     1. Chapter 3 Class 12 Matrices
2. Serial order wise
3. Miscellaneous

Transcript

Misc. 11 Find the matrix X so that X [ 8(1&2&3@4&5&6)] =[ 8( 7& 8& 9@2&4&6)] X [ 8(1&2&3@4&5&6)] = [ 8( 7& 8& 9@2&4&6)] X [ 8(1&2&3@4&5&6)]_(2 3) = [ 8( 7& 8& 9@2&4&6)]_(2 3) So X will be a matrix Let X =[ 8(u&w@v&x)]_(2 2) So, our equation becomes [ 8(u&w@v&x)]_(2 2) [ 8(1&2&3@4&5&6)]_(2 3) = [ 8( 7& 8& 9@2&4&6)] [ 8( (1)+ (4)& (2)+ (5)& (3)+ (6)@ (1)+ (4)& (2)+ (5)& (3)+ (6))] = [ 8( 7& 8& 9@2&4&6)] [ 8( +4 &2 +5 &3 +6 @ +4 &2 +5 &3 +6 )]_(2 3) = [ 8( 7& 8& 9@2&4&6)]_(2 3) Since the matrices are equal Corresponding elements are equal u + 4w = - 7 2u + 5w = - 8 3u + 6w = - 9 v + 4x = 2 2v + 5x = 4 3v + 6x = 6 Solving (1) u + 4w = -7 u = -7 4w Putting value of u in (2) 2u + 5w = - 8 2( -7 4w) + 5w = - 8 - 14 8w + 5w = - 8 - 14 3w = - 8 - 3w = - 8 + 14 - 3w = 6 w = 6/( 3) w = 2 Now, u = 7 4w Putting w = - 2 u = 7 4 (-2) u = 7 + 8 u = 1 Solving (4) v + 4x = 2 v = 2 4x Putting value of v in (5) 2v + 5x = 4 2 (2 4x) + 5x = 4 4 8x + 5x = 4 4 3x = 4 -3x = 4 4 -3x = 0 x = 0 Putting value of x = 0 in (4) v + 4x = 2 v + 4(0) =2 v + 0 = 2 v = 2 Hence u = 1 , v = 2 , w = 2 & x = 0 Hence matrix X = [ 8(u&w@v&x)] = [ 8(1& 2@2&0)]

Miscellaneous

Chapter 3 Class 12 Matrices
Serial order wise

About the Author Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.