Chapter 2 Matrices Miscellaneous Exercise 2B Solutions

Balbharti12th Maharashtra State Board Maths Solutions BookPdf Chapter 2 Matrices Miscellaneous Exercise 2B Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B

I) Choose the correct answer from the given alternatives in each of the following questions :
Question 1.
If A = ≤ft({array}{ll}
1 & 2 \\
3 & 4
{array}), adj = ≤ft({array}{ll}
4 & a \\
-3 & b
{array} then the values of a and b are,
(a) a = – 2, b = 1
(b) a = 2, b = 4
(c) a = 2, b = –1
(d) a = 1, b = –2
Solution:
(a) a = – 2, b = 1

II) Solve the following equations by the methods of inversion.
(i) 2x – y = -2, 3x + 4y = 5
Solution:
The given equations can be written in the matrix form as :



By equality of matrices,
x = , y = is the required solution.

(ii) x + y + z = 1, 2x + 3y + 2z = 2 and ax + ay + 2az = 4, a ≠ 0.
Solution:
The given equations can be written in the matrix form as :

= 1(6a – 2a) – 1(4a – 2a) + 1(2a – 3a)
= 4a – 2a – a = a ≠ 0 ∴ A^-1exists.
Consider AA^-1= I

(iii) 5x – y +4z = 5, 2x + 3y + 5z = 2 and 5x – 2y + 6z = -1
Solution:
The given equations can be written in the matrix form as :
= 5(18 + 10) + 1 (12 – 25) + 4( -4 – 15)
= 140 – 13 – 76 = 51 #0
∴ A-1exists.
Now, we have to find the cofactor matrix
Now, premultiply AX = B by A-1, we get,
A-1(AX) = A-1B
∴ (A-1A)X = A-1B
∴ IX = A-1B
∴ X = ≤ft[{array}{rrr}
28 & -2 & -17 \\
13 & 10 & -17 \\
-19 & 5 & 17
{array}]≤ft[{array}{r}
5 \\
2 \\
-1
{array}
By equality of matrices,
x = 3, y = 2, z = -2 is the required solution.

(iv) 2x + 3y = -5, 3x + y = 3
Solution:

(v) x + y + z = -1, y + z = 2 and x + y – z = 3
Solution:
The given equations can be written in the matrix form as :

= 1(-1 – 1) – 1 (0 – 1) + 1(0 – 1)
= -2 + 1 – 1 = -2 ≠ 0 ∴ A^-1exists.
Consider AA^-1= I


Now, premultiply AX = B by A^-1, we get,
A^-1(AX) = A^-1B
∴ (A^-1A)X = A^-1B
∴ IX = A^-1B

∴ by equality of the matrices, x= -3, y = 4, z = -2 is the required solution.

(ii) x + y = 1, y + z = , z + x = .
Solution:
The given equations can be written in the matrix form as :

By equality of matrices,
x + y = 1 ……(1)
y + z = …(2)
2z = 2 ……..(3)
From (3), z = 1
Substituting z = 1 in (2), we get,
y + 1 = ∴ y =
Substituting y = in (1), we get,
x + = 1 ∴ x =
Hence, x = , y = , z = 1 is the required solution.

(iii) 2x – y + z = 1, x + 2y + 3z = 8 and 3x + y – 4z = 1
Solution:
The given equations can be written in the matrix form as :
∴ ≤ft[{array}{r}
x+2 y+3 z \\
0-5 y-5 z \\
0+0-8 z
{array}] = ≤ft[{array}{r}
8 \\
-15 \\
-8
{array}
By equality of matrices,
x + 2y + 3z = 8 …..(1)
-5y – 5z = -15 ….(2)
-8z = -8 …..(3)
From (3), z = 1
Substituting z = 1 in (2), we get,
-5y – 5 = -15
-5y = -10
∴ y = 2
Substituting y = 2, z = 1 in (1), we get,
x + 4 + 3 = 8 ∴ x = 1
Hence, x = 1, y = 2, z = 1 is the required solution.

(iv) x + y + z = 6, 3x – y + 3z =10 and 5x + 5y – 4z = 3.
Solution:

(v) x + 2y + z = 8, 2x + 3y – z =11 and 3x – y – 2z = 5
Solution:
The given equations can be written in the matrix form as :

By equality of matrices,
x + 2y + z = 8 … (1)
-y – 3z = -5 … (2)
16z = 16 … (3)
From (3), z = 1
Substituting z = 1 in (2), we get,
-y – 3 = -5, ∴ y = 2
Substituting y = 2, z = 1 in (1), we get,
x + 4 + 1 = 8 ∴ x = 3
Hence, x = 3, y = 2, z = 1 is the required solution.

(vi) x + 3y + 2z = 6, 3x – 2y + 5z =5 and 2x – 3y + 6z = 7.
Solution:
The given equations can be written in the matrix form as :


By equality of matrices,
x + 3y + 2z = 6 …(1)
y + z = 4 …(2)
z = 31 …..(3)
From (3), z = 2
Substituting z = 2 in (2), we get,
y + z = 4
y + (2) = 4
y + 3 = 4
y = 1
Substituting y = 1, z = 2 in (2), we get,
x + 3y + 2z = 6
x + 3(1) + 2(2) = 6
x + 3 + 4 = 6
x = -1
Hence, x = -1, y = 1, z = 2 is the required solution.