Answer to Question #153655 in Linear Algebra for pallavi

Question

Determine the range, kernel, rank and nullity of the given matrix,

A = [2 3 1 2 0

0 3 −1 2 1

1 -3 2 4 3

2 3 0 3 0]

Answer 1

$Solution: By ~definition~ of~ kernel~ A, ker(A)= \{x \in R^5 |Ax=0 \}. \\ \therefore ~we ~solve~ the~ equation,~ Ax=0$

$\therefore \begin{bmatrix} 2 & 3 & 1 & 2 & 0 \\ 0 & 3 & -1 & 2 & 1 \\ 1 & -3 & 2 & 4 & 3 \\ 2&3&0&3&0 \end{bmatrix} \begin{bmatrix} x_1 \\x_2\\x_3\\x_4\\x_5\end{bmatrix} =\begin{bmatrix} 0 \\0\\0\\0\end{bmatrix} , \\where ~A=\begin{bmatrix} 2 & 3 & 1 & 2 & 0 \\ 0 & 3 & -1 & 2 & 1 \\ 1 & -3 & 2 & 4 & 3 \\ 2&3&0&3&0 \end{bmatrix}~,x=\begin{bmatrix} x_1 \\x_2\\x_3\\x_4\\x_5\end{bmatrix}~ and~ 0=\begin{bmatrix} 0 \\0\\0\\0\end{bmatrix}$

$Now ~we~ write ~ in ~ augmented ~ matrix ~ form ~ as$

$\therefore \begin{bmatrix} 2 & 3 & 1 & 2 & 0 &|0\\ 0 & 3 & -1 & 2 & 1&|0 \\ 1 & -3 & 2 & 4 & 3&|0 \\ 2 & 3 & 0 & 3 & 0 &|0 \end{bmatrix} \\Now ~solve ~ this ~ augmented ~ matrix ~ to ~ get~ solution ~ of ~Ax=0$

$Applying ~ R_1\Rightarrow(\frac{1}{2})R_1 \\ \begin{bmatrix} 1 & \frac{3}{2} & \frac{1}{2} & 1 & 0 &|0\\ 0 & 3 & -1 & 2 & 1&|0 \\ 1 & -3 & 2 & 4 & 3&|0 \\ 2 & 3 & 0 & 3 & 0 &|0 \end{bmatrix}$

$Applying ~ R_3\Rightarrow ~ R_3-R_1~ and ~R_4\Rightarrow ~ R_4-2R_1 \\ \begin{bmatrix} 1 & \frac{3}{2} & \frac{1}{2} & 1 & 0 &|0\\ 0 & 3 & -1 & 2 & 1&|0 \\ 0 & -\frac{9}{2} & \frac{3}{2} & 3 & 3&|0 \\ 0 & 0 & -1 & 1 & 0 &|0 \end{bmatrix}$

$Applying ~ R_2\Rightarrow ~ (\frac{1}{3})R_2 \\ \begin{bmatrix} 1 & \frac{3}{2} & \frac{1}{2} & 1 & 0 &|0\\ 0 & 1 & -\frac{1}{3} & \frac{2}{3}&\frac{1}{3}&|0 \\ 0 & -\frac{9}{2} & \frac{3}{2} & 3 & 3&|0 \\ 0 & 0 & -1 & 1 & 0 &|0 \end{bmatrix}$

$Applying ~ R_3\Rightarrow ~R_3+ (\frac{9}{2})R_2 \\ \begin{bmatrix} 1 & \frac{3}{2} & \frac{1}{2} & 1 & 0 &|0\\ 0 & 1 & -\frac{1}{3} & \frac{2}{3}&\frac{1}{3}&|0 \\ 0 & 0 & 0 & 6 & \frac{9}{2}&|0 \\ 0 & 0 & -1 & 1 & 0 &|0 \end{bmatrix}$

$Applying ~ R_3 \Leftrightarrow R_4 \\ \begin{bmatrix} 1 & \frac{3}{2} & \frac{1}{2} & 1 & 0 &|0\\ 0 & 1 & -\frac{1}{3} & \frac{2}{3}&\frac{1}{3}&|0 \\ 0&0&-1&1&0&|0\\ 0 & 0 & 0 & 6 & \frac{9}{2}&|0 \\ \end{bmatrix}$

