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CBSE Class 12 Math 2012 Solved Paper

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Question : 19 of 29
Marks: +1, -0
Using properties of determinants prove the following:
∣111abca3b3c3∣\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^3 & b^3 & c^3 \end{vmatrix} = (a - b) (b - c) (c - a) (a + b + c)
Solution:  
Δ = ∣111abca3b3c3∣\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^3 & b^3 & c^3 \end{vmatrix}
Applying C1C_1 → C1−C3C_1 - C_3 and C2C_2 → C2−C3C_2 - C_3, we have:
Δ = ∣1−11−11a−cb−cca3−c3b3−c3c3∣\begin{vmatrix} 1-1 & 1-1 & 1 \\ a-c & b-c & c \\ a^3-c^3 & b^3-c^3 & c^3 \end{vmatrix}
=
∣001a−cb−cc(a−c)(a2+ac+c2)(b−c)(c2+bc+c2)c3∣\begin{vmatrix} 0 & 0 & 1 \\ a-c & b-c & c \\ (a-c)(a^2+ac+c^2) & (b-c)(c^2+bc+c^2) & c^3 \end{vmatrix}
= (c - a) (b - c)
∣001−11c−a2+ac+c2b2+bc+c2c3∣\begin{vmatrix} 0 & 0 & 1 \\ -1 & 1 & c \\ -a^2+ac+c^2 & b^2+bc+c^2 & c^3 \end{vmatrix}
Applying C1C_1 → C1+C2C_1 + C_2, we have:
Δ = (c - a) (b - c)
∣00101cb2−a2+bc−acb2+bc+c2c3∣\begin{vmatrix} 0 & 0 & 1 \\ 0 & 1 & c \\ b^2-a^2+bc-ac & b^2+bc+c^2 & c^3 \end{vmatrix}
= (b - c) (c - a) (a - b)
∣00101c−a+b+cb2+bc+c2c3∣\begin{vmatrix} 0 & 0 & 1 \\ 0 & 1 & c \\ -a+b+c & b^2+bc+c^2 & c^3 \end{vmatrix}
= (a - b) (b - c) (c - a) (a + b + c)
∣011c∣\begin{vmatrix} 0 & 1 \\ 1 & c \end{vmatrix}
Expanding along C1C_1, we have:
Δ = (a - b) (b - c) (c - a) (a + b + c) - 1 ∣011c∣\left|\begin{array}{cc}0 & 1 \\ 1 & c\end{array}\right|
= (a - b) (b - c) (c - a) (a + b + c)
Hence proved.
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