If A = begin{bmatrix} sqrt{2020} & sqrt{2021} | vedclass

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(A) Given the matrix $A = \begin{bmatrix} \sqrt{2020} & \sqrt{2021} & \sqrt{2022} & \sqrt{2023} \ \sqrt{4040} & \sqrt{4042} & \sqrt{4044} & \sqrt{404

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(A) Given the matrix $A = \begin{bmatrix} \sqrt{2020} & \sqrt{2021} & \sqrt{2022} & \sqrt{2023} \ \sqrt{4040} & \sqrt{4042} & \sqrt{4044} & \sqrt{4046} \ \sqrt{6060} & \sqrt{6063} & \sqrt{6066} & \sqrt{6069} \ \sqrt{8080} & \sqrt{8084} & \sqrt{8088} & \sqrt{8092} \end{bmatrix}$.We can rewrite the rows as multiples of the first row:$R_2 = \sqrt{2} R_1$,$R_3 = \sqrt{3} R_1$,and $R_4 = 2 R_1$.Substituting these into the matrix:$A = \begin{bmatrix} \sqrt{2020} & \sqrt{2021} & \sqrt{2022} & \sqrt{2023} \ \sqrt{2}\sqrt{2020} & \sqrt{2}\sqrt{2021} & \sqrt{2}\sqrt{2022} & \sqrt{2}\sqrt{2023} \ \sqrt{3}\sqrt{2020} & \sqrt{3}\sqrt{2021} & \sqrt{3}\sqrt{2022} & \sqrt{3}\sqrt{2023} \ 2\sqrt{2020} & 2\sqrt{2021} & 2\sqrt{2022} & 2\sqrt{2023} \end{bmatrix}$.Applying row operations $R_2 ightarrow R_2 - \sqrt{2}R_1$,$R_3 ightarrow R_3 - \sqrt{3}R_1$,and $R_4 ightarrow R_4 - 2R_1$,we get:$A \sim \begin{bmatrix} \sqrt{2020} & \sqrt{2021} & \sqrt{2022} & \sqrt{2023} \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \end{bmatrix}$.Since there is only one non-zero row in the row-echelon form,the rank of $A$ is $1$.

If $A = \begin{bmatrix} \sqrt{2020} & \sqrt{2021} & \sqrt{2022} & \sqrt{2023} \\ \sqrt{4040} & \sqrt{4042} & \sqrt{4044} & \sqrt{4046} \\ \sqrt{6060} & \sqrt{6063} & \sqrt{6066} & \sqrt{6069} \\ \sqrt{8080} & \sqrt{8084} & \sqrt{8088} & \sqrt{8092} \end{bmatrix}$,then the rank of $A$ is

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