Q: Question 1: Evaluate: i. ii.

i. {aligned} ≤ft[{array}{l} 3 \\ 2 \\ 1 {array}]≤ft[{array}{lll} 2 & -4 & 3 {array}] &=≤ft[{array}{lll} 3(2) & 3(-4) & 3(3) \\ 2(2) & 2(-4) & 2(3) \\ 1(2) & 1(-4) & 1(3) {array}] \\ &=≤ft[{array}{ccc} 6 & -12 & 9 \\ 4 & -8 & 6 \\ 2 & -4 &am

Question 1

Question 1: Evaluate: i. ii.

Accepted Answer

i. {aligned} ≤ft[{array}{l} 3 \ 2 \ 1 {array}]≤ft[{array}{lll} 2 & -4 & 3 {array}] &=≤ft[{array}{lll} 3(2) & 3(-4) & 3(3) \ 2(2) & 2(-4) & 2(3) \ 1(2) & 1(-4) & 1(3) {array}] \ &=≤ft[{array}{ccc} 6 & -12 & 9 \ 4 & -8 & 6 \ 2 & -4 &am

Question 2

Question 2: If A = B = , = show that AB ≠ BA.

Accepted Answer

From (i) and (ii), we get AB ≠ BA

Question 3

Question 3: If A = ,B = state whether AB = BA? Justify your

Accepted Answer

Solution: From (i) and (ii), we get AB ≠ BA

Question 4

Question 4: Show that AB = BA, where i. A = , B = ii. A = , B =

Accepted Answer

From (i) and (ii), we get AB = BA

Question 5

Question 5: If A = , prove that A2 = 0.

Accepted Answer

A2 = A.A = = = = 0

Question 6

Question 6: Verify A(BC) = (AB)C in each of the following cases: i. A = , B = and C = ii. A = , B = and C =

Accepted Answer

From (i) and (ii), we get A(BC) = (AB)C.

Question 7

Question 7: Verify that A(B + C) = AB + AC in each of the following matrices: i. A = , B = and C = ii. A = , B = and C =

Accepted Answer

From (i) and (ii), we get A(B + C) = AB + AC. [Note: The question has been modified.]

Question 8

Question 8: If A = , B = , find AB – 2I, where I is unit matrix of order 2.

Accepted Answer

Step-by-step solution provided.

Question 9

Question 9: If A = , B = , show that matrix AB is non singular.

Accepted Answer

im ∴ AB is non-singular matrix.

Question 10

Question 10: If A = , find the product (A + I)(A – I).

Accepted Answer

Step-by-step solution provided.

Question 11

Question 11: If A = , B = , find α, if A2 = B.

Accepted Answer

A2 = B ∴ By equality of matrices, we get α2 = 1 and α + 1 = 2 ∴ α = ± 1 and α = 1 ∴ α = 1

Question 12

Question 12: If A = , show that A2 – 4A is scalar matrix.

Accepted Answer

A2 – 4A = A.A – 4A

Question 13

Question 13: If A = , find k so that A2 – 8A – kI = O, where I is a unit matrix and O is a null matrix of order 2.

Accepted Answer

A2 – 8A – kI = O ∴ A.A – 8A – kI = O ∴ by equality of matrices, we get 1 – 8 – k = 0 ∴ k = -7

Question 14

Question 14: If A = , B = , show that (A+B)2 = A2 + AB + B2.

Accepted Answer

We have to prove that (A + B)2 = A2 + AB + B2, i.e., to prove A2 + AB + BA + B2 = A2 + AB + B2, i.e., to prove BA = 0. BA =

Question 15

Question 15: If A = , prove that A2 – 5A + 7I = 0, where I is unit matrix of order 2.

Accepted Answer

A2 – 5A + 7I = 0 = A.A – 5A + 7I = 0

Question 16

Question 16: If A = and B = , show that (A + B)(A – B) = A2 – B2.

Accepted Answer

We have to prove that (A + B)(A – B) = A2 – B2, i.e., to prove A2 – AB + BA – B2 = A2 – B2, i.e., to prove – AB + BA = 0, i.e., to prove AB – BA. From (i) and (ii), we get AB = BA

Question 17

Question 17: If A = , B = and (A + B)2 = A2 + B2, find the values of a and b.

Accepted Answer

Given, (A + B)2 = A2 + B2 ∴ A2 + AB + BA + B2 = A2 + B2 ∴ AB + BA = 0 ∴ AB = -BA ∴ by equality of matrices, we get – 2 + a = 0 and 1 + b = 0 a = 2 and b = -1 [Note: The question has been modified.]

Question 18

Question 18: Find matrix X such that AX = B, where A = and B =

Accepted Answer

Let X = But AX = B ∴ ∴ By equality of matrices, we get a – 2b = -3 …(i) -2a + b = -l …(ii) By (i) x 2 + (ii), we get -3b =-7 ∴ b = Substituting b = in (i), we get a – 2 () = -3 ∴ a = -3 + ∴ X =

Question 19

Question 19: Find k, if A = and A2 = KA – 2I

Accepted Answer

A2 = kA – 2I ∴ AA + 2I = kA ∴ ∴ By equality of matrices, we get 3k = 3 ∴ k = 1

Question 20

Question 20: Find x, if = 0

Accepted Answer

∴ [6 + 12x + 14] =[0] ∴ By equality of matrices, we get ∴ 12x + 20 = 0 ∴ 12x =-20 ∴ x =

Question 21

Question 21: Find x and y, if

Accepted Answer

∴ By equality of matrices, we get x = 19 andy = 12

Question 22

Question 22: Find x, y, z if

Accepted Answer

∴ By equality of matrices, we get x – 3 = -6,y – 1 = 0, 2z = -2 ∴ x = – 3, y = 1, z = – 1

Question 23

Question 23: If A = show that A2 =

Accepted Answer

Step-by-step solution provided.

Question 24

Question 24: If A = , B = show that AB ≠ BA, but |AB| = |A|. |B|.

Accepted Answer

AB = Now, |AB| = = 28 – 20 = 8 |A| = = 5 – 6 = -1 |B| = = 0 – 8 = -8 ∴ |A|. |B| = (-1).(-8) = 8 = |AB| ∴ AB ≠ BA, but |AB| = |A|.|B|

Question 25

Question 25: Jay and Ram are two friends in a class. Jay wanted to buy 4 pens and 8 notebooks, Ram wanted to buy 5 pens and 12 notebooks. Both of them went to a shop. The price of a pen and a n

Accepted Answer

Let A be the matrix of pens and notebooks and B be the matrix of prices of one pen and one notebook. The total amount required for each one of them is obtained by matrix AB. ∴ Jay needs ₹ 104 and Ram needs ₹ 150.

Chapter 4 Determinants and Matrices Ex 4.6 Solutions