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Question Number 57619 by tanmay.chaudhury50@gmail.com last updated on 08/Apr/19

Answered by math1967 last updated on 09/Apr/19

 determinant (((q+r),(r+p),(p+q)),((y+z),(z+x),(x+y)))   determinant (((b+c−c−a−a−b),(c+a),(a+b)),((q+r−r−p−p−q),(r+p),(p+q)),((y+z−z−x−x−y),(z+x),(x+y)))C_1 −C_2 −C_3   −2 determinant ((a,(c+a),(a+b)),(p,(r+p),(p+q)),(x,(z+x),(x+y)))  =−2 determinant ((a,(c+a−a),(a+b−a)),(p,(r+p−p),(p+q−p)),(x,(z+x−x),(x+y−x)))C_2 −C_1 ,C_3 −C_1   =−2 determinant ((a,c,b),(p,r,q),(x,z,(y )))=2 determinant ((a,b,c),(p,q,r),(x,y,z))C_2 ⇔C_3

|q+rr+pp+qy+zz+xx+y||b+ccaabc+aa+bq+rrppqr+pp+qy+zzxxyz+xx+y|C1C2C32|ac+aa+bpr+pp+qxz+xx+y|=2|ac+aaa+bapr+ppp+qpxz+xxx+yx|C2C1,C3C1=2|acbprqxzy|=2|abcpqrxyz|C2C3

Commented by tanmay.chaudhury50@gmail.com last updated on 09/Apr/19

thank you sir

thankyousir

Commented by math1967 last updated on 09/Apr/19

You are welcome sir

Youarewelcomesir

Answered by tanmay.chaudhury50@gmail.com last updated on 09/Apr/19

another way  ∣b+c    c+a   a+b ∣  ∣q+r    r+p    p+q∣  ∣y+z    z+x    x+y∣  =∣b  c  a∣ +∣b  c  b∣ +∣b   a  a∣+∣b  a b∣      ∣q  r   p∣     ∣q   r   q∣  ∣q    p    p∣   ∣q   p  q∣  +      ∣ y z x∣      ∣y   z   y∣  ∣y   x    x∣   ∣y  x  y∣       ∣c   c  a∣+∣c  c  b ∣+∣c  a  a∣+∣c  a  b∣    ∣r   r   p∣   ∣r   r  q∣    ∣r   p  p∣   ∣r   p   q∣     ∣z   z   x∣ ∣z    z  y∣   ∣z   x   x∣  ∣z   x   y∣    △=△_1 +△_2 +△_3 +△_4 +△_5 +△_6 +△_7 +△_8   now △_2 =△_3 =△_4 =△_5 =△_6 =△_7 =0  reason two identical collumn    so △=△_1 +△_8   now inter change of collumn keep value of   determinant unchanged  So △=2∣a b c∣                    ∣p  q  r∣                     ∣x   y  z∣ proved

anotherwayb+cc+aa+bq+rr+pp+qy+zz+xx+y=∣bca+bcb+baa+babqrpqrqqppqpq+yzxyzyyxxyxycca+ccb+caa+cabrrprrqrpprpqzzxzzyzxxzxy=1+2+3+4+5+6+7+8now2=3=4=5=6=7=0reasontwoidenticalcollumnso=1+8nowinterchangeofcollumnkeepvalueofdeterminantunchangedSo=2abcpqrxyzproved

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