Maharashtra State Board Class 12 (HSC Board) Commerce Mathematics & Statistics Solutions Chapter 2 Matrices Ex 2.1 - Digest Answers | Maharashtra Board Solutions

Q: Question 1: Construct a matrix A = [aij]3×2 whose elements aij isgiven by (i) aij =

Step-by-step solution provided.

Q: Question 3: Which of the following matrices are singular or non-singular: (i)

Step-by-step solution provided.

Balbharati Maharashtra State Board12th Commerce Maths Solution Book PdfChapter 2 Matrices Ex 2.1 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.1

Question 1 Maharashtra Board Solution

Construct a matrix A = [aij]3×2 whose elements aij isgiven by (i) aij =

Solution & Step-by-Step Answer:

(ii) aij= i – 3j
Solution:
aij= i – 3j
∴ a11= 1 – 3(1) = 1 – 3 = -2
a12= 1 – 3(2) = 1 – 6 = -5
a21= 2 – 3(1) = 2 – 3 = -1
a22= 2 – 3(2) = 2 – 6 = -4
a31= 3 – 3(1) = 3 – 3 = 0
a32= 3 – 3(2) = 3 – 6 = -3
∴ A = ≤ft[{array}{cc}
-2 & -5 \\
-1 & -4 \\
0 & -3
{array}

(iii) a_ij=
Solution:

Question 2 Maharashtra Board Solution

Classify each of the following matrices as a row, a column, a square, a diagonal, a scalar, a unit, an upper triangular, a lower triangular matrix:

Solution & Step-by-Step Answer:

(i) Since, all the elements below the diagonal are zero, it is an upper triangular matrix. (ii) This matrix has only one column, it is a column matrix. (iii) This matrix has only one row, it is a row matrix. (iv) Since, diagonal elements are equal and non-diagonal elements are zero, it is a scalar matrix. (v) Since, all the elements above the diagonal are zero, it is a lower triangular matrix. (vi) Since, all the non-diagonal elements are zero, it is a diagonal matrix. (vii) Since, diagonal elements are 1 and non-diagonal elements are 0, it is an identity (or unit) matrix.

Question 3 Maharashtra Board Solution

Which of the following matrices are singular or non-singular: (i)

Solution & Step-by-Step Answer:

(ii) ≤ft[{array}{ccc}
5 & 0 & 5 \\
1 & 99 & 100 \\
6 & 99 & 105
{array}
Solution:

(iii) ≤ft[{array}{ccc}
3 & 5 & 7 \\
-2 & 1 & 4 \\
3 & 2 & 5
{array}
Solution:
Let C = ≤ft[{array}{ccc}
3 & 5 & 7 \\
-2 & 1 & 4 \\
3 & 2 & 5
{array}
∴ |C| = ≤ft|{array}{rrr}
3 & 5 & 7 \\
-2 & 1 & 4 \\
3 & 2 & 5
{array}
= 3(5 – 8) – 5(-10 – 12) + 7(-4 – 3)
= -9 + 110 – 49
= 52 ≠ 0
∴ C is a non-singular matrix.

(iv) ≤ft[{array}{cc}
7 & 5 \\
-4 & 7
{array}
Solution:
Let D = ≤ft[{array}{cc}
7 & 5 \\
-4 & 7
{array}
∴ |D| = ≤ft|{array}{rr}
7 & 5 \\
-4 & 7
{array}
= 49 – (-20)
= 69 ≠ 0
∴ D is a non-singular matrix.

Question 4 Maharashtra Board Solution

Find k, if the following matrices are singular: (i)

Solution & Step-by-Step Answer:

Let A = Since, A is a singular matrix, |A| = 0 ∴ = 0 ∴ 7k – (-6) = 0 ∴ 7k = -6 ∴ k =

(ii) ≤ft[{array}{ccc}
4 & 3 & 1 \\
7 & ~K & 1 \\
10 & 9 & 1
{array}
Solution:
Let B = ≤ft[{array}{ccc}
4 & 3 & 1 \\
7 & ~K & 1 \\
10 & 9 & 1
{array}
Since, B is a singular matrix, |B| = 0
∴ ≤ft|{array}{rrr}
4 & 3 & 1 \\
7 & k & 1 \\
10 & 9 & 1
{array} = 0
∴ 4(k – 9) – 3(7 – 10) + 1(63 – 10k) = 0
∴ 4k – 36 + 9 + 63 – 10k = 0
∴ -6k + 36 = 0
∴ 6k = 36
∴ k = 6.

(iii) ≤ft[{array}{ccc}
K-1 & 2 & 3 \\
3 & 1 & 2 \\
1 & -2 & 4
{array}
Solution:
Let C = ≤ft[{array}{ccc}
K-1 & 2 & 3 \\
3 & 1 & 2 \\
1 & -2 & 4
{array}
Since, C is a singular matrix, |C| = 0
∴ ≤ft|{array}{crr}
k-1 & 2 & 3 \\
3 & 1 & 2 \\
1 & -2 & 4
{array} = 0
∴ (k – 1)(4 + 4) – 2(12 – 2) + 3(-6 – 1) = 0
∴ 8k – 8 – 20 – 21 = 0
∴ 8k = 49
∴ k =

Chapter 2 Matrices Ex 2.1 Solutions