Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 2 Matrices Ex 2.1 Questions and Answers.

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

Question 1.

Construct a matrix A = [a_{ij}]_{3×2} whose elements aij isgiven by

(i) a_{ij} = \(\frac{(i-j)^{2}}{5-i}\)

Solution:

(ii) a_{ij} = i – 3j

Solution:

a_{ij} = i – 3j

∴ a_{11} = 1 – 3(1) = 1 – 3 = -2

a_{12} = 1 – 3(2) = 1 – 6 = -5

a_{21} = 2 – 3(1) = 2 – 3 = -1

a_{22} = 2 – 3(2) = 2 – 6 = -4

a_{31} = 3 – 3(1) = 3 – 3 = 0

a_{32} = 3 – 3(2) = 3 – 6 = -3

∴ A = \(\left[\begin{array}{cc}

-2 & -5 \\

-1 & -4 \\

0 & -3

\end{array}\right]\)

(iii) a_{ij} = \(\frac{(i+j)^{3}}{5}\)

Solution:

Question 2.

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:

(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.

Which of the following matrices are singular or non-singular:

(i) \(\left[\begin{array}{ccc}

a & b & c \\

p & q & r \\

2 a-p & 2 b-q & 2 c-r

\end{array}\right]\)

Solution:

(ii) \(\left[\begin{array}{ccc}

5 & 0 & 5 \\

1 & 99 & 100 \\

6 & 99 & 105

\end{array}\right]\)

Solution:

(iii) \(\left[\begin{array}{ccc}

3 & 5 & 7 \\

-2 & 1 & 4 \\

3 & 2 & 5

\end{array}\right]\)

Solution:

Let C = \(\left[\begin{array}{ccc}

3 & 5 & 7 \\

-2 & 1 & 4 \\

3 & 2 & 5

\end{array}\right]\)

∴ |C| = \(\left|\begin{array}{rrr}

3 & 5 & 7 \\

-2 & 1 & 4 \\

3 & 2 & 5

\end{array}\right|\)

= 3(5 – 8) – 5(-10 – 12) + 7(-4 – 3)

= -9 + 110 – 49

= 52 ≠ 0

∴ C is a non-singular matrix.

(iv) \(\left[\begin{array}{cc}

7 & 5 \\

-4 & 7

\end{array}\right]\)

Solution:

Let D = \(\left[\begin{array}{cc}

7 & 5 \\

-4 & 7

\end{array}\right]\)

∴ |D| = \(\left|\begin{array}{rr}

7 & 5 \\

-4 & 7

\end{array}\right|\)

= 49 – (-20)

= 69 ≠ 0

∴ D is a non-singular matrix.

Question 4.

Find k, if the following matrices are singular:

(i) \(\left[\begin{array}{cc}

7 & 3 \\

-2 & K

\end{array}\right]\)

Solution:

Let A = \(\left[\begin{array}{cc}

7 & 3 \\

-2 & K

\end{array}\right]\)

Since, A is a singular matrix, |A| = 0

∴ \(\left|\begin{array}{rr}

7 & 3 \\

-2 & k

\end{array}\right|\) = 0

∴ 7k – (-6) = 0

∴ 7k = -6

∴ k = \(-\frac{6}{7}\)

(ii) \(\left[\begin{array}{ccc}

4 & 3 & 1 \\

7 & \mathrm{~K} & 1 \\

10 & 9 & 1

\end{array}\right]\)

Solution:

Let B = \(\left[\begin{array}{ccc}

4 & 3 & 1 \\

7 & \mathrm{~K} & 1 \\

10 & 9 & 1

\end{array}\right]\)

Since, B is a singular matrix, |B| = 0

∴ \(\left|\begin{array}{rrr}

4 & 3 & 1 \\

7 & k & 1 \\

10 & 9 & 1

\end{array}\right|\) = 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) \(\left[\begin{array}{ccc}

K-1 & 2 & 3 \\

3 & 1 & 2 \\

1 & -2 & 4

\end{array}\right]\)

Solution:

Let C = \(\left[\begin{array}{ccc}

K-1 & 2 & 3 \\

3 & 1 & 2 \\

1 & -2 & 4

\end{array}\right]\)

Since, C is a singular matrix, |C| = 0

∴ \(\left|\begin{array}{crr}

k-1 & 2 & 3 \\

3 & 1 & 2 \\

1 & -2 & 4

\end{array}\right|\) = 0

∴ (k – 1)(4 + 4) – 2(12 – 2) + 3(-6 – 1) = 0

∴ 8k – 8 – 20 – 21 = 0

∴ 8k = 49

∴ k = \(\frac{49}{8}\)