Test Index

CBSE Class 12 Maths 2010 Solved Paper

© examsnet.com
Question : 26 of 29
Marks: +1, -0
Using properties of determinants show the following:
∣(b+c)2abcaab(a+c)2bcacbc(a+b)2∣\begin{vmatrix} (b+c)^2 & ab & ca \\ ab & (a+c)^2 & bc \\ ac & bc & (a+b)^2 \end{vmatrix}
= 2 abc (a+b+c)3(a+b+c)^3
Solution:  
Consider,
Δ =
∣(b+c)2abcaab(a+c)2bcacbc(a+b)2∣\begin{vmatrix} (b+c)^2 & ab & ca \\ ab & (a+c)^2 & bc \\ ac & bc & (a+b)^2 \end{vmatrix}
= 2 abc (a+b+c)3(a+b+c)^3
By performing R1R_1 → aR1,R2aR_1, R_2 → bR2,R3bR_2, R_3 → cR3cR_3 and dividing the determinant by abc, we get
Δ =
1abc∣a(b+c)2a2ba2cab2b(a+c)2b2cac2bc2c(a+b)2∣\frac{1}{abc} \begin{vmatrix} a(b+c)^2 & a^2 b & a^2 c \\ a b^2 & b(a+c)^2 & b^2 c \\ a c^2 & b c^2 & c(a+b)^2 \end{vmatrix}
Now, taking a, b, c common from C1,C2C_1, C_2 and C3C_3
Δ =
abcabc∣(b+c)2a2a2b2(a+c)2b2c2c2(a+b)2∣\frac{abc}{abc} \begin{vmatrix} (b+c)^2 & a^2 & a^2 \\ b^2 & (a+c)^2 & b^2 \\ c^2 & c^2 & (a+b)^2 \end{vmatrix}
⇒ Δ =
∣(b+c)2a2a2b2(c+a)2b2c2c2(a+b)∣\begin{vmatrix} (b+c)^2 & a^2 & a^2 \\ b^2 & (c+a)^2 & b^2 \\ c^2 & c^2 & (a+b) \end{vmatrix}
Applying C1C_1 → C1−C2,C2C_1 - C_2, C_2 → C2−C3C_2 - C_3
Δ = (a+b+c)2(a+b+c)^2
∣b+c−a0a2b−c−ac+a−bb20c−a−b(a+b)2∣\begin{vmatrix} b+c-a & 0 & a^2 \\ b-c-a & c+a-b & b^2 \\ 0 & c-a-b & (a+b)^2 \end{vmatrix}
Applying R3R_3 → R3−(R1+R2)R_3 - (R_1 + R_2)
Δ = (a+b+c)2(a+b+c)^2
∣b+c−a0a2b−c−ac+a−bb22a−2b−2a2ab∣\begin{vmatrix} b+c-a & 0 & a^2 \\ b-c-a & c+a-b & b^2 \\ 2a-2b & -2a & 2ab \end{vmatrix}
Applying C1C_1 → C1+C2C_1 + C_2
Δ = (a+b+c)2(a+b+c)^2
∣b+c−a0a20c+a−bb2−2b−2a2ab∣\begin{vmatrix} b+c-a & 0 & a^2 \\ 0 & c+a-b & b^2 \\ -2b & -2a & 2ab \end{vmatrix}
Applying C3C_3 → C3+bC2C_3 + bC_2
Δ = (a+b+c)2(a+b+c)^2
∣b+c−a0a20c+a−bbc+ab−2b−2a0∣\begin{vmatrix} b+c-a & 0 & a^2 \\ 0 & c+a-b & bc+ab \\ -2b & -2a & 0 \end{vmatrix}
Applying C1C_1 → aC1aC_1 and C2C_2 → bC2bC_2
Δ = (a+b+c)2ab\frac{(a+b+c)^2}{ab}
∣ab+ac−a20a20bc+ab−b2bc+ab−2ab−2ab0∣\begin{vmatrix} ab+ac-a^2 & 0 & a^2 \\ 0 & bc+ab-b^2 & bc+ab \\ -2ab & -2ab & 0 \end{vmatrix}
Applying C1C_1 → C1−C2C_1 - C_2
Δ = (a+b+c)2ab\frac{(a+b+c)^2}{ab}
∣ab+ac−a20a2−bc−ab+b2bc+ab−b2bc+ab0−2ab0∣\begin{vmatrix} ab+ac-a^2 & 0 & a^2 \\ -bc-ab+b^2 & bc+ab-b^2 & bc+ab \\ 0 & -2ab & 0 \end{vmatrix}
Expanding along R3R_3
= (a+b+c)2ab\frac{(a+b+c)^2}{ab}
2ab(ab2c+a2b2+abc2+a2bc−a2bc−a3b+a2bc+a3b−a2b2)2ab(ab^2c+a^2b^2+abc^2+a^2bc-a^2bc-a^3b+a^2bc+a^3b-a^2b^2)
= 2 (a+b+c)2(a+b+c)^2 ab2c+abc2+a2bcab^2c+abc^2+a^2bc
= 2 (a+b+c)3(a+b+c)^3 abc = R.H.S.
© examsnet.com
Go to Question:
Prev Question
Next Question