If A = begin{bmatrix} 1 + a^2 + a^4 & 1 + ab | vedclass

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(A) The given matrix $A$ can be expressed as the product of two matrices: $A = \begin{bmatrix} 1 & a & a^2 \ 1 & b & b^2 \ 1 & c & c^2 \end{bmatrix}

Vedclass · Accepted Answer

$-24$

Vedclass · Answer

(A) The given matrix $A$ can be expressed as the product of two matrices: $A = \begin{bmatrix} 1 & a & a^2 \ 1 & b & b^2 \ 1 & c & c^2 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \ a & b & c \ a^2 & b^2 & c^2 \end{bmatrix}$.Thus,$\det(A) = \det \begin{bmatrix} 1 & a & a^2 \ 1 & b & b^2 \ 1 & c & c^2 \end{bmatrix} 	imes \det \begin{bmatrix} 1 & 1 & 1 \ a & b & c \ a^2 & b^2 & c^2 \end{bmatrix} = [(a-b)(b-c)(c-a)]^2$.Given $\det(4I) = 4^3 \det(I) = 64 	imes 1 = 64$.Therefore,$[(a-b)(b-c)(c-a)]^2 = 64$,which implies $(a-b)(b-c)(c-a) = \pm 8$.We know that if $x+y+z=0$,then $x^3+y^3+z^3 = 3xyz$.Let $x=(a-b)$,$y=(b-c)$,and $z=(c-a)$. Then $x+y+z = (a-b)+(b-c)+(c-a) = 0$.Thus,$(a-b)^3 + (b-c)^3 + (c-a)^3 = 3(a-b)(b-c)(c-a) = 3(\pm 8) = \pm 24$.

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