D = ∣ ( 1 + α ) 2 ( 1 + 2 α ) 2 ( 1 + 3 α ) 2 ( 2 + α ) 2 ( 2 + 2 α ) 2 ( 2 + 3 α ) 2 ( 3 + α ) 2 ( 3 + 2 α ) 2 ( 3 + 3 α ) 2 ∣ D = \begin{vmatrix} (1+\alpha)^{2} & (1+2\alpha)^{2} & (1+3\alpha)^{2} \\ (2+\alpha)^{2} & (2+2\alpha)^{2} & (2+3\alpha)^{2} \\ (3+\alpha)^{2} & (3+2\alpha)^{2} & (3+3\alpha)^{2} \end{vmatrix} D = ( 1 + α ) 2 ( 2 + α ) 2 ( 3 + α ) 2 ( 1 + 2 α ) 2 ( 2 + 2 α ) 2 ( 3 + 2 α ) 2 ( 1 + 3 α ) 2 ( 2 + 3 α ) 2 ( 3 + 3 α ) 2 entry ( i , j ) = ( i + j α ) 2 , D = − 648 α \text{entry } (i,j) = (i + j\alpha)^{2}, \qquad D = -648\,\alpha entry ( i , j ) = ( i + j α ) 2 , D = − 648 α Find α with D ( α ) = − 648 α , from the options \text{Find } \alpha \text{ with } D(\alpha) = -648\alpha, \text{ from the options} Find α with D ( α ) = − 648 α , from the options ( i + j α ) 2 = i 2 + 2 i j α + j 2 α 2 (i + j\alpha)^2 = i^2 + 2ij\alpha + j^2\alpha^2 ( i + j α ) 2 = i 2 + 2 ij α + j 2 α 2 col j = a ⃗ + 2 j α b ⃗ + j 2 α 2 e ⃗ \text{col}_j = \vec a + 2j\alpha\,\vec b + j^2\alpha^2\,\vec e col j = a + 2 j α b + j 2 α 2 e a ⃗ = ( 1 , 4 , 9 ) T , b ⃗ = ( 1 , 2 , 3 ) T , e ⃗ = ( 1 , 1 , 1 ) T \vec a = (1,4,9)^{T},\quad \vec b = (1,2,3)^{T},\quad \vec e = (1,1,1)^{T} a = ( 1 , 4 , 9 ) T , b = ( 1 , 2 , 3 ) T , e = ( 1 , 1 , 1 ) T M = X P , X = [ 1 1 1 4 2 1 9 3 1 ] M = X P, \qquad X = \begin{bmatrix} 1 & 1 & 1 \\ 4 & 2 & 1 \\ 9 & 3 & 1 \end{bmatrix} M = X P , X = 1 4 9 1 2 3 1 1 1 P = [ 1 1 1 2 α 4 α 6 α α 2 4 α 2 9 α 2 ] P = \begin{bmatrix} 1 & 1 & 1 \\ 2\alpha & 4\alpha & 6\alpha \\ \alpha^2 & 4\alpha^2 & 9\alpha^2 \end{bmatrix} P = 1 2 α α 2 1 4 α 4 α 2 1 6 α 9 α 2 det X = ∣ 1 1 1 4 2 1 9 3 1 ∣ = − 2 \det X = \begin{vmatrix} 1 & 1 & 1 \\ 4 & 2 & 1 \\ 9 & 3 & 1 \end{vmatrix} = -2 det X = 1 4 9 1 2 3 1 1 1 = − 2 det P = α ⋅ α 2 ∣ 1 1 1 2 4 6 1 4 9 ∣ = α 3 ( 4 ) = 4 α 3 \det P = \alpha \cdot \alpha^{2} \begin{vmatrix} 1 & 1 & 1 \\ 2 & 4 & 6 \\ 1 & 4 & 9 \end{vmatrix} = \alpha^{3}\,(4) = 4\alpha^{3} det P = α ⋅ α 2 1 2 1 1 4 4 1 6 9 = α 3 ( 4 ) = 4 α 3 D = det X ⋅ det P = ( − 2 ) ( 4 α 3 ) = − 8 α 3 D = \det X \cdot \det P = (-2)(4\alpha^{3}) = -8\alpha^{3} D = det X ⋅ det P = ( − 2 ) ( 4 α 3 ) = − 8 α 3 − 8 α 3 = − 648 α ⇒ α 3 = 81 α -8\alpha^{3} = -648\alpha \Rightarrow \alpha^{3} = 81\alpha − 8 α 3 = − 648 α ⇒ α 3 = 81 α α ( α 2 − 81 ) = 0 ⇒ α = 0 , ± 9 \alpha(\alpha^{2} - 81) = 0 \Rightarrow \alpha = 0,\ \pm 9 α ( α 2 − 81 ) = 0 ⇒ α = 0 , ± 9 α = 9 : − 8 ( 729 ) = − 5832 = − 648 ( 9 ) \alpha = 9:\ -8(729) = -5832 = -648(9) α = 9 : − 8 ( 729 ) = − 5832 = − 648 ( 9 ) α = − 9 : − 8 ( − 729 ) = 5832 = − 648 ( − 9 ) \alpha = -9:\ -8(-729) = 5832 = -648(-9) α = − 9 : − 8 ( − 729 ) = 5832 = − 648 ( − 9 ) ⇒ ( 2 ) and ( 3 ) satisfy the equation \Rightarrow (2) \text{ and } (3) \text{ satisfy the equation} ⇒ ( 2 ) and ( 3 ) satisfy the equation D = − 8 α 3 ; α = 0 , ± 9 ⇒ options ( 2 ) and ( 3 ) D = -8\alpha^{3}; \quad \alpha = 0, \pm 9 \Rightarrow \text{options } (2) \text{ and } (3) D = − 8 α 3 ; α = 0 , ± 9 ⇒ options ( 2 ) and ( 3 )
