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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(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

Similar Questions

Let $f(x) = \left| \begin{array}{ccc} x^3 & \sin x & \cos x \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{array} \right|$,where $p$ is a constant. Then $\frac{d^3}{dx^3} \{f(x)\}$ at $x = 0$ is

Let $A = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & x \end{bmatrix}$ and $A^2 = A$. If $r$ is the rank of $A$,then $r + x =$

The rank of the matrix $\begin{bmatrix} 1 & -1 & 1 \\ 1 & 1 & -1 \\ -1 & 1 & 1 \end{bmatrix}$ is

Consider a homogeneous system of three linear equations in three unknowns represented by $AX=O$. If $X=\left[\begin{array}{c}l \\ m \\ 0\end{array}\right]$,where $l \neq 0, m \neq 0, l, m \in \mathbb{R}$,represents an infinite number of solutions of this system,then the rank of $A$ is:

If $y(x) = \left| \begin{array}{ccc} \sin x & \cos x & \sin x + \cos x + 1 \\ 27 & 28 & 27 \\ 1 & 1 & 1 \end{array} \right|$,$x \in R$,then $\frac{d^2 y}{d x^2} + y$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module