Ex 3.2, 7 (i) - Chapter 3 Class 12 Matrices (Term 1)
Last updated at Aug. 16, 2021 by Teachoo
Ex 3.2
Ex 3.2, 1
Ex 3.2, 2 (i)
Ex 3.2, 2 (ii) Important
Ex 3.2, 2 (iii)
Ex 3.2, 2 (iv)
Ex 3.2, 3 (i)
Ex 3.2, 3 (ii) Important
Ex 3.2, 3 (iii)
Ex 3.2, 3 (iv) Important
Ex 3.2, 3 (v)
Ex 3.2, 3 (vi) Important
Ex 3.2, 4
Ex 3.2, 5
Ex 3.2, 6
Ex 3.2, 7 (i) You are here
Ex 3.2, 7 (ii) Important
Ex 3.2, 8
Ex 3.2, 9
Ex 3.2, 10
Ex 3.2, 11
Ex 3.2, 12 Important
Ex 3.2, 13 Important
Ex 3.2, 14
Ex 3.2, 15
Ex 3.2, 16 Important
Ex 3.2, 17 Important
Ex 3.2, 18
Ex 3.2, 19 Important
Ex 3.2, 20 Important
Ex 3.2, 21 (MCQ) Important
Ex 3.2, 22 (MCQ) Important
Chapter 3 Class 12 Matrices (Term 1)
Serial order wise
Transcript
Ex 3.2, 7 Find X and Y, if (i) X + Y = [■8(7&0@2&5)] and X – Y = [■8(3&0@0&3)] Let X + Y = [■8(7&0@2&5)] X – Y = [■8(3&0@0&3)] Adding (1) and (2) X + Y + X – Y = [■8(7&0@2&5)] + [■8(3&0@0&3)] X + Y + X – Y = [■8(7+3&0+0@2+0&5+3)] 2X + 0 = [■8(10&0@2&8)] …(1) …(2) 2X = [■8(10&0@2&8)] X = 1/2 [■8(10&0@2&8)] X = [■8(10/2&0/2@2/2&8/2)] X = [■8(5&0@1&4)] Thus, X = [■8(5&0@1&4)] Putting value of X in (1) X + Y = [■8(7&0@2&5)] Y = [■8(7&0@2&5)] – X Y = [■8(7&0@2&5)] – [■8(5&0@1&4)] Y = [■8(7−5&0−0@2−1&5−4)] Y = [■8(2&0@1&1)] Hence, X = [■8(𝟓&𝟎@𝟏&𝟒)] & Y = [■8(𝟐&𝟎@𝟏&𝟏)]
