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HCU MCA 2013 Paper

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Question : 51 of 75

Marks: +1, -0

Suppose a, b and c are all > 0, and if D is the determinant below, which of the following is true about D ?

\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}

I

II

III

.D = 0

Only $I$
Both $I$ and $II$
Only $II$
Both $III$ and $III$

Solution:

(a) Given that, a, b,c > 0

and

\; \; \; \; \; \; \; \;

D =

\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}

Using operation

R_2 \rightarrow C_1 + C_2 + C_3,

\; \; \; \; \; \; \; \; \; \; \; \; \; \;

D =

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

\; \; \; \; \; \; \; \; \; \; \; \; \; \;

D =

(a+b+c) \begin{vmatrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{vmatrix}

Using operation

R_2 \rightarrow R_2 - R_1, R_3 \rightarrow R_3 - R_1

\; \; \; \; \; \; \; \; \; \; \; \; \; \;

D =

(a+b+c) \begin{vmatrix} 1 & b & c \\ 0 & c-b & a-c \\ 0 & a-b & b-c \end{vmatrix}

Expanding along

C_1

D =

(a + b + c) \{(c - b)(b - c) - (a - c)(a - b)\}

= (a + b + c) \{bc - b^2 - c^2 + bc - a^2 + ac + ba - bc\}

= -\frac{1}{2}(a + b + c) \{(a - b)^2 + (b - c)^2 + (c - a)^2\}

Since, a ,b ,c > 0

Then, D > 0

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