Home
Class 12
MATHS
|[2a1b1, a1b2+a2b1, a1b3+a3b1] , [a1b2+a...

`|[2a_1b_1, a_1b_2+a_2b_1, a_1b_3+a_3b_1] , [a_1b_2+a_2b_1, 2a_2b_2, a_2b_3+a_3b_2] , [a_1b_3+a_3b_1, a_3b_2+a_2b_3, 2a_3b_3]|=`

Text Solution

Verified by Experts

`Delta = |{:(2a_(1)b_(1),,a_(1)b_(2)+a_(2)b_(1),,a_(1)b_(3)+a_(3)b_(1)),( a_(1)b_(2)+a_(2)b_(1),,2a_(2)b_(2),,a_(2)b_(3)+a_(3)b_(2)),(a_(2)b_(3)+a_(3)b_(1),,a_(3)b_(2)+a_(2)b_(3),,2a_(3)b_(3)):}|.`
`=|{:(a_(1),,b_(1),,0),( a_(2),,b_(2),,0),(a_(3),,b_(3),,0):}| xx |{:(b^(1),,a_(1),,0),( b_(2),,a_(2),,0),(b_(3),,a_(3),,0):}|=0`
Now `ax^(2) +2hxy +by^(2) +2gy +2fy+c`
`=(lx+ my+n) (l'x+m'y+n')`
`=ll'x^(2) +(lm +ml') xy+ mm'y^(2) +(ln'+l'n)x`
`+(mn'+m'n)y+nn'`
Comparing the coefficient we get
`a=ll' ,h =(1)/(2) (lm'+ml ),b =mm`
`g=(1)/(2) (ln' +nl) ,f =(1)/(2) (mn' +nn) ,c=nn`
`:. |{:(a,,h,,g),( h,,b,,f),(f,,f,,c):}|`
`=|{:(ll,,(1)/(2)(lm'+ml'),,(1)/(2)(ln'+nl')),((1)/(2)(lm'+l'n),,mm,,(1)/(2)(mn'+m'n)),((1)/(2)(ln'+l'n),,(1)/(2)(mn'+m'n),,n n):}|`
` =(1)/(8) |{:(2ll',,lm'+l'm,,ln'+l'n),( lm'+l'm,,2mm',,mn'+m'n),(ln'+l'n,,mn'+m'n,,2n n):}|`
`=0`
Promotional Banner

Similar Questions

Explore conceptually related problems

Let =|2a_1b_1a_1b_2+a_2b_1a_1b_3+a_3b_1a_1b_2+a_2b_1 2a_2b_2a_2b_3+a_3b_2a_1b_3+a_3b_1a_3b_2+a_2b_3 2a_3b_3| . Expressing as the product of two determinants, show that =0. Hence, show that if a x^2+2h x y+b y^2+2gx+2fy+c=(l x+m y+n)(l^(prime)x+m^(prime)y+n),t h e n|a hgh bfgfc|=0.

det[[2a_(1)b_(1),a_(1)b_(2)+a_(2)b_(1),a_(1)b_(3)+a_(3)b_(1)a_(1)b_(2)+a_(2)b_(1),2a_(2)b_(2),a_(2)b_(3)+a_(3)b_(2)a_(1)b_(3)+a_(3)b_(1),a_(3)b_(2)+a_(2)b_(3),2a_(3)b_(3)]]=

Show that | (a_1l_1+b_1m_1, a_1l_2+b_1m_2,a_1l_3+b_1m_3),(a_2l_1+b_2m_1, a_2l_2+b_2m_2, a_2l_3+b_2m_3),(a_3l_1+b_3m_1, a_3l_2+b_3m_2,a_3l_3+b_3m_3)| = 0

Let = |(2a_(1)b_(1),a_(1)b_(2)+a_(2)b_(1),a_(1)b_(3)+a_(3)b_(1)),(a_(1)b_(2)+a_(2)b_(1),2a_(2)b_(2),a_(2)b_(3)+a_(3)b_(2)),(a_(1)b_(3)+a_(3)b_(1),a_(3)b_(2)+a_(2)b_(3),2a_(3)b_(3))| Express the determinant D as a product of two determinants. Hence or otherwise show that D = 0.

Prove that if alpha, beta, gamma !=0 then |(alpha+a_1b_1, a_1b_2, a_1b_3), (a_2b_1, beta+a_2b_2, a_2b_3), (a_3b_1, a_3b_2, gamma+a_3b_3)|=alpha beta gamma [1+(a_1b_1)/alpha + (a_2b_2)/beta+(a_3b_3)/gamma]

If |[a_1,b_1,c_1] , [a_2,b_2,c_2] ,[a_3,b_3,c_3]|=5; then the value of |[b_2c_3-b_3c_2,c_2a_3-c_3a_2,a_2b_3-a_3b_2] , [b_3c_1-b_1c_3,c_3a_1-c_1a_3,a_3b_1-a_1b_3] , [b_1c_2-b_2c_1,a_2c_1-a_1c_2,b_2a_1-b_1a_2]|

Show that if x_1, x_2, x_3!=0 |x_1+a_1b1a_1b_2a_1b_3a_2b_1x_2+a_2b_2a_2b_3a_3b_1a_3b_2x_3+a_3b_3|=x_1x_2x_3(1+(a_1b_1)/(x_1)+(a_2b_2)/(x_2)+(a_3b_3)/(x_3)) .

Show that if x_(1),x_(2),x_(3) ne 0 |{:(x_(1) +a_(1)b_(1),,a_(1)b_(2),,a_(1)b_(3)),(a_(2)b_(1),,x_(2)+a_(2)b_(2),,a_(2)b_(3)),(a_(3)b_(1),,a_(3)b_(2),,x_(3)+a_(3)b_(3)):}| =x_(1)x_(2)x_(3) (1+(a_(1)b_(1))/(x_(1))+(a_(2)b_(2))/(x_(2))+(a_(3)b_(3))/(x_(3)))