Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 2 Matrices Ex 2.2 Questions and Answers.

## Maharashtra State Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.2

Question 1.

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

2 & -3 \\

5 & -4 \\

-6 & 1

\end{array}\right]\), B = \(\left[\begin{array}{cc}

-1 & 2 \\

2 & 2 \\

0 & 3

\end{array}\right]\) and C = \(\left[\begin{array}{cc}

4 & 3 \\

-1 & 4 \\

-2 & 1

\end{array}\right]\) show that

(i) A + B = B + A

(ii) (A + B) + C = A + (B + C)

Solution:

From (1) and (2), we get

(A + B) + C = A + (B + C).

Question 2.

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

1 & -2 \\

5 & 3

\end{array}\right]\), B = \(\left[\begin{array}{ll}

1 & -3 \\

4 & -7

\end{array}\right]\), then find the matrix A – 2B + 6I, where I is the unit matrix of order 2.

Solution:

Question 3.

If A = \(\left[\begin{array}{ccc}

1 & 2 & -3 \\

-3 & 7 & -8 \\

0 & -6 & 1

\end{array}\right]\), B = \(\left[\begin{array}{ccc}

9 & -1 & 2 \\

-4 & 2 & 5 \\

4 & 0 & -3

\end{array}\right]\), then find the matrix C such that A + B + C is a zero matrix.

Solution:

A + B + C = 0

∴ C = -A – B

Question 4.

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

1 & -2 \\

3 & -5 \\

-6 & 0

\end{array}\right]\), B = \(\left[\begin{array}{cc}

-1 & -2 \\

4 & 2 \\

1 & 5

\end{array}\right]\) and C = \(\left[\begin{array}{cc}

2 & 4 \\

-1 & -4 \\

-3 & 6

\end{array}\right]\), find the matrix X such that 3A – 4B + 5X = C.

Solution:

3A – 4B + 5X = C

∴ 5X = C – 3A + 4B

Question 5.

If A = \(\left[\begin{array}{ccc}

5 & 1 & -4 \\

3 & 2 & 0

\end{array}\right]\), find (A^{T})^{T}.

Solution:

A = \(\left[\begin{array}{ccc}

5 & 1 & -4 \\

3 & 2 & 0

\end{array}\right]\)

∴ A^{T} = \(\left[\begin{array}{rr}

5 & 3 \\

1 & 2 \\

-4 & 0

\end{array}\right]\)

∴ (A^{T})^{T} = \(\left[\begin{array}{ccc}

5 & 1 & -4 \\

3 & 2 & 0

\end{array}\right]\) = A

Question 6.

If A = \(\left[\begin{array}{ccc}

7 & 3 & 1 \\

-2 & -4 & 1 \\

5 & 9 & 1

\end{array}\right]\), find (A^{T})^{T}.

Solution:

A = \(\left[\begin{array}{ccc}

7 & 3 & 1 \\

-2 & -4 & 1 \\

5 & 9 & 1

\end{array}\right]\)

∴ A^{T} = \(\left[\begin{array}{rrr}

7 & -2 & 5 \\

3 & -4 & 9 \\

1 & 1 & 1

\end{array}\right]\)

∴ (A^{T})^{T} = \(\left[\begin{array}{ccc}

7 & 3 & 1 \\

-2 & -4 & 1 \\

5 & 9 & 1

\end{array}\right]\) = A

Question 7.

Find a, b, c if \(\left[\begin{array}{ccc}

1 & \frac{3}{5} & a \\

b & -5 & -7 \\

-4 & c & 0

\end{array}\right]\) is a symetric matrix.

Solution:

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

1 & \frac{3}{5} & a \\

b & -5 & -7 \\

-4 & c & 0

\end{array}\right]\)

Since, A is a symmetric matrix, a_{ij} = a_{ji} for all i and j

Question 8.

Find x, y, z if \(\left[\begin{array}{ccc}

0 & -5 i & x \\

y & 0 & z \\

\frac{3}{2} & -\sqrt{2} & 0

\end{array}\right]\) is a skew symmetric matrix.

Solution:

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

0 & -5 i & x \\

y & 0 & z \\

\frac{3}{2} & -\sqrt{2} & 0

\end{array}\right]\)

Since, A is skew-symmetric matrix,

Question 9.

For each of the following matrices, find its transpose and state whether it is symmetric, skew-symmetric or neither:

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

1 & 2 & -5 \\

2 & -3 & 4 \\

-5 & 4 & 9

\end{array}\right]\)

Solution:

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

1 & 2 & -5 \\

2 & -3 & 4 \\

-5 & 4 & 9

\end{array}\right]\)

Then A^{T} = \(\left[\begin{array}{rrr}

1 & 2 & -5 \\

2 & -3 & 4 \\

-5 & 4 & 9

\end{array}\right]\)

Since, A = A^{T}, A is a symmetric matrix.

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

2 & 5 & 1 \\

-5 & 4 & 6 \\

-1 & -6 & 3

\end{array}\right]\)

Solution:

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

2 & 5 & 1 \\

-5 & 4 & 6 \\

-1 & -6 & 3

\end{array}\right]\)

Then B^{T} = \(\left(\begin{array}{rrr}

2 & -5 & -1 \\

5 & 4 & -6 \\

1 & 6 & 3

\end{array}\right)\)

∴ B ≠ B^{T}

Also,

-B^{T} = \(\left(\begin{array}{rrr}

2 & -5 & -1 \\

5 & 4 & -6 \\

1 & 6 & 3

\end{array}\right)=\left(\begin{array}{rrr}

-2 & 5 & 1 \\

-5 & -4 & 6 \\

-1 & -6 & -3

\end{array}\right)\)

∴ B ≠ -B^{T}

Hence, B is neither symmetric nor skew-symmetric matrix.

