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Matrices & Determinants: Square Matrix Order Diagonal Elements Sum Roots Equation

JEE Maths question with a full step-by-step solution.

Question
In a square matrix AA of order 3, the diagonal elements aiia_{ii} are the sum of the roots of the equation x2−(a+b)x+ab=0x^2 - (a+b)x + ab = 0; ai,i+1a_{i,i+1} are the product of the roots of the same equation; ai+1,i=1a_{i+1,i} = 1 for all valid ii; and the remaining elements are zero. The value of det⁡(A)\det(A) is:
A00
B(a+b)3(a+b)^3
Ca3−b3a^3 - b^3
D(a2+b2)(a+b)(a^2+b^2)(a+b)correct
Solution
Step 1: Identify the elements The roots of x2−(a+b)x+ab=0x^2-(a+b)x+ab=0 are aa and bb, so their sum is a+ba+b and their product is abab. Therefore:
A=(a+bab01a+bab01a+b)A = \begin{pmatrix}a+b & ab & 0\\ 1 & a+b & ab\\ 0 & 1 & a+b\end{pmatrix}
 Step 2: Compute the determinant by expanding along R1R_1
det⁡(A)=(a+b)∣a+bab1a+b∣−ab∣1ab0a+b∣\det(A) = (a+b)\begin{vmatrix}a+b & ab\\ 1 & a+b\end{vmatrix} - ab\begin{vmatrix}1 & ab\\ 0 & a+b\end{vmatrix}
=(a+b)[(a+b)2−ab]−ab(a+b)= (a+b)\big[(a+b)^2 - ab\big] - ab(a+b)
=(a+b)[(a+b)2−ab−ab]=(a+b)[(a+b)2−2ab]= (a+b)\big[(a+b)^2 - ab - ab\big] = (a+b)\big[(a+b)^2 - 2ab\big]
=(a+b)(a2+b2)= (a+b)(a^2+b^2)
 Answer: (4)
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