Question 5, Exercise 2.2

Solutions of Question 5 of Exercise 2.2 of Unit 02: Matrices and Determinants. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Show that |abclmnxyz|=|alxbmycnz|

L.H.S.=|abclmnxyz|=|abclmnxyz|t|A|=|At|=|alxbmycnz|=R.H.S.

Show that |abc13a23b33c456|=|abc123456|.

L.H.S.=|abc13a23b33c456|=|abc13a+3a23b+3b33c+3c456| by R23R1=|abc123456|=R.H.S.

Show that |111abcb+cc+aa+b|=0

L.H.S.=|111abcb+cc+aa+b|=|111abca+b+cb+c+ac+a+b| by R3+R2=(a+b+c)|111abc111|taking common from R3=(a+b+c)0 by R1R3=0=R.H.S.

Show that |bccaababca2b2c2|=|111a2b2c2a3b3c3|

L.H.S.=|bccaababca2b2c2|=1abc|abcbcaabca2b2c2a3b3c3| by aR1,bR2,cR3=abcabc|111a2b2c2a3b3c3| by taking abc common from R1=|111a2b2c2a3b3c3|=R.H.S.