a) In order to find Nul A, we must solve

\(\displaystyle{A}{x}={0}\)

Since A is \(\displaystyle{2}\times{5}\) matrix, it can multyplied from left only by vector with 5 rows. That means x is in \(\displaystyle{\mathbb{{{R}}}}^{{5}}\), i.e. k=5

b) As we saw in part a) \(\displaystyle{x}\in{\mathbb{{{R}}}}^{{5}}\), so we know

Col \(\displaystyle{A}={\left\lbrace{y}:{y}={A}{x}\ \text{ for some }\ {x}\in{\mathbb{{{R}}}}^{{5}}\right\rbrace}\)

So we have a multiplication of \(\displaystyle{2}\times{5}\) matrix A and \(\displaystyle{5}\times{1}\) vector x. From definition of matrix muplication, y is \(\displaystyle{2}\times{1}\) vector, i.e k=2

Result: a) 5, b) 2

\(\displaystyle{A}{x}={0}\)

Since A is \(\displaystyle{2}\times{5}\) matrix, it can multyplied from left only by vector with 5 rows. That means x is in \(\displaystyle{\mathbb{{{R}}}}^{{5}}\), i.e. k=5

b) As we saw in part a) \(\displaystyle{x}\in{\mathbb{{{R}}}}^{{5}}\), so we know

Col \(\displaystyle{A}={\left\lbrace{y}:{y}={A}{x}\ \text{ for some }\ {x}\in{\mathbb{{{R}}}}^{{5}}\right\rbrace}\)

So we have a multiplication of \(\displaystyle{2}\times{5}\) matrix A and \(\displaystyle{5}\times{1}\) vector x. From definition of matrix muplication, y is \(\displaystyle{2}\times{1}\) vector, i.e k=2

Result: a) 5, b) 2