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Question 83 of 110

Q.The factor of the determinant ∣x+abcax+bcabx+c∣\begin{vmatrix} x+a & b & c \\ a & x+b & c \\ a & b & x+c \end{vmatrix} is:

(a) x+cx + c
(b) xx
(c) x−a+b+cx - a + b + c
(d) x+bx + b
Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Board 2018MCQ· 1mImportance★★★★★
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Column and row reduction shows the determinant equals x2(x+a+b+c)x^2(x+a+b+c), so among the listed options, xx is the factor present.

Let D=∣x+abcax+bcabx+c∣D = \begin{vmatrix} x+a & b & c \\ a & x+b & c \\ a & b & x+c \end{vmatrix}.

Apply C1→C1+C2+C3C_1 \to C_1+C_2+C_3. Each row sums to (x+a)+b+c=a+(x+b)+c=a+b+(x+c)=x+a+b+c(x+a)+b+c = a+(x+b)+c = a+b+(x+c) = x+a+b+c, so the new first column is (x+a+b+c)(x+a+b+c) in every row:

D=(x+a+b+c)∣1bc1x+bc1bx+c∣D = (x+a+b+c)\begin{vmatrix} 1 & b & c \\ 1 & x+b & c \\ 1 & b & x+c \end{vmatrix}

Now apply R2→R2−R1R_2 \to R_2-R_1 and R3→R3−R1R_3\to R_3-R_1:

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