In a square matrix A of order 3,a_{ii} are th | vedclass

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(D) The roots of the equation $x^2 - (a + b)x + ab = 0$ are $a$ and $b$. The sum of the roots is $a + b$ and the product of the roots is $ab$. Given t

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$(a^2 + ab + b^2)(a + b)$

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(D) The roots of the equation $x^2 - (a + b)x + ab = 0$ are $a$ and $b$. The sum of the roots is $a + b$ and the product of the roots is $ab$. Given the matrix $A$ of order $3 	imes 3$: $a_{11} = a_{22} = a_{33} = a + b$ $a_{12} = a_{23} = ab$ $a_{21} = a_{32} = 1$ All other elements are $0$. Thus,$A = \begin{bmatrix} a+b & ab & 0 \ 1 & a+b & ab \ 0 & 1 & a+b \end{bmatrix}$. Expanding the determinant along the first row: $\det(A) = (a+b) \begin{vmatrix} a+b & ab \ 1 & a+b \end{vmatrix} - ab \begin{vmatrix} 1 & ab \ 0 & a+b \end{vmatrix} + 0$ $\det(A) = (a+b)((a+b)^2 - ab) - ab(a+b)$ $\det(A) = (a+b)(a^2 + 2ab + b^2 - ab) - ab(a+b)$ $\det(A) = (a+b)(a^2 + ab + b^2) - ab(a+b)$ $\det(A) = (a+b)(a^2 + ab + b^2 - ab) = (a+b)(a^2 + b^2)$.

In a square matrix $A$ of order $3$,$a_{ii}$ are the sum of the roots of the equation $x^2 - (a + b)x + ab = 0$; $a_{i, i+1}$ are the product of the roots,$a_{i, i-1}$ are all unity,and the rest of the elements are all zero. The value of the determinant of $A$ is equal to

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