CBSE Class 12 Sample Paper for 2019 Boards

Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Question 15

Using properties of determinants, prove that:

|a (b - c) (c + b) (a + c) b (c - a) (a - b) (b + a) c| = (a + b + c) (a 2 + b 2 + c 2 )

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Question 15 Using properties of determinants, prove that: |β 8(π&πβπ&π+π@π+π&π&πβπ@πβπ&π+π&π)| = (π+π+π) (π^2+π^2+π^2) Solving LHS |β 8(π&πβπ&π+π@π+π&π&πβπ@πβπ&π+π&π)| Since we cannot solve by adding or subtracting rows We multiply a with C1, b with C2 and c with C3 = πππ/πππ |β 8(π&πβπ&π+π@π+π&π&πβπ@πβπ&π+π&π)| = 1/πππ |β 8(π(π)&π(πβπ)&π(π+π)@π(π+π)&π(π)&π(πβπ)@π(πβπ)&π(π+π)&π(π))| = 1/πππ |β 8(π^2&π^2βππ&π^2+ππ@π^2+ππ&π^2&π^2βππ@π^2βππ&π^2+ππ&π^2 )| C1 β C1 + C2 + C3 = 1/πππ |β 8(π^2+π^2βππ+π^2+ππ&π^2βππ&π^2+ππ@π^2+ππ+π^2+π^2βππ&π^2&π^2βππ@π^2βππ+π^2+ππ+π^2&π^2+ππ&π^2 )| = 1/πππ |β 8(π^2+π^2+π^2&π^2βππ&π^2+ππ@π^2+π^2+π^2&π^2&π^2βππ@π^2+π^2+π^2&π^2+ππ&π^2 )| Taking π^2+π^2+π^2 common from C1 = ((π^2 + π^2+ γ πγ^2))/πππ |β 8(1&π^2βππ&π^2+ππ@1&π^2&π^2βππ@1&π^2+ππ&π^2 )| R2 β R2 β R1 = ((π^2 + π^2+ γ πγ^2))/πππ |β 8(1&π^2βππ&π^2+ππ@1β1&π^2β(π^2βππ)&π^2βππβ(π^2+ππ)@1&π^2+ππ&π^2 )| = ((π^2 + π^2+ γ πγ^2))/πππ |β 8(1&π^2βππ&π^2+ππ@0&ππ&βππβππ@1&π^2+ππ&π^2 )| R3 β R3 β R1 = ((π^2 + π^2+ γ πγ^2))/πππ |β 8(1&π^2βππ&π^2+ππ@0&ππ&βππβππ@1β1&π^2+ππβ(π^2βππ)&π^2β(π^2+ππ))| = ((π^2 + π^2+ γ πγ^2))/πππ |β 8(1&π^2βππ&π^2+ππ@0&ππ&βππβππ@0&ππ+ππ&βππ)| = ((π^2 + π^2+ γ πγ^2))/πππ |β 8(1&π(πβπ)&π(π+π)@0&ππ&βπ(π+π)@0&π(π+π)&βππ)| Taking b common from C2, and c common from C3 = ((π^2 + π^2+ γ πγ^2))/πππΓππ|β 8(1&(πβπ)&(π+π)@0&π&β(π+π)@0&(π+π)&βπ)| = ((π^2 + π^2+ γ πγ^2))/π |β 8(1&(πβπ)&(π+π)@0&π&β(π+π)@0&(π+π)&βπ)| = ((π^2 + π^2+ γ πγ^2))/π |β 8(1&(πβπ)&(π+π)@0&π&β(π+π)@0&(π+π)&βπ)| Finding determinant along C1 = ((π^2 + π^2+ γ πγ^2))/π Γ [1 (c Γ (-b) β (β(a + b))(a + c) + 0 + 0] = ((π^2 + π^2+ γ πγ^2))/π Γ [βbc + (a + b)(a + c)] = ((π^2 + π^2+ γ πγ^2))/π Γ [βbc + a(a + c) + b(a + c)] = ((π^2 + π^2+ γ πγ^2))/π Γ [βbc + a2 + ac + ab + bc] = ((π^2 + π^2+ γ πγ^2))/π Γ [a2 + ac + ab] = ((π^2 + π^2+ γ πγ^2))/π Γ a[a + c + b] = (π+π+π) (π^2+π^2+π^2)