Determinant

**sreesuja** · July 9th 2012, 08:54 PM

using property of determinant and without expanding prove that
[1 1+p 1+p+q ]
[2 3+2p 1+3p+2q ] =i
[3 6+3p 1+6p+3q ]

**Soroban** · July 12th 2012, 03:16 PM

Hello, sreesuja!

$\text{Using property of determinants and without expanding,}$

. . $\text{prove that: }\:\begin{vmatrix}1 & 1+p & 1+p+q \\ 2 & 3+2p & 1+3p+2q \\ 3 & 6+3p & 1+6p+3q \end{vmatrix} \;=\;1$

$\text{We have: }\: \begin{vmatrix}1 & 1+p & 1+p+q \\ 2 & 3+2p & 1+3p+2q \\ 3 & 6+3p & 1+6p+3q \end{vmatrix}$

$\begin{array}{c} \\ R_2-2R_1 \\ R_3 - 3R_1 \end{array}\begin{vmatrix}1 & 1+p & 1+p+q \\ 0 & 1 & \text{-}1 + p \\ 0 & 3 & \text{-}2+3p \end{vmatrix}$

$\begin{array}{c}\\ \\ R_3-3R_2 \end{array} \begin{vmatrix}1 & 1+p & 1+p+q \\ 0&1&\text{-}1+p \\ 0&0&1\end{vmatrix}$

The determinant is in diagonalized form; its value is 1.

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