Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 2 Matrices Ex 2.3 Questions and Answers.

## Maharashtra State Board 12th Maths Solutions Chapter 2 Matrices Ex 2.3

Question 1.

Solve the following equations by the inversion method.

(i) x + 2y = 2, 2x + 3y = 3

Solution:

The given equations can be written in the matrix form as :

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

1 & 2 \\

2 & 3

\end{array}\right]\left[\begin{array}{l}

x \\

y

\end{array}\right]\) = \(\left[\begin{array}{l}

2 \\

3

\end{array}\right]\)

This is of the form AX = B, where

∴ A^{-1} = \(\left[\begin{array}{rr}

-3 & 2 \\

2 & -1

\end{array}\right]\)

Now, premultiply AX = B by A^{-1}, we get,

A^{-1}(AX) = A^{-1}B

∴ (A^{-1}A)X = A^{-1}B

∴ IX = A^{-1}B

∴ X = \(=\left[\begin{array}{rr}

-3 & 2 \\

2 & -1

\end{array}\right]\left[\begin{array}{l}

2 \\

3

\end{array}\right]\)

∴ \(\left[\begin{array}{l}

x \\

y

\end{array}\right]\) = \(=\left[\begin{array}{r}

-6+6 \\

4-3

\end{array}\right]\) = \(=\left[\begin{array}{l}

0 \\

1

\end{array}\right]\)

By equality of matrices,

x = 0, y = 1 is the required solution.

(ii) x + y = 4, 2x – y = 5

Solution:

x + y = 4, 2x – y = 5

The given equations can be written in the matrix form as:

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

1 & 1 \\

2 & -1

\end{array}\right]\left[\begin{array}{l}

x \\

y

\end{array}\right]\) = \(\left[\begin{array}{l}

4 \\

5

\end{array}\right]\)

This is of the form AX = B ⇒ X ⇒ A^{-1}B

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

1 & 1 \\

2 & -1

\end{array}\right]\)

|A| = -1 – 2 = -3 ≠ 0

By equality of matrices.

x = 3, y = 1

(iii) 2x + 6y = 8, x + 3y = 5

Solution:

The given equations can be written in the matrix form as :

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

2 & 6 \\

1 & 3

\end{array}\right]\left[\begin{array}{l}

x \\

y

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

8 \\

5

\end{array}\right]\)

This is of the form AX = B, where

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

2 & 6 \\

1 & 3

\end{array}\right]\), X = \(\left[\begin{array}{l}

x \\

y

\end{array}\right]\) and B = \(\left[\begin{array}{l}

8 \\

5

\end{array}\right]\)

Let us find A^{-1}.

|A| = \(\left|\begin{array}{ll}

2 & 6 \\

1 & 3

\end{array}\right|\) = 6 – 6 = 0

∴ A^{-1} does not exist.

Hence, x and y do not exist.

Question 2.

Solve the following equations by reduction method.

(i) 2x + y = 5, 3x + 5y = -3

Solution:

The given equations can be written in the matrix form as :

By equality of matrices,

2x + y = 5 …(1)

7y = -21 …(2)

From (2), y = -3

Substituting y = -3 in (1), we get,

2x – 3 = 5

∴ 2x = 8 ∴ x = 4

Hence, x = 4, y = -3 is the required solution.

(ii) x + 3y = 2, 3x + 5y = 4.

Solution:

The given equations can be written in the matrix form as :

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

1 & 3 \\

3 & 5

\end{array}\right]\left[\begin{array}{l}

x \\

y

\end{array}\right]\) = \(\left[\begin{array}{l}

2 \\

4

\end{array}\right]\)

By R_{2} – 3R_{1}, we get

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

1 & 3 \\

0 & -4

\end{array}\right]\left[\begin{array}{l}

x \\

y

\end{array}\right]\) = \(\left(\begin{array}{r}

2 \\

-2

\end{array}\right)\)

∴ \(\left[\begin{array}{l}

x+3 \\

0-4 y

\end{array}\right]\) = \(\left[\begin{array}{r}

2 \\

-2

\end{array}\right]\)

By equality of matrices,

x + 3y = 2 …(1)

-4y = -2

From (2), y = \(\frac{1}{2}\)

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

x + \(\frac{3}{2}\) = 2

∴ x = 2 – \(\frac{3}{2}=\frac{1}{2}\)

Hence, x = \(\frac{1}{2}\), y = \(\frac{1}{2}\) is the required solution.

(iii) 3x – y = 1, 4x + y = 6

Solution:

The given equations can be written in the matrix form as :

By equality of matrices,

12x – 4y = 4 … (1)

7y = 14 … (2)

From (2), y = 2

Substituting y = 2 in (1), we get,

12x – 8 = 4

∴ 12x = 12 ∴ x = 1

Hence, x = 1, y = 2 is the required solution.

