A = \(\begin{bmatrix}-1 & -3 & 2\\ -5 & -9 & 1\end{bmatrix}\), and v = \(\begin{bmatrix}5\\ -3\\ -2\end{bmatrix}\)

Then show that v is in the null A

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Login NowLet A be a m x n matrix. Then, the null space of A, denoted by nulA, is the set of all vectors x ∈ R^{n}, such that Ax = 0 i.e.

Nul A = {x ∈ R^{n} : Ax = 0}

Note: Nul A ≠ Φ

**Solution:**

Given,

A = \(\begin{bmatrix}-1 & -3 & 2\\ -5 & -9 & 1\end{bmatrix}\), and v = \(\begin{bmatrix}5\\ -3\\ -2\end{bmatrix}\)

From the definition of Null Space, If v is the null of A then AV must be zero.

so, Av = 0

= \(\begin{bmatrix}-1 & -3 & 2\\ -5 & -9 & 1\end{bmatrix}\)\(\begin{bmatrix}5\\ -3\\ -2\end{bmatrix}\)

= \(\begin{bmatrix}-5+9-4\\ -25+27-2\end{bmatrix}\)

= \(\begin{bmatrix}0\\ 0\end{bmatrix}\)

Since Av = 0. So, v is in the null A.

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