Answer:
The rank of the matrix is 3.
Step-by-step explanation:
Consider the prided information.
[tex]x=\begin{bmatrix}x_1\\ x_2\\ x_3\end{bmatrix}=\begin{bmatrix}9&2&6&5&8\\ 12&8&6&4&10\\ 3&4&0&2&1\end{bmatrix}[/tex]
Reduce the matrix in row echelon form as shown:
[tex]R_1\:\leftrightarrow \:R_2\begin{bmatrix}12&8&6&4&10\\ 9&2&6&5&8\\ 3&4&0&2&1\end{bmatrix}\\[/tex]
[tex]R_2\:\leftarrow \:R_2-\frac{3}{4}\cdot \:R_1\ and\ R_3\:\leftarrow \:R_3-\frac{1}{4}\cdot \:R_1\\\\\begin{bmatrix}12&8&6&4&10\\ 0&-4&\frac{3}{2}&2&\frac{1}{2}\\ 0&2&-\frac{3}{2}&1&-\frac{3}{2}\end{bmatrix}[/tex]
[tex]R_3\:\leftarrow \:R_3+\frac{1}{2}\cdot \:R_2\\\begin{bmatrix}12&8&6&4&10\\ 0&-4&\frac{3}{2}&2&\frac{1}{2}\\ 0&0&-\frac{3}{4}&2&-\frac{5}{4}\end{bmatrix}[/tex]
[tex]R_2\:\leftarrow \:R_2-\frac{3}{2}\cdot \:R_3\ and\ R_1\:\leftarrow \:R_1-6\cdot \:R_3\\\begin{bmatrix}12&8&0&20&0\\ 0&-4&0&6&-2\\ 0&0&1&-\frac{8}{3}&\frac{5}{3}\end{bmatrix}\\R_2\:\leftarrow \:-\frac{1}{4}\cdot \:R_2\ , \ R_1\:\leftarrow \:R_1-8\cdot \:R_2\ and\ \:R_1\:\leftarrow \frac{1}{12}\cdot \:R_1\\\begin{bmatrix}1&0&0&\frac{8}{3}&-\frac{1}{3}\\ 0&1&0&-\frac{3}{2}&\frac{1}{2}\\ 0&0&1&-\frac{8}{3}&\frac{5}{3}\end{bmatrix}[/tex]
The rank of a matrix is the number of non zeros rows.
Thus, the rank of the matrix is 3.
