Question 6, Exercise 2.2
Solutions of Question 6 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.
Questiopn 6(i)
Prov that |a−bb−cc−ab−cc−aa−bc−aa−bb−c|=0
Solution
Let L.H.S=|a−bb−cc−ab−cc−aa−bc−aa−bb−c|=|a−cb−ac−bb−cc−aa−bc−aa−bb−c|R1+R2=−|c−aa−bb−cb−cc−aa−bc−aa−bb−c|−R1=0R1≅R3=R.H.S.
Question 6(ii)
Prov that |1aa31bb31cc3|=(a−b)(b−c)(c−a)(a+b+c)
Solution
Let L.H.S.=|1aa31bb31cc3| Subtract first row from second row and second row from third row. We have, =|0a−ba3−b30b−cb3−c31cc3| =|0(a−b)(a−b)(a2+ab+b2)0(b−c)(b−c)(b2+bc+c2)1cc3| =(a−b)(b−c)|01(a2+ab+b2)01(b2+bc+c2)1cc3| =(a−b)(b−c)|1(a2+ab+b2)1(b2+bc+c2)| =(a−b)(b−c)[(b2+bc+c2)−(a2+ab+b2)] =(a−b)(b−c)(b2+bc+c2−a2−ab−b2) =(a−b)(b−c)(bc+c2−a2−ab) =(a−b)(b−c)(b(c−a)+(c−a)(c+a)) =(a−b)(b−c)(c−a)(a+b+c) =R.H.S.
Question 6(iii)
Prov that |1aa21bb21cc2|=(a−b)(b−c)(c−a)
Solution
Let L.H.S.=|1aa21bb21cc2| Subtract first row from second row and second row from third row, we have, =|0a−ba2−b20b−cb2−c21cc2| =|0(a−b)(a−b)(a+b)0(b−c)(b−c)(b+c)1cc2| =(a−b)(b−c)|01(a+b)01(b+c)1cc2| =(a−b)(b−c)|1(a+b)1(b+c)| =(a−b)(b−c)(b+c−b−a) =(a−b)(b−c)(c−a) =R.H.S.
Question 6(iv)
Prove that |−a2abacab−b2bcacbc−c2|=4a2b2c2
Solution
L.H.S.=|−a2abacab−b2bcacbc−c2|=−a2(b2c2−b2c2)−ab(−abc2−abc2)+ac(ab2c+ab2c)=0+2a2b2c2+2a2b2c2=4a2b2c2=R.H.S.
Question 6(v)
Prov that |bca31acab31babc31c|=0a≠0,b≠0,c≠0.
Solution
Let L.H.S.=|bca31acab31babc31c| Multiply first row by a, second row by b and third row by c. We have, =1abc|abca4a1abcab4b1bcabc4c1c| =abcabc|1a411b411c41| =|1a411b411c41| Two identical columns, so determinant is zero. =0 =R.H.S.
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