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## Determinant of a matrix

Linear Algebra
Tolaso J Kos
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### Determinant of a matrix

Let $A$ be an $n \times n$ matrix that is defined as: $$A=a_{ij}=\left\{\begin{matrix} 5\,, & i=j \\ 2\,, & i<j\\ -2\,, &i>j \end{matrix}\right.$$ 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.
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Papapetros Vaggelis
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### 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, $\displaystyle{D_{n}=\dfrac{7^{n}+3^{n}}{2}\,,n\in\mathbb{N}}$ .