Chapter 2 Matrices Miscellaneous Exercise 2 Solutions

Balbharati Maharashtra State Board12th Commerce Maths Solution Book PdfChapter 2 Matrices Miscellaneous Exercise 2 Questions and Answers.

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

(I) Choose the correct alternative.

(II) Fill in the blanks:

(III) State whether each of the following is True or False:

(IV) Solve the following:

(ii) ≤ft[{array}{ll}
1 & 1 \\
1 & 1
{array}
Solution:
Let A = ≤ft[{array}{ll}
1 & 1 \\
1 & 1
{array}
Then |A| = ≤ft|{array}{ll}
1 & 1 \\
1 & 1
{array}
= 1 – 1
= 0
∴ A is a singular matrix.
Hence, A-1does not exist.

(iii) ≤ft[{array}{lll}
3 & 4 & 3 \\
1 & 1 & 0 \\
1 & 4 & 5
{array}
Solution:
Let A = ≤ft[{array}{lll}
3 & 4 & 3 \\
1 & 1 & 0 \\
1 & 4 & 5
{array}
Then |A| = ≤ft|{array}{lll}
3 & 4 & 3 \\
1 & 1 & 0 \\
1 & 4 & 5
{array}
= 3(5 – 0) – 4(5 – 0) + 3(4 – 1)
= 15 – 20 + 9
= 4 ≠ 0
∴ A is a non-singular matrix.
Hence, A-1exists.

(iv) ≤ft[{array}{lll}
1 & 2 & 3 \\
2 & 4 & 5 \\
2 & 4 & 6
{array}
Solution:
Let A = ≤ft[{array}{lll}
1 & 2 & 3 \\
2 & 4 & 5 \\
2 & 4 & 6
{array}
Then |A| = ≤ft|{array}{lll}
1 & 2 & 3 \\
2 & 4 & 5 \\
2 & 4 & 6
{array}
= 1(24 – 20) – 2(12 – 10) + 3(8 – 8)
= 4 – 4 + 0
= 0
∴ A is a singular matrix.
Hence, A-1does not exist.

(ii) ≤ft[{array}{ll}
2 & 1 \\
7 & 4
{array}
Solution:

(iii) ≤ft[{array}{ccc}
2 & -3 & 3 \\
2 & 2 & 3 \\
3 & -2 & 2
{array}
Solution:

(iv) ≤ft[{array}{ccc}
2 & 0 & -1 \\
5 & 1 & 0 \\
0 & 1 & 3
{array}
Solution:

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




By equality of matrices,
x = 3, y = 1, z = 2 is the required solution.

(iii) x – y + z = 4, 2x + y – 3z = 0 and x + y + z = 2
Solution:

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

By equality of matrices,
x + 2y + z = 3 …….(1)
-7y – z = -8 …….(2)
2z = 4 …….(3)
From (3), z = 2
Substituting z = 2 in (2), we get
-7y – 2 = -8
∴ -7y = -6
∴ y =
Substituting y = , z = 2 in (1), we get
x + 2() + 2 = 3
x = 3 – 2 – =
Hence, x = , y = and z = 2 is the required solution.

(iii) x – 3y + z = 2, 3x + y + z = 1 and 5x + y + 3z = 3.
Solution: