Appropriately matching the information given in the three columns of the following table.
Column 1Column 2Column 3(I)If a, b, c ϵR−{0} such that a≠b≠cand 1a+1b+1c=0 and A=⎡⎢⎣1+a1111+b1111+c⎤⎥⎦,then(i)A is singular matrix(P)|adj A|=|A|2(II)If α, β, γ ϵ R, andA=⎡⎢⎣1cos(α−β)cos(α−γ)cos(β−α)1cos(β−γ)cos(γ−α)cos(γ−β)1⎤⎥⎦,then(ii)A is singular matrix(Q)adj(adj A)=|A|A(III)If ω≠1 be cube root of unity and(iii)A is non-singular matrix(R)|A| is equal toA=⎡⎢⎣1+2ω100+ω200ω2111+ω101+2ω202ωωω22+ω100+2ω200⎤⎥⎦is equal to minimum value of,thencos−1(x−1x)+cos−1(y2y+1)+cos−1(z2+z+1)(where x, y, z are real numbers)(IV)If a, b, c ϵR−{0} such that a≠b≠c,andA=⎡⎢
⎢⎣0(a−b)3(a−c)3(b−a)30(b−c)3(c−a)3(c−b)30⎤⎥
⎥⎦,then(iv)Invertible(S)|A−1|=1|A|
Which of the following is only incorrect combination ?
Appropriately matching the information given in the three columns of the following table.
Column 1Column 2Column 3(I)If a, b, c ϵR−{0} such that a≠b≠cand 1a+1b+1c=0 and A=⎡⎢⎣1+a1111+b1111+c⎤⎥⎦,then(i)A is singular matrix(P)|adj A|=|A|2(II)If α, β, γ ϵ R, andA=⎡⎢⎣1cos(α−β)cos(α−γ)cos(β−α)1cos(β−γ)cos(γ−α)cos(γ−β)1⎤⎥⎦,then(ii)A is singular matrix(Q)adj(adj A)=|A|A(III)If ω≠1 be cube root of unity and(iii)A is non-singular matrix(R)|A| is equal toA=⎡⎢⎣1+2ω100+ω200ω2111+ω101+2ω202ωωω22+ω100+2ω200⎤⎥⎦is equal to minimum value of,thencos−1(x−1x)+cos−1(y2y+1)+cos−1(z2+z+1)(where x, y, z are real numbers)(IV)If a, b, c ϵR−{0} such that a≠b≠c,andA=⎡⎢ ⎢⎣0(a−b)3(a−c)3(b−a)30(b−c)3(c−a)3(c−b)30⎤⎥ ⎥⎦,then(iv)Invertible(S)|A−1|=1|A|
Which of the following is only incorrect combination ?
The correct option is D (IV)(iii)(Q)
(I)Δ=abc∣∣
∣
∣
∣∣1a+1 1b 1c1a 1b+1 1c1a 1b 1c+1∣∣
∣
∣
∣∣=(1a+1b+1c+1)abc∣∣
∣
∣
∣∣1 1b 1c1 1b+1 1c1 1b 1c+1∣∣
∣
∣
∣∣=abc≠0
∴ Non-singular, so invertible
Also it is symmetric
(II) ⎡⎢⎣1 cos(α−β) cos(α−γ)cos(β−α) 1 cos(β−γ)cos(γ−α) cos(γ−β) 1⎤⎥⎦=∣∣
∣∣cos α sin α 0cos β sin β 0cos γ sin γ 0∣∣
∣∣∣∣
∣∣cos α cos β cos γsin α sin β sin γ0 0 0∣∣
∣∣
(III) ⎡⎢⎣1+2ω100+ω200 ω2 11 1+w101+2ω202 ωω ω2 2+ω100+2ω200⎤⎥⎦=∣∣
∣
∣∣ω ω2 11 ω ωω ω2 −ω∣∣
∣
∣∣=∣∣
∣∣0 0 1+ω1 ω ωω ω2 −ω∣∣
∣∣=(1+ω)×0=0∴ Singuler
(IV) This is a skew - symmetric matrix of odd order
⇒ singular.
(IV)(iii)(Q)
(I)Δ=abc∣∣ ∣ ∣ ∣∣1a+1 1b 1c1a 1b+1 1c1a 1b 1c+1∣∣ ∣ ∣ ∣∣=(1a+1b+1c+1)abc∣∣ ∣ ∣ ∣∣1 1b 1c1 1b+1 1c1 1b 1c+1∣∣ ∣ ∣ ∣∣=abc≠0
∴ Non-singular, so invertible
Also it is symmetric
(II) ⎡⎢⎣1 cos(α−β) cos(α−γ)cos(β−α) 1 cos(β−γ)cos(γ−α) cos(γ−β) 1⎤⎥⎦=∣∣ ∣∣cos α sin α 0cos β sin β 0cos γ sin γ 0∣∣ ∣∣∣∣ ∣∣cos α cos β cos γsin α sin β sin γ0 0 0∣∣ ∣∣
(III) ⎡⎢⎣1+2ω100+ω200 ω2 11 1+w101+2ω202 ωω ω2 2+ω100+2ω200⎤⎥⎦=∣∣ ∣ ∣∣ω ω2 11 ω ωω ω2 −ω∣∣ ∣ ∣∣=∣∣ ∣∣0 0 1+ω1 ω ωω ω2 −ω∣∣ ∣∣=(1+ω)×0=0∴ Singuler
(IV) This is a skew - symmetric matrix of odd order
⇒ singular.
