Q45 / 160 · AP EAMCET 2021 (23 Aug Shift 2) · Mathematics

The value of the determinant $\left|\begin{array}{ccc}a+b & a+2 b & a+3 b \\ a+2 b & a+3 b & a+4 b \\ a+4 b & a+5 b & a+6 b\end{array}\right|$ is

  • A
    a
  • B
    b
  • Correct answer
    C
    $0$
  • D
    $a+b$

Solution

To find $\Delta=\left|\begin{array}{ccc}a+b & a+2 b & a+3 b \\ a+2 b & a+3 b & a+4 b \\ a+4 b & a+5 b & a+6 b\end{array}\right|$ $\Delta=\left|\begin{array}{ccc}a+b & (a+b)+b & a+3 b \\ a+2 b & (a+2 b)+b & a+4 b \\ a+4 b & (a+4 b)+b & a+6 b\end{array}\right|$ $\Delta=\left|\begin{array}{ccc}a+b & a+b & a+3 b \\ a+2 b & a+2 b & a+4 b \\ a+4 b & a+4 b & a+6 b\end{array}\right|$ $+\left|\begin{array}{lll}a+b & b & a+3 b \\ a+2 b & b & a+4 b \\ a+4 b & b & a+6 b\end{array}\right|$ [The value of determinant is zero if it is two rows/columns are same \} $\Delta=0+b\left|\begin{array}{ccc}a+b & 1 & a+3 b \\ a+2 b & 1 & a+4 b \\ a+4 b & 1 & a+6 b\end{array}\right|$ Applying $R_2 \rightarrow R_2-R_1$ and $R_3 \rightarrow R_3-R_1$, $\Delta=b\left|\begin{array}{ccc}a+b & 1 & a+3 b \\ b & 0 & b \\ 3 b & 0 & 3 b\end{array}\right|$ $\begin{aligned} & \Delta=b\{b(3 b)-b(3 b)\} \\ & \Delta=b \cdot 0 \Rightarrow \Delta=0\end{aligned}$

Hence, the correct answer is option: 'C'.

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