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Manipal Entrance Test 2014 Math Paper

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Question : 77 of 80

Marks: +1, -0

If

\begin{vmatrix} x + y + 2z & x & y \\ z & y + z + 2x & y \\ z & x & z + x + 2y \end{vmatrix}

=k(x+y+z)^{3},

then the value of

k

is

1
2
4
8

Solution:

\Delta = \begin{vmatrix} x + y + 2z & x & y \\ z & y + z + 2x & y \\ z & x & z + x + 2y \end{vmatrix}

= \begin{vmatrix} 2( x + y + z) & x & y \\ 2( x + y + z) & y + z + 2 x & y \\ 2( x + y + z) & x & z + x + 2 y \end{vmatrix}

\ ( \text{using } C_{1} \rightarrow C_{1}+C_{2}+C_{3}\ )

Take out

2(x+y+z)

common from

C_{1},

we get

\Delta = 2(x+y+z)

\begin{vmatrix} 1 & x & y \\ 1 & y + z + 2x & y \\ 1 & x & z + x + 2y \end{vmatrix}

=2(x+y+z)

\begin{vmatrix} 1 & x & y \\ 0 & y + z + 2x & 0 \\ 0 & 0 & z + x + 2y \end{vmatrix}

(using

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

Take out

(x+y+z)

common from

R_{2}

and

R_{3}

we get

\Delta = 2(x+y+z)(x+y+z)(x+y+z)

\times \begin{vmatrix} 1 & x & y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}

Expanding along

R_{3}

, we get

\Delta = 2(x+y+z)^{3}[(1)(1-0)]

=2(x+y+z)^{3}=k(x+y+z)^{3}

(given)

∴

k=2

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