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Question

(A) Let $\Delta = \left| {\begin{array}{*{20}{c}}{1 + a + x}&{a + y}&{a + z}\{b + x}&{1 + b + y}&{b + z}\{c + x}&{c + y}&{1 + c + z}\end{array}} i

Vedclass · Accepted Answer

$1 + a + b + c + x + y + z$

Vedclass · Answer

(A) Let $\Delta = \left| {\begin{array}{*{20}{c}}{1 + a + x}&{a + y}&{a + z}\{b + x}&{1 + b + y}&{b + z}\{c + x}&{c + y}&{1 + c + z}\end{array}} ight|$.Applying the column operation $C_1 	o C_1 + C_2 + C_3$:$\Delta = \left| {\begin{array}{*{20}{c}}{1 + a + b + c + x + y + z}&{a + y}&{a + z}\{1 + a + b + c + x + y + z}&{1 + b + y}&{b + z}\{1 + a + b + c + x + y + z}&{c + y}&{1 + c + z}\end{array}} ight|$.Let $S = 1 + a + b + c + x + y + z$. Then $\Delta = S \left| {\begin{array}{*{20}{c}}1&{a + y}&{a + z}\1&{1 + b + y}&{b + z}\1&{c + y}&{1 + c + z}\end{array}} ight|$.Applying $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_1$:$\Delta = S \left| {\begin{array}{*{20}{c}}1&{a + y}&{a + z}\0&{1 + b - a}&{b - a}\0&{c - a}&{1 + c - a}\end{array}} ight|$.Expanding along $C_1$:$\Delta = S 	imes [1 	imes ((1 + b - a)(1 + c - a) - (b - a)(c - a))]$.$\Delta = S 	imes [1 + c - a + b + bc - ab - a - ac + a^2 - (bc - ab - ac + a^2)]$.$\Delta = S 	imes [1 + c - a + b - c + a] = S 	imes 1 = 1 + a + b + c + x + y + z$.

The determinant $\left| {\begin{array}{*{20}{c}}{1 + a + x}&{a + y}&{a + z}\\{b + x}&{1 + b + y}&{b + z}\\{c + x}&{c + y}&{1 + c + z}\end{array}} \right|$ is equal to:

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