Question

Aljabar determinan matriks. Carilah hasil pengoperasian nya
[tex]\displaystyle \begin{vmatrix}n & n+1 & n+2\\ n+1 & n+2 & n+3\\ n+2 & n+3 & n+4\end{vmatrix}[/tex]

Answer 1

Verified answer

[tex]\begin{aligned}&\begin{vmatrix}n & n+1 & n+2\\ n+1 & n+2 & n+3\\ n+2 & n+3 & n+4\end{vmatrix}=\boxed{\bf0}\end{aligned}[/tex]
 

Penjelasan dengan langkah-langkah:

Determinan Matriks

Kita akan menghitung nilai:

[tex]\begin{aligned}\begin{vmatrix}n & n+1 & n+2\\ n+1 & n+2 & n+3\\ n+2 & n+3 & n+4\end{vmatrix}\end{aligned}[/tex]

Metode Ekspansi Kofaktor

[tex]\begin{aligned}&\begin{vmatrix}n & n+1 & n+2\\ n+1 & n+2 & n+3\\ n+2 & n+3 & n+4\end{vmatrix}\\&{=\ }n\begin{vmatrix}n+2 & n+3\\n+3 & n+4\end{vmatrix}-(n+1)\begin{vmatrix}n+1 &n+3\\ n+2&n+4\end{vmatrix}+(n+2)\begin{vmatrix}n+1 & n+2\\ n+2 & n+3\end{vmatrix}\\&{=\ }n\left[(n+2)(n+4)-(n+3)^2\right]-(n+1)\left[(n+1)(n+4)-(n+2)(n+3)\right]\ddots\\&\quad+(n+2)\left[(n+1)(n+3)-(n+2)^2\right]\\&{=\ }n(-1)-(n+1)(-2)+(n+2)(-1)\\&{=\ }-n+2n+2-n-2\\&{=\ }2n-2n+2-2\\&{=\ }\boxed{\bf0}\end{aligned}[/tex]

Metode Reduksi Baris

Kita lakukan reduksi baris pada matriksnya sehingga menjadi bentuk eselon baris.

[tex]\begin{aligned}&\begin{pmatrix}n & n+1 & n+2\\ n+1 & n+2 & n+3\\ n+2 & n+3 & n+4\end{pmatrix}\\R_2-\left(\frac{n+1}{n}\right)R_1\rightarrow R_2\ \Rightarrow &\begin{pmatrix}n & n+1 & n+2\\ 0 &\vphantom{\Bigg|} -\dfrac{1}{n} & -\dfrac{2}{n}\\ n+2 & n+3 & n+4\end{pmatrix}\\R_3-\left(\frac{n+2}{n}\right)R_1\rightarrow R_3\ \Rightarrow&\begin{pmatrix}n & n+1 & n+2\\ 0 &\vphantom{\Bigg|} -\dfrac{1}{n} & -\dfrac{2}{n}\\ 0 & -\dfrac{2}{n} & -\dfrac{4}{n}\end{pmatrix}\end{aligned}[/tex]
[tex]\begin{aligned}R_3-2R_1\rightarrow R_3\ \Rightarrow&\begin{pmatrix}n & n+1 & n+2\\ 0 &\vphantom{\Bigg|} -\dfrac{1}{n} & -\dfrac{2}{n}\\ 0 & 0 & 0\end{pmatrix}\\\end{aligned}[/tex]

Maka:

[tex]\begin{aligned}&\begin{vmatrix}n & n+1 & n+2\\ n+1 & n+2 & n+3\\ n+2 & n+3 & n+4\end{vmatrix}=\begin{vmatrix}n & n+1 & n+2\\ 0 &\vphantom{\Bigg|} -\dfrac{1}{n} & -\dfrac{2}{n}\\ 0 & 0 & 0\end{vmatrix}=\boxed{\bf0}\end{aligned}[/tex]