Q:

Let A be the matrix: [130 024 154 11-4] Find a basis for the nullspace of A.

Accepted Solution

A:
Answer:The basis for the null space of A is [tex]{\left[\begin{array}{c}-1&-1&1&0\end{array}\right],\left[\begin{array}{c}-1&1&0&1\end{array}\right]}[/tex]Step-by-step explanation:The first step is to find the reduced row echelon form of the matrix:[tex]\left[\begin{array}{cccc}1&0&1&1\\3&2&5&1\\0&4&4&-4\end{array}\right][/tex]Make zeros in column 1 except the entry at row 1, column 1. Subtract row 1 multiplied by 3 from row 2 [tex]\left(R_2=R_2-\left(3\right)R_1\right)[/tex][tex]\left[\begin{array}{cccc}1&0&1&1\\0&2&2&-2\\0&4&4&-4\end{array}\right][/tex]Make zeros in column 2 except the entry at row 2, column 2. Subtract row 2 multiplied by 2 from row 3 [tex]\left(R_3=R_3-\left(2\right)R_2\right)[/tex][tex]\left[\begin{array}{cccc}1&0&1&1\\0&2&2&-2\\0&0&0&0\end{array}\right][/tex]Multiply the second row by 1/2 [tex]\left(R_2=\left(1/2\right)R_2\right)[/tex][tex]\left[\begin{array}{cccc}1&0&1&1\\0&1&1&-1\\0&0&0&0\end{array}\right][/tex] Β  Β  2. Convert the matrix equation back to an equivalent system and solve the matrix equation[tex]1x_{1} +x_{3} +1x_{4}=0\\ 1x_{2} +x_{3} -1x_{4}=0\\0=0[/tex][tex]\left[\begin{array}{cccc}1&0&1&1\\0&1&1&-1\\0&0&0&0\end{array}\right] \left[\begin{array}{c}x_{1} &x_{2} &x_{3}&x_{4} \end{array}\right]=\left[\begin{array}{c}0&0&0\end{array}\right][/tex]If we take [tex]x_{3}=t, x_{4}=s[/tex] then [tex]x_{1}=-s-t,x_{2}=s-t,x_{3}=t,x_{4}=s[/tex]Therefore, [tex]\boldsymbol{x}=\left[\begin{array}{c}-s-t&s-t&t&s\end{array}\right]=\left[\begin{array}{c}-1&-1&1&0\end{array}\right]t+\left[\begin{array}{c}-1&1&0&1\end{array}\right]s\\\boldsymbol{x}=\left[\begin{array}{c}-1&-1&1&0\end{array}\right]x_{3} +\left[\begin{array}{c}-1&1&0&1\end{array}\right]x_{4}[/tex]The null space has a basis formed by the set {[tex]{\left[\begin{array}{c}-1&-1&1&0\end{array}\right],\left[\begin{array}{c}-1&1&0&1\end{array}\right]}[/tex]}