2x + y – 3z = 0
x + y + z = 2
Solution:
Matrix form of the given equation is AX = B
i.e
∴ |A| =
∴ Solution is unique.
Now, X = A-1B =
(adj.A)B
Therefore, x = 2, y = -1, z = 1
Question 13. 2x + 3y +3 z = 5
x – 2y + z = – 4
3x – y – 2z = 3
Solution:
Matrix form of given equation is AX = B
i.e.
∴ |A| =
∴ Solution is unique.
Now, X = A-1B =
(adj.A)B
Therefore, x = 1, y = 2, z = -1
Question 14. x – y + 2z = 7
3x + 4y – 5z = – 5
2x – y + 3z = 12
Solution:
Matrix form of given equation is AX = B
i.e.
∴ |A| =
∴ Solution is unique.
Now, X = A-1B =
(adj.A)B
Therefore, x = 2, y = 1, z = 3
Question 15. If A=
, find A–1. Using A–1 solve the system of equations 2x – 3y + 5z = 11
3x + 2y – 4z = – 5
x + y – 2z = – 3
Solution:
Given: A=
Now, |A|=
∴ |A|=
Means, A-1 exists.
And A-1 =
(adj.A)……(1) Now,
∴ adj. A =
From eq. (1),
A-1=
Now, Matrix form of given equation is AX = B
i.e.
∵ Solution is unique.
∴ X=A-1B
⇒
Therefore, x = 1, y = 2, z = 3
Question 16. The cost of 4 kg onion, 3 kg wheat and 2 kg rice is 60 rupees. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is 90 rupees. The cost of 6 kg onion 2 kg wheat and 3 kg rice is 70 rupees. Find cost of each item per kg by matrix method.
Solution:
Let Rs x, Rs y, Rs z per kg be the prices of onion, wheat and rice respectively.
A.T.Q.
4x+3y+2z=60
2x+4y+6z=90
6x+2y+3z=70
Matrix form of equation is AX = B
where, A=
,B= and X= =>
Now, |A|=
∴ Solution is unique
Now, X=A-1B=
(adj. A)B……(1) Now,
∴ (adj.A)=
From eqn.(1)
Therefore, x = 5, y = 8, z = 8
Hence, the cost of onion, wheat and rice are Rs. 5, Rs 8 and Rs 8 per kg.