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UPSC NDA I 2023 Mathematics Solved Paper

Question : 13 of 120

Marks: +1, -0

\begin{vmatrix} x+1 & x+2 & x+3 \\ x+2 & x+3 & x+4 \\ x+a & x+b & x+c \end{vmatrix}

equal to ?

Solution:

Concept:

a,b,c

are in arithmetic progression (AP), then

2b = a + c

Explanation:

We need to evaluate the determinant:

\begin{vmatrix} x+1 & x+2 & x+3 \\ x+2 & x+3 & x+4 \\ x+a & x+b & x+c \end{vmatrix}

Apply the column operation

C_1 \rightarrow C_1 + C_3 - 2C_2

This gives:

\begin{vmatrix} 0 & x+2 & x+3 \\ 0 & x+3 & x+4 \\ a+c-2b & x+b & x+c \end{vmatrix}

Since

a,b,c

are in AP, we have

a + c = 2b

, so

a+c-2b = 0

Thus the first column becomes all zeros:

\begin{vmatrix} 0 & x+2 & x+3 \\ 0 & x+3 & x+4 \\ 0 & x+b & x+c \end{vmatrix}

If any entire row or column of a matrix is zero, the determinant is zero.

Hence the value of the given determinant is

0

Answer:

0

(option B).

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