What is the value of the determinant {vmatrix} 1 & 1 & 1 \\ 1 & 1+xyz & 1 \\ 1 & 1 & 1+xyz {vmatrix} ?

Correct Answer is : x2y2z2x²y²z²x2y2z2

Correct Answer is : x2y2z2x²y²z²x2y2z2

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UPSC NDA I 2017 Mathematics Solved Paper

Question : 36 of 120

Marks: +1, -0

\begin{vmatrix} 1 & 1 & 1 \\ 1 & 1+xyz & 1 \\ 1 & 1 & 1+xyz \end{vmatrix} ?

Solution:

Concept:

Simplifying a determinant using elementary row operations reduces the matrix to an upper triangular form, making expansion easy.

Explanation:

Let

\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1+xyz & 1 \\ 1 & 1 & 1+xyz \end{vmatrix}

Subtract the first row from the second and third rows:

R_2 \rightarrow R_2 - R_1

R_3 \rightarrow R_3 - R_1

This yields

\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 0 & xyz & 0 \\ 0 & 0 & xyz \end{vmatrix}

Now expand the determinant along the first row:

\Delta = 1 \cdot \begin{vmatrix} xyz & 0 \\ 0 & xyz \end{vmatrix} - 1 \cdot 0 + 1 \cdot 0

The remaining

2 \times 2

determinant is

(xyz)(xyz) - 0 = x^{2}y^{2}z^{2}

Therefore,

\Delta = x^{2}y^{2}z^{2}

Answer:

x^{2}y^{2}z^{2}

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