Test Index

AP EAMCET Engineering 2013 Solved Paper

Section: Mathematics

© examsnet.com

Question : 96 of 160

Marks: +1, -0

\begin{vmatrix} x+2 & x+3 & x+5 \\ x+4 & x+6 & x+9 \\ x+8 & x+11 & x+15 \end{vmatrix}

is equal to

$3 x^{2}+4 x+5$
$x^{3}+8 x+2$
0
$-2$

Solution:

Let

\Delta=\begin{vmatrix} x+2 & x+3 & x+5 \\ x+4 & x+6 & x+9 \\ x+8 & x+11 & x+15 \end{vmatrix}

Apply operations

R_{2} \rightarrow R_{2}-R_{1}, R_{3} \rightarrow R_{3}-R_{1}

, we get

\Delta=\begin{vmatrix} x+2 & x+3 & x+5 \\ 2 & 3 & 4 \\ 6 & 8 & 10 \end{vmatrix}

Again, apply operation

C_{2} \rightarrow C_{2}-C_{1}

,

C_{3} \rightarrow C_{3}-C_{1}

, we get

\Delta=\begin{vmatrix} x+2 & 1 & 3 \\ 2 & 1 & 2 \\ 6 & 2 & 4 \end{vmatrix}

Expand along

R_{1}

, we get

\Delta=(x+2)(4-4)-1(8-12)

+3(4-6)

=0+4+3(-2)=4-6=-2

© examsnet.com

Go to Question:

Prev Question

Next Question