Balbharati Maharashtra State Board12th Commerce Maths Solution Book PdfChapter 2 Matrices Ex 2.3 Questions and Answers.
Maharashtra State Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.3
Question 1
Maharashtra Board Solution
Evaluate: (i)
Solution & Step-by-Step Answer:
=
(ii) ≤ft[{array}{lll}
2 & -1 & 3
{array}]≤ft[{array}{l}
4 \\
3 \\
1
{array}
Solution:
≤ft[{array}{lll}
2 & -1 & 3
{array}]≤ft[{array}{l}
4 \\
3 \\
1
{array} = [8 – 3 + 3] = [8]
Question 2
Maharashtra Board Solution
If A = , B = . State whether AB = BA? Justify your
Solution & Step-by-Step Answer:
Solution: From (1) and (2), AB ≠ BA.
Question 3
Maharashtra Board Solution
Show that AB = BA, where A = , B =
Solution & Step-by-Step Answer:
From (1) and (2), AB = BA.
Question 4
Maharashtra Board Solution
Verify A(BC) = (AB)C, if A = , B = , and C =
Solution & Step-by-Step Answer:
From (1) and (2), A(BC) = (AB)C.
Question 5
Maharashtra Board Solution
Verify that A(B + C) = AB + AC, if A = , B = and C =
Solution & Step-by-Step Answer:
From (1) and (2), A(B + C) = AB + AC.
Question 6
Maharashtra Board Solution
If A = , B = , show that matrix AB is non-singular.
Solution & Step-by-Step Answer:
Hence, AB is a non-singular matrix.
Question 7
Maharashtra Board Solution
If A + I = , find the product (A + I)(A – I).
Solution & Step-by-Step Answer:
Question 8
Maharashtra Board Solution
If A = , show that A2 – 4A is a scalar matrix.
Solution & Step-by-Step Answer:
which is a scalar matrix.
Question 9
Maharashtra Board Solution
If A = , find k so that A2 – 8A – kI = O, where I is a 2 × 2 unit matrix and O is null matrix of order 2.
Solution & Step-by-Step Answer:
By equality of matrices, -k – 7 = 0 ∴ k = -7.
Question 10
Maharashtra Board Solution
If A = , prove that A2 – 5A + 7I = 0, where I is a 2 × 2 unit matrix.
Solution & Step-by-Step Answer:
Question 11
Maharashtra Board Solution
If A = , B = and if(A + B)2 = A2 + B2, find values of a and b.
Solution & Step-by-Step Answer:
(A + B)2 = A2 + B2 ∴ (A + B)(A + B) = A2 + B2 ∴ A2 + AB + BA + B2 = A2 + B2 ∴ AB + BA = 0 ∴ AB = -BA By the equality of matrices, we get 0 = a – 2 ……..(1) 0 = 1 + b ……..(2) a + 2b = 2a – 4 ……..(3) -a – 2b = 2 + 2b ……..(4) From equations (1) and (2), we get a = 2 and b = -1 The values of a and b satisfy equations (3) and (4) also. Hence, a = 2 and b = -1.
Question 12
Maharashtra Board Solution
Find k, if A = and A2 = kA – 2I.
Solution & Step-by-Step Answer:
By equality of matrices, 1 = 3k – 2 ……..(1) -2 = -2k ……..(2) 4 = 4k ……..(3) -4 = -2k – 2 ……..(4) From (2), k = 1. k = 1 also satisfies equation (1), (3) and (4). Hence, k = 1.
Question 13
Maharashtra Board Solution
Find x and y, if
Solution & Step-by-Step Answer:
By equality of matrices, x = 19 and y = 12.
Question 14
Maharashtra Board Solution
Find x, y, z, if
Solution & Step-by-Step Answer:
By equality of matrices, -6 = x – 3, 0 = y – 1 and -2 = 2z ∴ x = -3, y = 1 and z = -1.
Question 15
Maharashtra Board Solution
Jay and Ram are two friends. Jay wants to buy 4 pens and 8 notebooks. Ram wants to buy 5 pens and 12 notebooks. The price of one pen and one notebook was ₹ 6 and ₹ 10 respectively. Using matrix multiplication, find the amount each one of them requires for buying the pens and notebooks.
Solution & Step-by-Step Answer:
The given data can be written in matrix form as: Number of Pens and Notebooks For finding the amount each one of them requires to buy the pens and notebook, we require the multiplication of the two matrices A and B. Hence, Jay requires ₹ 104 and Ram requires ₹ 150 to buy the pens and notebooks.