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

0 & 1+2 i & i-2 \\

-1-2 i & 0 & -7 \\

2-i & 7 & 0

\end{array}\right]\)

Solution:

Hence, C is a skew-symmetric matrix.

Question 10.

Construct the matrix A = [a_{ij}]_{3×3}, where a_{ij} = i – j. State whether A is symmetric or skew-symmetric.

Solution:

Question 11.

Solve the following equations for X and Y, if 3X – Y = \(\left[\begin{array}{cc}

1 & -1 \\

-1 & 1

\end{array}\right]\) and X – 3Y = \(\left[\begin{array}{ll}

0 & -1 \\

0 & -1

\end{array}\right]\)

Solution:

Question 12.

Find matrices A and B, if 2A – B = \(\left[\begin{array}{ccc}

6 & -6 & 0 \\

-4 & 2 & 1

\end{array}\right]\) and A – 2B = \(\left[\begin{array}{ccc}

3 & 2 & 8 \\

-2 & 1 & -7

\end{array}\right]\)

Solution:

Question 13.

Find x and y, if \(\left[\begin{array}{ccc}

2 x+y & -1 & 1 \\

3 & 4 y & 4

\end{array}\right]+\left[\begin{array}{ccc}

-1 & 6 & 4 \\

3 & 0 & 3

\end{array}\right]=\left[\begin{array}{ccc}

3 & 5 & 5 \\

6 & 18 & 7

\end{array}\right]\)

Solution:

By equality of matrices, we get

2x + y – 1 = 3 ……..(1)

and 4y = 18 ……….(2)

From (2), y = \(\frac{9}{2}\)

Substituting y = \(\frac{9}{2}\) in (1), we get

2x + \(\frac{9}{2}\) – 1 = 3

∴ 2x = 3 – \(\frac{7}{2}\) = \(\frac{-1}{2}\)

∴ x = \(\frac{-1}{4}\)

Hence, x = \(\frac{-1}{4}\) and y = \(\frac{9}{2}\).

Question 14.

If \(\left[\begin{array}{cc}

2 a+b & 3 a-b \\

c+2 d & 2 c-d

\end{array}\right]=\left[\begin{array}{cc}

2 & 3 \\

4 & -1

\end{array}\right]\), find a, b, c and d.

Solution:

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

2 a+b & 3 a-b \\

c+2 d & 2 c-d

\end{array}\right]=\left[\begin{array}{cc}

2 & 3 \\

4 & -1

\end{array}\right]\)

By equality of matrices,

2a + b = 2 ….. (1)

3a – b = 3 …… (2)

c + 2d = 4 …… (3)

2c – d = -1 …… (4)

Adding (1) and (2), we get

5a = 5

∴ a = 1

Substituting a = 1 in (1), we get

2(1) + b = 2

∴ b = 0

Multiplying equation (4) by 2, we get

4c – 2d = -2 …… (5)

Adding (3) and (5), we get

5c = 2

∴ c = \(\frac{2}{5}\)

Substituting c = \(\frac{2}{5}\) in (4), we get

2(\(\frac{2}{5}\)) – d = -1

∴ d = \(\frac{4}{5}\) + 1 = \(\frac{9}{5}\)

Hence, a = 1, b = 0, c = \(\frac{2}{5}\) and d = \(\frac{9}{5}\).

Question 15.

There are two book shops owned by Suresh and Ganesh. Their sales (in Rupees) for books in three subjects – Physics, Chemistry and Mathematics for two months, July and August 2017 are given by two matrices A and B:

July sales (in Rupees), Physics, Chemistry, Mathematics

A = \(\left[\begin{array}{lll}

5600 & 6750 & 8500 \\

6650 & 7055 & 8905

\end{array}\right]\) First Row Suresh / Second Row Ganesh

August Sales (in Rupees), Physics, Chemistry, Mathematics

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

6650 & 7055 & 8905 \\

7000 & 7500 & 10200

\end{array}\right]\) First Row Suresh / Second Row Ganesh

(i) Find the increase in sales in Z from July to August 2017.

(ii) If both book shops get 10% profit in the month of August 2017,

find the profit for each bookseller in each subject in that month.

Solution:

The sales for July and August 2017 for Suresh and Ganesh are given by the matrices A and B as:

(i) The increase in sales (in ₹) from July to August 2017 is obtained by subtracting the matrix A from B.

Hence, the increase in sales (in ₹) from July to August 2017 for:

Suresh book shop: ₹ 1050 in Physics, ₹ 305 in Chemistry, and ₹ 405 in Mathematics.

Ganesh book shop: ₹ 350 in Physics, ₹ 445 in Chemistry, and ₹ 1295 in Mathematics.

(ii) Both the book shops get 10% profit in August 2017,

the profit for each bookseller in each subject in August 2017 is obtained by the scalar multiplication of matrix B by 10%,

i.e. \(\frac{10}{100}=\frac{1}{10}\)

Hence, the profit for Suresh book shop are ₹ 665 in Physics, ₹ 705.50 in Chemistry and ₹ 890.50 in Mathematics and for Ganesh book shop are ₹ 700 in Physics, ₹ 750 in Chemistry and ₹ 1020 in Mathematics.