(iv) 5x + 2y = 4, 7x + 3y = 5

Solution:

5x + 2y = 4 ………..(1)

7x + 3y = 5 …………(2)

Multiplying Eq. (1) with 7 and Eq. (2) with 5

Put y = -3 into Eq. (1)

5x + 2y = 4

5x + 2(-3) = 4

5x – 6 = 4

5x = 4 + 6

5x = 10

x = \(\frac{10}{5}\)

x = 2

Hence, x = 2, y = -3 is the required solution.

Question 3.

The cost of 4 pencils, 3 pens and 2 erasers is ₹ 60. The cost of 2 pencils, 4 pens and 6 erasers is ₹ 90, whereas the cost of 6 pencils, 2 pens and 3 erasers is ₹ 70. Find the cost of each item by using matrices.

Solution:

Let the cost of 1 pencil, 1 pen and 1 eraser be ₹ x, ₹ y and ₹ z respectively.

Then, from the given conditions,

4x + 3y + 2z = 60

2x + 4y + 6z = 90, i.e., x + 2y + 3z = 45

6x + 2y + 3z = 70

These equations can be written in the matrix form as :

By equality of matrices,

x + 2y + 3z = 45 …….(1)

– 5y – 10z = – 120 …….(2)

5z = 40

From (3), z = 8

Substituting z = 8 in (2), we get,

– 5y – 80 = -120

∴ – 5y = -40 ∴ y = 8

Substituting y = 8, z = 8 in (1), we get,

x + 16 + 24 = 45

∴ x + 40 = 45 ∴ x = 5

∴ x = 5, y = 8, z = 8

Hence, the cost is ₹ 5 for a pencil, ₹ 8 for a pen and ₹ 8 for an eraser.

Question 4.

If three numbers are added, their sum is 2. If 2 times the second number is subtracted from the sum of first and third numbers, we get 8 and if three times the first number is added to the sum of second and third numbers, we get 4. Find the numbers using matrices.

Solution:

Let the three numbers be x, y and z. According to the given conditions,

x + y + z = 2

x + z – 2y = 8, i.e., x – 2y + 2 = 8

and y + z + 3x = 4, i.e., 3x + y + z = 4

Hence, the system of linear equations is

x + y + z = 2

x – 2y + z = 8

3x + y + z = 4

These equations can be written in the matrix form as :

By equality of matrices,

x + y + z = 2 ……(1)

-3y = 6 ……(2)

– 2y – 2z = -2 ……..(3)

From (2), y = -2

Substituting y = -2 in (3), we get,

-2(-2) – 2z = -2

∴ -2z = -6 ∴ z = 3

Substituting y = -2, z = 3 in (1), we get,

x – 2 + 3 = 2 ∴ x = 1

Hence, the required numbers are 1, -2 and 3.

Question 5.

The total cost of 3 T.V. sets and 2 V.C.R.s is ₹ 35000. The shop-keeper wants profit of ₹ 1000 per television and ₹ 500 per V.C.R. He can sell 2 T. V. sets and 1 V.C.R. and get the total revenue as ₹ 21,500. Find the cost price and the selling price of a T.V. sets and a V.C.R.

Solution:

Let the cost of each T.V. set be ₹ x and each V.C.R. be ₹ y. Then the total cost of 3 T.V. sets and 2 V.C.R.’s is ₹ (3x + 2y) which is given to be ₹ 35,000.

∴ 3x + 2y = 35000

The shopkeeper wants profit of ₹ 1000 per T.V. set and of ₹ 500 per V.C.R.

∴ the selling price of each T.V. set is ₹ (x + 1000) and of each V.C.R. is ₹ (y + 500).

∴ selling price of 2 T.V. set and 1 V.C.R. is

₹ [2(x + 1000) + (y + 500)] which is given to be ₹ 21,500.

∴ 2(x + 1000) + (y + 500) = 21500

∴ 2x + 2000 + y + 500 = 21500

∴ 2x + y = 19000

Hence, the system of linear equations is

3x + 2y = 35000

2x + y = 19000

These equations can be written in the matrix form as :

By equality of matrices,

2x + y = 19000 ……….(1)

-x = -3000 ……….(2)

From (2), x = 3000

Substituting x = 3000 in (1), we get,

2(3000) + y = 19000

∴ y = 13000

∴ the cost price of one T.V. set is ₹ 3000 and of one V.C.R. is ₹ 13000 and the selling price of one T.V. set is ₹ 4000 and of one V.C.R. is ₹ 13500.