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CBSE Class 12 Math 2020 Outside Delhi Set 1 Solved Paper

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Section - D
Q. Nos. 33 to 36 carry 6 marks each.
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Question : 33 of 36
Marks: +1, -0
Using properties of determinates prove that:
∣a−bb+cab−cc+abc−aa+bc∣=a3+b3+c3−3abc\begin{vmatrix} a-b & b+c & a \\ b-c & c+a & b \\ c-a & a+b & c \end{vmatrix}=a^3+b^3+c^3-3 a b c
OR
If A=[13220−1123]A=\begin{bmatrix} 1 & 3 & 2 \\ 2 & 0 & -1 \\ 1 & 2 & 3 \end{bmatrix}, then show that A3−4A2−3A+11I=OA^3-4 A^2-3 A+11 I=O. Hence find A−1A^{-1}
Solution:     👈: Video Solution
∣a−bb+cab−cc+abc−aa+bc∣\begin{vmatrix} a-b & b+c & a \\ b-c & c+a & b \\ c-a & a+b & c \end{vmatrix}
=∣−bb+c+aa−cc+a+bb−aa+b+cc∣[  Applying  C1→C1−C3  and  C2→C2+C3]=\begin{vmatrix} -b & b+c+a & a \\ -c & c+a+b & b \\ -a & a+b+c & c \end{vmatrix} \left[\;\text{Applying}\; C_1\rightarrow C_1-C_3 \;\text{and}\; C_2\rightarrow C_2+C_3\right]
=(−1)(a+b+c)∣b1ac1ba1c∣=(-1)(a+b+c)\begin{vmatrix} b & 1 & a \\ c & 1 & b \\ a & 1 & c \end{vmatrix}
[Taking (−1)(-1) common from C1C_1 and (a+b+c)(a+b+c) common from C2C_2 ]
=(−1)(a+b+c)∣b1ac−b0b−aa−b0c−a∣[  Applying  R2→R2−R1  and  R3→R3−R1]=(-1)(a+b+c)\begin{vmatrix} b & 1 & a \\ c-b & 0 & b-a \\ a-b & 0 & c-a \end{vmatrix}\left[\;\text{Applying}\; R_2\rightarrow R_2-R_1 \;\text{and}\; R_3\rightarrow R_3-R_1\right]
=(−1)(a+b+c)[−(c−b)(c−a)+(b−a)(a−b)]=(-1)(a+b+c)[-(c-b)(c-a)+(b-a)(a-b)]
=(−1)(a+b+c)[−c2+ac+bc−ab+ba−b2−a2+ab]=(-1)(a+b+c)[-c^2+a c+b c-a b+b a-b^2-a^2+a b]
=(−1)(a+b+c)(−a2−b2−c2+ab+bc+ac)=(-1)(a+b+c)(-a^2-b^2-c^2+a b+b c+a c)
=(a+b+c)(a2+b2+c2−ab−bc−ac)=(a+b+c)(a^2+b^2+c^2-a b-b c-a c)
=a3+ab2+ac2−a2b−abc−a2c+ba2+b3+bc2−ab2−b2c−abc+ca2+c=a^3+a b^2+a c^2-a^2 b-a b c-a^2 c+b a^2+b^3+b c^2-a b^2-b^2 c-a b c+c a^2+c
b2+c3−acb−bc2−ac2b^2+c^3-a c b-b c^2-a c^2
=a3+b3+c3−3abc=a^3+b^3+c^3-3 a b c
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