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Question 145 of 146

Q.Using properties of determinants, prove that ∣111+3x1+3y1111+3z1∣=9(3xyz+xy+yz+zx)\begin{vmatrix} 1 & 1 & 1+3x \\ 1+3y & 1 & 1 \\ 1 & 1+3z & 1 \end{vmatrix} = 9(3xyz + xy + yz + zx)

Assam AhsecCBSE Class XII Board 2018Subjective· 4mImportance★★★★★
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The determinant equals 9(3xyz+xy+yz+zx)9(3xyz+xy+yz+zx).

Concept. A determinant can be expanded along a row/column; the resulting polynomial is then factorised.

Why this method. Direct expansion here leads cleanly to the target expression.

Working. Expand along the first row:

Δ=1[1−(1+3z)]−1[(1+3y)−1]+(1+3x)[(1+3y)(1+3z)−1].\Delta=1\big[1-(1+3z)\big]-1\big[(1+3y)-1\big]+(1+3x)\big[(1+3y)(1+3z)-1\big]. …

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