Find the basis of the null space of A.

[tex]\left[\begin{array}{cccc}-1\\1\\1\\0\end{array}\right][/tex], [tex]\left[\begin{array}{cccc}0\\2\\0\\1\end{array}\right][/tex] Are two vectors which are basis of null space of A

Step-by-step explanation:

Basis of a null space is a vector matrix which when multiplied by the reduced echelon form of the matrix gives null matrix or 0 matrix

So, to find the reduced echelon form we apply the row operations :

R₁ = R₁/2

[tex]\left[\begin{array}{cccc}1&-2&3&4\\2&-1&3&2\\4&-5&9&10\\0&-1&1&2\end{array}\right][/tex]

R₂ = R₂ - 2R₁

[tex]\left[\begin{array}{cccc}1&-2&3&4\\0&3&-3&-6\\4&-5&9&10\\0&-1&1&2\end{array}\right][/tex]

R₃ = R₃ - 4R₁

[tex]\left[\begin{array}{cccc}1&-2&3&4\\0&3&-3&-6\\0&3&-3&-6\\0&-1&1&2\end{array}\right][/tex]

R₂ = R₂/3

[tex]\left[\begin{array}{cccc}1&-2&3&4\\0&1&-1&-2\\0&3&-3&-6\\0&-1&1&2\end{array}\right][/tex]

R₁ = R₁ + 2R₂

[tex]\left[\begin{array}{cccc}1&0&1&0\\0&1&-1&-2\\0&3&-3&-6\\0&-1&1&2\end{array}\right][/tex]

R₃ = R₃ - 3R₂

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

R₄ = R₄ + R₂

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

which is a reduced echelon form of the matrix A

now ,

[tex]\left[\begin{array}{cccc}1&0&1&0\\0&1&-1&-2\\0&0&0&0\\0&0&0&0\end{array}\right][/tex] [tex]\left[\begin{array}{cccc}x1\\x2\\x3\\x4\end{array}\right][/tex] [tex]=\left[\begin{array}{cccc}0\\0\\0\\0\end{array}\right][/tex]

therefore,

x₁ + x₃ = 0

x₁ = -x₃

x₂ - x₃ - 2x₄ = 0

x₂ = x₃ + 2x₄

so now replacing values of x₁ , x₂ in the vector matrix

[tex]\left[\begin{array}{cccc}-x3\\x3 + 2x4\\x3\\x4\end{array}\right][/tex]

we can write the above matrix as,

[tex]x3\left[\begin{array}{cccc}-1\\1\\1\\0\end{array}\right][/tex] [tex]+ x4\left[\begin{array}{cccc}0\\2\\0\\1\end{array}\right][/tex]

therefore we can say that above mentioned vector matrices are the basis of the null space of matrix A