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.
Question 5(i)
Show that |abclmnxyz|=|alxbmycnz|
Solution
L.H.S.=|abclmnxyz|=|abclmnxyz|t∵|A|=|At|=|alxbmycnz|=R.H.S.
Question 5(ii)
Show that |abc1−3a2−3b3−3c456|=|abc123456|.
Solution
L.H.S.=|abc1−3a2−3b3−3c456|=|abc1−3a+3a2−3b+3b3−3c+3c456| by R2−3R1=|abc123456|=R.H.S.
Question 5(iii)
Show that |111abcb+cc+aa+b|=0
Solution
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 R1≃R3=0=R.H.S.
Question 5(iv)
Show that |bccaababca2b2c2|=|111a2b2c2a3b3c3|
Solution
L.H.S.=|bccaababca2b2c2|=1abc|abcbcaabca2b2c2a3b3c3| by aR1,bR2,cR3=abcabc|111a2b2c2a3b3c3| by taking abc common from R1=|111a2b2c2a3b3c3|=R.H.S.
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