Let A = begin{bmatrix} 1 & 0 & -1 & -3 0 & 1 | vedclass

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(B) The given matrix is $A = \begin{bmatrix} 1 & 0 & -1 & -3 \ 0 & 1 & 1 & k-1 \ 0 & 0 & k-1 & 1 \end{bmatrix}$.The rank of a matrix is the number o

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Does not exist

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(B) The given matrix is $A = \begin{bmatrix} 1 & 0 & -1 & -3 \ 0 & 1 & 1 & k-1 \ 0 & 0 & k-1 & 1 \end{bmatrix}$.The rank of a matrix is the number of non-zero rows in its row-echelon form.For the rank of $A$ to be $2$,the third row must become a zero row.This requires the elements of the third row to be zero,i.e.,$k-1 = 0$ and $1 = 0$.Since $1 = 0$ is a contradiction,it is impossible for the third row to be a zero row for any value of $k$.Therefore,the rank of $A$ will always be $3$ for any $k \in R$ where $k 
eq 1$.If $k = 1$,the matrix becomes $A = \begin{bmatrix} 1 & 0 & -1 & -3 \ 0 & 1 & 1 & 0 \ 0 & 0 & 0 & 1 \end{bmatrix}$,which also has rank $3$.Thus,there is no value of $k$ for which the rank of $A$ is $2$.

Let $A = \begin{bmatrix} 1 & 0 & -1 & -3 \\ 0 & 1 & 1 & k-1 \\ 0 & 0 & k-1 & 1 \end{bmatrix}$ and $k \in R$. Then,the value of $k$,if it exists,for which the rank of $A$ is $2$,is

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