$Applying ~ R_3 \Rightarrow(-1) R_3~ and ~ R_4 \Rightarrow(\frac{1}{6}) R_4 \\ \begin{bmatrix} 1 & \frac{3}{2} & \frac{1}{2} & 1 & 0 &|0\\ 0 & 1 & -\frac{1}{3} & \frac{2}{3}&\frac{1}{3}&|0 \\ 0&0&1&-1&0&|0\\ 0 & 0 & 0 & 1 & \frac{3}{4}&|0 \\ \end{bmatrix}$

$which ~ is ~ row ~ echelon~ form. ~Now~ we ~can ~ find ~ the ~ solution.\\Here~ x_5 ~has~ no ~ leading~ 1. \\ \therefore x_5 ~ ~ is ~ free ~ variable.\\ \therefore we~ put ~ x_5 =t ~~~~\{where~ t\in R\} \\ By ~ back ~ substitution~ we ~ get ~ the ~ equations~$

$x_4+(\frac{3}{4})x_5=0\Rightarrow x_4=-\frac{3}{4}x_5=-\frac{3}{4}t \\x_3-x_4=0\Rightarrow x_3=x_4=-\frac{3}{4}t$

$x_2-\frac{1}{3}x_3+\frac{2}{3}x_4+\frac{1}{3}x_5=0\Rightarrow x_2=\frac{1}{3}x_3-\frac{2}{3}x_4-\frac{1}{3}x_5=\frac{1}{3}(-\frac{3}{4}t)-\frac{2}{3}(-\frac{3}{4}t)-\frac{1}{3}t=(-\frac{3}{12})t+\frac{6}{12}t-\frac{1}{3}t=t(-\frac{3}{12}+\frac{6}{12}-\frac{1}{3})=t(\frac{3}{12}-\frac{1}{3})=t(\frac{1}{4}-\frac{1}{3})=t(\frac{3-4}{12})=t(-\frac{1}{12})=-\frac{1}{12}t$

$\therefore x_2=-\frac{1}{12}t$

$and ~last~ equation~ x_1+\frac{3}{2}x_2+\frac{1}{2}x_3+x_4=0$

$\therefore x_1=-\frac{3}{2}x_2-\frac{1}{2}x_3-x_4=-\frac{3}{2}(-\frac{1}{12}t)-\frac{1}{2}(-\frac{3}{4}t)-(-\frac{3}{4}t)=\frac{3}{24}t+\frac{3}{8}t+\frac{3}{4}t$

$x_1=t(\frac{3}{24}+\frac{3}{8}+\frac{3}{4})=t(\frac{1}{8}+\frac{3}{8}+\frac{3}{4})=t(\frac{4}{8}+\frac{3}{4})=t(\frac{4}{8}+\frac{3}{4})=t(\frac{1}{2}+\frac{3}{4})=\frac{5}{4}t$

$\therefore ker(A)=\begin{bmatrix} x_1 \\x_2\\x_3\\x_4\\x_5\end{bmatrix}=\{\begin{bmatrix} \frac{5}{4}t \\-\frac{1}{12}t \\-\frac{3}{4}t \\-\frac{3}{4}t\\ 1 \end{bmatrix} | t\in R\}=\{\begin{bmatrix} \frac{5}{4}\\-\frac{1}{12} \\-\frac{3}{4}\\-\frac{3}{4}\\ 1 \end{bmatrix} t ~~| t\in R\}$

$Range(A)=~columns~ of~leading ~1's ~ in ~ row ~ echelon~ form\\ \therefore~ Range(A)=\{ \begin{bmatrix} 2 \\0\\1\\2\end{bmatrix},\begin{bmatrix} 3 \\3\\-3\\3\end{bmatrix},\begin{bmatrix} 1 \\-1\\2\\0\end{bmatrix},\begin{bmatrix} 2 \\2\\4\\3\end{bmatrix}\}$

$Rank (A)= number ~ of ~ non~ zero~ rows~in ~ row ~ echelon~ form \\ rank(A)=4$

$By ~rank ~ nullity~ theorem,\\rank(A)+nullity(A)=n, ~ where ~ n ~ is ~ number ~ of ~columns ~ in ~matrix ~A$

$\therefore 4+nullity(A)=5\Rightarrow~nullity(A)=5-4=1\\ \therefore nullity(A)=1$