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  • Using properties of determinants, prove that det 1 1+p 1+p+q 2 3+2p 4+3p+2q 3 6+3p 10+6p+3q =1

Q :14        Using properties  of determinants, prove that

                \begin{vmatrix} 1 &1+p &1+p+q \\ 2&3+2p &4+3p+2q \\ 3&6+3p & 10+6p+3q \end{vmatrix}=1

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Given determinant \triangle = \begin{vmatrix} 1 &1+p &1+p+q \\ 2&3+2p &4+3p+2q \\ 3&6+3p & 10+6p+3q \end{vmatrix}

Applying the row transformation; R_{2} \rightarrow R_{2}-2R_{1}  and R_{3} \rightarrow R_{3}-3R_{1} we have then;

\triangle = \begin{vmatrix} 1 &1+p &1+p+q \\ 0&1 &2+p \\ 0&3 & 7+3p \end{vmatrix}

Now, applying another row transformation R_{3}\rightarrow R_{3}-3R_{2} we have;

\triangle = \begin{vmatrix} 1 &1+p &1+p+q \\ 0&1 &2+p \\ 0&0 & 1 \end{vmatrix}

We can expand the remaining determinant along C_{1}, we have;

\triangle = 1\begin{vmatrix} 1 & 2+p\\ 0 &1 \end{vmatrix} = 1(1-0) =1

Hence the result is proved.

 

Posted by

Divya Prakash Singh

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