## Elementary Linear Algebra 7th Edition

(a) The nullspace of $A$ consists of the vectors on the following form x= \left[\begin{aligned} x_{1}\\ x_{2}\\x_3\\x_4 \end{aligned}\right]= \left[\begin{aligned}2t\\5t\\t\\t \end{aligned}\right] =t\left[\begin{aligned}2 \\5\\1\\1 \end{aligned}\right]+t\left[\begin{aligned}-1\\-2 \\0\\1 \end{aligned}\right]. (b) The nullity is $1$. (c) The rank of $A$ is $3$. Since $A$ has $4$ columns, one can see that $$\text{rank}(A)+\text{nullity}(A)=3+1=4.$$
Given the matrix $$A=\left[ \begin {array}{cccc} 1&0&-2&0\\ 4&-2&4&-2 \\ -2&0&1&3\end {array} \right] .$$ The reduced row echelon form is $$\left[ \begin {array}{cccc} 1&0&0&-2\\ 0&1&0&-5 \\ 0&0&1&-1\end {array} \right] .$$ The corresponding system is \begin{aligned} x_{1} -2x_4 &=0\\ x_{2} -5x_4 &=0\\ x_{3} - x_4 &=0 \end{aligned}. The solution of the above system is $x_1= 2t$, $x_2=5t$,$x_3=t$, $x_4=t$. (a) The nullspace of $A$ consists of the vectors on the following form x= \left[\begin{aligned} x_{1}\\ x_{2}\\x_3\\x_4 \end{aligned}\right]= \left[\begin{aligned}2t\\5t\\t\\t \end{aligned}\right] =t\left[\begin{aligned}2 \\5\\1\\1 \end{aligned}\right]+t\left[\begin{aligned}-1\\-2 \\0\\1 \end{aligned}\right]. (b) The nullity is $1$. (c) The rank of $A$ is $3$. Since $A$ has $4$ columns, one can see that $$\text{rank}(A)+\text{nullity}(A)=3+1=4.$$