Balbharti12th Maharashtra State Board Maths Solutions BookPdf Chapter 2 Matrices Ex 2.3 Questions and Answers.
Maharashtra State Board 12th Maths Solutions Chapter 2 Matrices Ex 2.3
Solution & Step-by-Step Answer:
The given equations can be written in the matrix form as : = This is of the form AX = B, where ∴ A-1 = Now, premultiply AX = B by A-1, we get, A-1(AX) = A-1B ∴ (A-1A)X = A-1B ∴ IX = A-1B ∴ X = ∴ = = 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:
≤ft[{array}{cc}
1 & 1 \\
2 & -1
{array}]≤ft[{array}{l}
x \\
y
{array}] = ≤ft[{array}{l}
4 \\
5
{array}
This is of the form AX = B ⇒ X ⇒ A-1B
A = ≤ft[{array}{cc}
1 & 1 \\
2 & -1
{array}
|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 :
≤ft[{array}{ll}
2 & 6 \\
1 & 3
{array}]≤ft[{array}{l}
x \\
y
{array}]=≤ft[{array}{l}
8 \\
5
{array}
This is of the form AX = B, where
A = ≤ft[{array}{ll}
2 & 6 \\
1 & 3
{array}], X = ≤ft[{array}{l}
x \\
y
{array}] and B = ≤ft[{array}{l}
8 \\
5
{array}
Let us find A-1.
|A| = ≤ft|{array}{ll}
2 & 6 \\
1 & 3
{array} = 6 – 6 = 0
∴ A-1does not exist.
Hence, x and y do not exist.
Solution & Step-by-Step Answer:
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 :
≤ft[{array}{ll}
1 & 3 \\
3 & 5
{array}]≤ft[{array}{l}
x \\
y
{array}] = ≤ft[{array}{l}
2 \\
4
{array}
By R2– 3R1, we get
≤ft[{array}{rr}
1 & 3 \\
0 & -4
{array}]≤ft[{array}{l}
x \\
y
{array}] = ≤ft({array}{r}
2 \\
-2
{array}
∴ ≤ft[{array}{l}
x+3 \\
0-4 y
{array}] = ≤ft[{array}{r}
2 \\
-2
{array}
By equality of matrices,
x + 3y = 2 …(1)
-4y = -2
From (2), y =
Substituting y = in (1), we get,
x + = 2
∴ x = 2 –
Hence, x = , y = 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 =
x = 2
Hence, x = 2, y = -3 is the required solution.
Solution & Step-by-Step Answer:
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.
Solution & Step-by-Step Answer:
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.
Solution & Step-by-Step Answer:
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.
