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UPSC NDA 2 2019 Mathematics Solved Paper

Question : 13 of 120

Marks: +1, -0

\begin{vmatrix} x+a & b & c \\ a & x+b & c \\ a & b & x+c \end{vmatrix} ?

Solution:

Concept:

Using elementary row operations that do not change the determinant value, we simplify the matrix.

Explanation:

Let the given determinant be

\Delta = \begin{vmatrix} x+a & b & c \\ a & x+b & c \\ a & b & x+c \end{vmatrix}

Apply

R_1 \to R_1 - R_2

\Delta = \begin{vmatrix} x & -x & 0 \\ a & x+b & c \\ a & b & x+c \end{vmatrix}

Apply

R_2 \to R_2 - R_3

\Delta = \begin{vmatrix} x & -x & 0 \\ 0 & x & -x \\ a & b & x+c \end{vmatrix}

Expand the determinant along the first row:

\Delta = x \cdot \begin{vmatrix} x & -x \\ b & x+c \end{vmatrix} - (-x) \cdot \begin{vmatrix} 0 & -x \\ a & x+c \end{vmatrix} + 0 \cdot \begin{vmatrix} 0 & x \\ a & b \end{vmatrix}

Compute the

2 \times 2

determinants:

First term:

x \cdot [x(x+c) - (-x)b] = x \cdot (x^2 + cx + bx)

Second term:

+x \cdot [0\cdot(x+c) - (-x)a] = x \cdot (ax)

\Delta = x(x^2 + cx + bx) + x(ax) = x(x^2 + cx + bx + ax) = x^2(x + a + b + c)

Given

x + a + b + c = 0

, therefore

\Delta = x^2 \cdot 0 = 0

Answer:

0

(Option A).

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