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AP EAPCET 23-Aug-2021 Shift 2 Paper

Section: Mathematics

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Question : 5 of 160

Marks: +1, -0

The value of the determinant

\begin{vmatrix} a+b & a+2 b & a+3 b \\ a+2 b & a+3 b & a+4 b \\ a+4 b & a+5 b & a+6 b \end{vmatrix}

is

a
b
0
a + b

Solution:

\Delta = \begin{vmatrix} a+b & a+2 b & a+3 b \\ a+2 b & a+3 b & a+4 b \\ a+4 b & a+5 b & a+6 b \end{vmatrix}

\Rightarrow \Delta = \begin{vmatrix} a+b & (a+ b)+b & a+3 b \\ a+2 b & (a+ 2b)+b & a+4 b \\ a+4 b & (a+ 4b)+b & a+6 b \end{vmatrix}

\Delta = \begin{vmatrix} a+b & a+ b & a+3 b \\ a+2 b & a+2 b & a+4 b \\ a+4 b & a+4b & a+6 b \end{vmatrix}

+ \begin{vmatrix} a+b & b & a+3 b \\ a+2 b & b & a+4 b \\ a+4 b & b & a+6 b \end{vmatrix}

[The value of determinant is zero if it is two rows/columns are same]

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

Applying

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

and

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

,

\Delta = b \begin{vmatrix} a+b & 1 & a+3b \\ b & 0 & b \\ 3b & 0 & 3b \end{vmatrix}

\Delta = b \{ b(3 b) - b(3 b) \}

\Delta = b . 0 \Rightarrow \Delta = 0

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