If A = [a_{ij}],1 leq i, j leq n with n geq 2 | vedclass

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

Question

(C) Given that $n \geq 2$ and $a_{ij} = i + j$.Case-$1$: Let $n = 2$.$A = \begin{bmatrix} 2 & 3 \ 3 & 4 \end{bmatrix} \Rightarrow |A| = (2)(4) - (3)(

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) Given that $n \geq 2$ and $a_{ij} = i + j$.Case-$1$: Let $n = 2$.$A = \begin{bmatrix} 2 & 3 \ 3 & 4 \end{bmatrix} \Rightarrow |A| = (2)(4) - (3)(3) = 8 - 9 = -1 
eq 0$.Since the determinant is non-zero,the rank of $A$ is $2$.Case-$2$: Let $n = 3$.$A = \begin{bmatrix} 2 & 3 & 4 \ 3 & 4 & 5 \ 4 & 5 & 6 \end{bmatrix}$.Applying row operations $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_2$,we get:$A \sim \begin{bmatrix} 2 & 3 & 4 \ 1 & 1 & 1 \ 1 & 1 & 1 \end{bmatrix}$.Since $R_2$ and $R_3$ are identical,the rank is $2$.For any $n > 2$,the rows $R_i$ follow the pattern $R_i = (i+1, i+2, \dots, i+n)$.Note that $R_3 - R_2 = R_2 - R_1 = (1, 1, \dots, 1)$.Thus,$R_3 - 2R_2 + R_1 = 0$,which means the rows are linearly dependent for $n \geq 3$.Therefore,the rank of $A$ is $2$ for all $n \geq 2$.

If $A = [a_{ij}]$,$1 \leq i, j \leq n$ with $n \geq 2$ and $a_{ij} = i + j$ is a matrix,then the rank of $A$ is

Similar Questions

The rank of the matrix $\left[\begin{array}{cccc}3 & 2 & 1 & -4 \\ 2 & 3 & 0 & -1 \\ 1 & -6 & 3 & -8\end{array}\right]$ is

In a matrix $A$,if all the sub-matrices of order $k$ are singular and there is at least one non-singular sub-matrix of order $r$ $(r < k)$,then the rank $(\rho)$ of the matrix $A$:

The rank of the matrix $\left[\begin{array}{ccc}1 & 0 & 2 \\ 0 & 1 & -2 \\ 1 & -1 & 4 \\ 2 & 2 & 8\end{array}\right]$ is

If $f(x) = \left| \begin{array}{ccc} x^3 - x & a + x & b + x \\ x - a & x^2 - x & c + x \\ x - b & x - c & 0 \end{array} \right|$,then:

If $f(x) = \left| \begin{array}{ccc} 2 \cos x & 1 & 0 \\ x - \frac{\pi}{2} & 2 \cos x & 1 \\ 0 & 1 & 2 \cos x \end{array} \right|$, then $f^{\prime}(\pi)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module