Algebra

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sreesuja

New member
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Jul 9, 2012
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4
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 ]
 
I'm going to assume that "without expanding" just means the usual expansion by minors and that you are allowed to use "row reduction" to reduce to a triangular matrix- the determinant of a triangular matrix is the product of the numbers on the main diagonal. Start by reducing the first column by subtracting twice the first row from the second row and subtracting three times the first row from the third row. Then reduce the second column.

I will point out that the determinant is NOT "i". Did you mean "1"?
 
Hello, sreesuja!

\(\displaystyle \text{Prove: }\:\begin{vmatrix}1 & p+1 & p+q+1 \\ 2 & 2p +3 & 3p+2q+1 \\ 3 & 3p+6 & 6p+3q+1 \end{vmatrix}\)
Click to expand...

. . . . . . . . . \(\displaystyle \begin{array}{c} \\ \\ \\ R_2-2R_1 \\ R_3-3R_1 \end{array}\;\begin{vmatrix} 1 & p+1 & p+q+1 \\ 0 & 1 & p - 1 \\ 0 & 3 & 3p-2 \end{vmatrix}\)


. . . . . .\(\displaystyle \begin{array}{c}R_1 - (p\!+\!1)R_2 \\ \\ R_3 - 3R_2 \end{array}\;\begin{vmatrix}1 & 0 & p+q-p^2+2 \\ 0 & 1 & p-1 \\ 0 & 0 & 1 \end{vmatrix}\)


\(\displaystyle \begin{array}{c}R_1 - (p\!+\!q\!-\!p^2\!+\!2)R_3 \\ R_2 - (p\!-\!1)R_3 \\ \\ \end{array}\;\begin{vmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}\)


\(\displaystyle \text{Therefore: }\:\begin{vmatrix}1&0&0\\0&1&0\\0&0&1\end{vmatrix}\;=\;1\)
 
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