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 Post subject: Determinant of a matrix
PostPosted: Sat Jun 25, 2016 7:53 am 
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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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PostPosted: Sat Jun 25, 2016 7:55 am 
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\(\displaystyle{D_{1}=\left|5\right|=5}\)

\(\displaystyle{D_{2}=\begin{vmatrix}
5 & 2\\
-2 & 5
\end{vmatrix}=25+4=29}\)

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, \(\displaystyle{D_{n}=\dfrac{7^{n}+3^{n}}{2}\,,n\in\mathbb{N}}\) .


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PostPosted: Thu Apr 19, 2018 9:21 am 

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