Question

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

Answer 1

The basis for the null space of A is
`{\left[\begin{array}{c}-1&-1&1&0\end{array}\right],\left[\begin{array}{c}-1&1&0&1\end{array}\right]}`

Explanation:

1. The first step is to find the reduced row echelon form of the matrix:

`\left[\begin{array}{cccc}1&0&1&1\\3&2&5&1\\0&4&4&-4\end{array}\right]`

• Make zeros in column 1 except the entry at row 1, column 1. Subtract row 1 multiplied by 3 from row 2
  `\left(R_2=R_2-\left(3\right)R_1\right)`

`\left[\begin{array}{cccc}1&0&1&1\\0&2&2&-2\\0&4&4&-4\end{array}\right]`

• Make zeros in column 2 except the entry at row 2, column 2. Subtract row 2 multiplied by 2 from row 3
  `\left(R_3=R_3-\left(2\right)R_2\right)`

`\left[\begin{array}{cccc}1&0&1&1\\0&2&2&-2\\0&0&0&0\end{array}\right]`

• Multiply the second row by 1/2
  `\left(R_2=\left(1/2\right)R_2\right)`

`\left[\begin{array}{cccc}1&0&1&1\\0&1&1&-1\\0&0&0&0\end{array}\right]`

2. Convert the matrix equation back to an equivalent system and solve the matrix equation

`1x_(1) +x_(3) +1x_(4)=0\\ 1x_(2) +x_(3) -1x_(4)=0\\0=0`

`\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]`

If we take
`x_(3)=t, x_(4)=s` then
`x_(1)=-s-t,x_(2)=s-t,x_(3)=t,x_(4)=s`

Therefore,

`\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)`

The null space has a basis formed by the set {
`{\left[\begin{array}{c}-1&-1&1&0\end{array}\right],\left[\begin{array}{c}-1&1&0&1\end{array}\right]}`}