Consider an algorithm that solves Avec$(x)=$vec$(b)$ in $O(n^4)$ operations where $A$ is an $nn$ matrix. Which of the following is true?

Solving Avec$(x)=$vec$(b)$ for a $1010$ matrix will take exactly $1,000$ operations.

Solving Avec$(x)=$vec$(b)$ for a $100100$ matrix will take about $1,000$ times as many operations as solving Avec$(x)=$vec$(b)$ for a $1010$ matrix.

Solving Avec$(x)=$vec$(b)$ for a $100100$ matrix will take about $10$ times as many operations as solving Avec$(x)=$vec$(b)$ for a $1010$ matrix.

Solving Avec$(x)=$vec$(b)$ for a $1010$ matrix will take exactly $100,000$ operations.

Solving Avec$(x)=$vec$(b)$ for a $100100$ matrix will take about $10,000$ times as many operations as solving Avec$(x)=$vec$(b)$ for a $1010$ matrix.

Solving Avec$(x)=$vec$(b)$ for a $1010$ matrix will take exactly $100$ operations.

Solving Avec$(x)=$vec$(b)$ for a $100100$ matrix will take about $100$ times as many operations as solving Avec$(x)=$vec$(b)$ for a $1010$ matrix.