The rank of the matrix begin{bmatrix} 4 & 2 & | vedclass

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(A) Let the matrix be $A = \begin{bmatrix} 4 & 2 & 1-x \ 5 & k & 1 \ 6 & 3 & 1+x \end{bmatrix}$.For the rank of the matrix to be $1$,all rows must b

Vedclass · Accepted Answer

$k = \frac{5}{2}, x = \frac{1}{5}$

Vedclass · Answer

(A) Let the matrix be $A = \begin{bmatrix} 4 & 2 & 1-x \ 5 & k & 1 \ 6 & 3 & 1+x \end{bmatrix}$.For the rank of the matrix to be $1$,all rows must be proportional to each other.Comparing $R_1$ and $R_3$:$R_3 = c R_1 \Rightarrow 6 = 4c \Rightarrow c = \frac{6}{4} = \frac{3}{2}$.Checking the second element: $3 = 2 	imes \frac{3}{2} = 3$ (This holds).Checking the third element: $1+x = (1-x) 	imes \frac{3}{2}$.$2(1+x) = 3(1-x) \Rightarrow 2 + 2x = 3 - 3x \Rightarrow 5x = 1 \Rightarrow x = \frac{1}{5}$.Now,comparing $R_1$ and $R_2$:$R_2 = d R_1 \Rightarrow 5 = 4d \Rightarrow d = \frac{5}{4}$.Checking the second element: $k = 2 	imes \frac{5}{4} = \frac{5}{2}$.Checking the third element: $1 = (1-x) 	imes \frac{5}{4}$.Substituting $x = \frac{1}{5}$: $1 = (1 - \frac{1}{5}) 	imes \frac{5}{4} = \frac{4}{5} 	imes \frac{5}{4} = 1$ (This holds).Thus,the rank is $1$ when $k = \frac{5}{2}$ and $x = \frac{1}{5}$.

The rank of the matrix $\begin{bmatrix} 4 & 2 & 1-x \\ 5 & k & 1 \\ 6 & 3 & 1+x \end{bmatrix}$ is $1$,then

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