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 Author: Tolaso J Kos [ Sat Jun 25, 2016 7:53 am ] Post subject: Determinant of a matrix Let $$A$$ be an $$n \times n$$ matrix that is defined as: $$A=a_{ij}=\left\{\begin{matrix}5\,, & i=j \\ 2\,, & ij \end{matrix}\right.$$ If $$D_n$$ is its determinant then prove that $$D_n=10D_{n-1}-21D_{n-2}$$ and in continunation evaluate the det of the matrix.

 Author: Papapetros Vaggelis [ Sat Jun 25, 2016 7:55 am ] Post subject: Re: Determinant of a matrix $$\displaystyle{D_{1}=\left|5\right|=5}$$$$\displaystyle{D_{2}=\begin{vmatrix}5 & 2\\ -2 & 5\end{vmatrix}=25+4=29}$$ Let $$\displaystyle{n\geq 3}$$ .If $$\displaystyle{\Gamma_{i}\,,\Sigma_{i}\,,1\leq i\leq n}$$ are the lines and the columns, repsectively, of the matrix, then by using the operation $$\displaystyle{\Sigma_{1}\to \Sigma_{1}+\Sigma_{n}}$$ we get :$$\displaystyle{D_{n}=\begin{vmatrix}5 & 2 & 2 & ... &2 &2 \\ -2 &5 &2 &... & 2 &2 \\ -2 &-2 &5 &... &2 &2 \\ -2 &-2 &-2 &5 &... &2 \\ ... &... &... &... &... &... \\ -2 &-2 &-2 &... &-2 &5 \end{vmatrix}=\begin{vmatrix}7 & 2 & 2 & ... &2 &2 \\ 0 &5 &2 &... & 2 &2 \\ 0 &-2 &5 &... &2 &2 \\ 0 &-2 &-2 &5 &... &2 \\ ... &... &... &... &... &... \\ 3 &-2 &-2 &... &-2 &5 \end{vmatrix}}$$and then, by using $$\displaystyle{\Gamma_{1}\to \Gamma_{1}+\Gamma_{n}}$$, we have that :\displaystyle{\begin{aligned} D_{n}&=\begin{vmatrix}10 & 0 & 0 & ... &0 &7 \\ 0 &5 &2 &... & 2 &2 \\ 0 &-2 &5 &... &2 &2 \\ 0 &-2 &-2 &5 &... &2 \\ ... &... &... &... &... &... \\ 3 &-2 &-2 &... &-2 &5 \end{vmatrix}\\&=10\cdot \begin{vmatrix} 5 &2 &... & 2 &2 \\ -2 &5 &... &2 &2 \\ -2 &-2 &5 &... &2 \\ ... &... &... &... &... \\ -2 &-2 &... &-2 &5 \end{vmatrix}+3\,(-1)^{n+1}\cdot \begin{vmatrix}0 & 0 & ... &0 &7 \\ 5 &2 &... & 2 &2 \\ -2 &5 &... &2 &2 \\ -2 &-2 &5 &... &2 \\ ... &... &... &... &... \\ -2 &-2 &... &-2 &5 \end{vmatrix}\\&=10\,D_{n-1}+3\,(-1)^{n+1}\,7\,(-1)^{1+n-1}\cdot \begin{vmatrix} 5 &2 &... & 2 \\ -2 &5 &... &2 \\ -2 &-2 &5 &... \\ ... &... &... &... \\ -2 &-2 &... &5 \end{vmatrix}\\&=10\,D_{n-1}-21\,D_{n-2}\end{aligned}} We observe that $$\displaystyle{D_{1}=5=\dfrac{7+3}{2}\,,D_{2}=29=\dfrac{7^2+3^2}{2}}$$ .Suppose that $$\displaystyle{D_{k}=\dfrac{7^{k}+3^{k}}{2}\,,\forall\,k\in\left\{1,...,n-1\right\}\,(I)}$$ .$$\displaystyle{\bullet\,k=n}$$ \displaystyle{\begin{aligned} D_{n}&=10\,D_{n-1}-21\,D_{n-2}\\&\stackrel{(I)}{=}10\cdot \dfrac{7^{n-1}+3^{n-1}}{2}-21\cdot \dfrac{7^{n-2}+3^{n-2}}{2}\\&=5\,7^{n-1}+5\,3^{n-1}-\dfrac{21}{2}\,7^{n-2}-\dfrac{21}{2}\,3^{n-2}\\&=7^{n-2}\,\left[5\cdot 7-\dfrac{21}{2}\right]+3^{n-2}\,\left[5\cdot 3-\dfrac{21}{2}\right]\\&=\left(35-\dfrac{21}{2}\right)\,7^{n-2}+\left(15-\dfrac{21}{2}\right)\,3^{n-2}\\&=\dfrac{49}{2}\,7^{n-2}+\dfrac{9}{2}\,3^{n-2}\\&=\dfrac{1}{2}\,\left(7^2\cdot 7^{n-2}+3^2\cdot 3^{n-2}\right)\\&=\dfrac{7^{n}+3^{n}}{2}\end{aligned}} .So, by induction, $$\displaystyle{D_{n}=\dfrac{7^{n}+3^{n}}{2}\,,n\in\mathbb{N}}$$ .

 Author: Skist1989 [ Thu Apr 19, 2018 9:21 am ] Post subject: Re: Determinant of a matrix That's kinda interesting but could you explain more detailed please? I'm, doing a complete guide where i want to include the most detailed solutions that i can find. Service like write paper for me will provide students with the most fresh and detailed information on their subject of choice!

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