If the rank of the matrix A=begin{bmatrix} 1 | vedclass

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(B) Given matrix $A = \begin{bmatrix} 1 & 2 & 1 & -1 \ -1 & 2 & 3 & 5 \ 0 & 1 & k & k \end{bmatrix}$.We perform row operations to reduce the matrix

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$x^2+x-2=0$

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(B) Given matrix $A = \begin{bmatrix} 1 & 2 & 1 & -1 \ -1 & 2 & 3 & 5 \ 0 & 1 & k & k \end{bmatrix}$.We perform row operations to reduce the matrix to row echelon form.Applying $R_2 ightarrow R_2 + R_1$:$A \sim \begin{bmatrix} 1 & 2 & 1 & -1 \ 0 & 4 & 4 & 4 \ 0 & 1 & k & k \end{bmatrix}$.Applying $R_2 ightarrow \frac{1}{4}R_2$:$A \sim \begin{bmatrix} 1 & 2 & 1 & -1 \ 0 & 1 & 1 & 1 \ 0 & 1 & k & k \end{bmatrix}$.Applying $R_3 ightarrow R_3 - R_2$:$A \sim \begin{bmatrix} 1 & 2 & 1 & -1 \ 0 & 1 & 1 & 1 \ 0 & 0 & k-1 & k-1 \end{bmatrix}$.For the rank of the matrix to be $2$,the third row must be a zero row. Thus,$k-1 = 0$,which implies $k = 1$.Now,we check which quadratic equation has $k=1$ as a root:For option $B$: $x^2+x-2 = (1)^2 + (1) - 2 = 1 + 1 - 2 = 0$.Thus,$k=1$ is a root of $x^2+x-2=0$.

If the rank of the matrix $A=\begin{bmatrix} 1 & 2 & 1 & -1 \\ -1 & 2 & 3 & 5 \\ 0 & 1 & k & k \end{bmatrix}$ is $2$ and $k$ is a real number,then $k$ is a root of the following quadratic equation:

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