If the rank of the matrix $A=\left[\begin{array}{cccc}1 & 2 & 1 & -1 \\ -1 & 2 & 3 & 5 \\ 0 & 1 & k &…
If the rank of the matrix $A=\left[\begin{array}{cccc}1 & 2 & 1 & -1 \\ -1 & 2 & 3 & 5 \\ 0 & 1 & k & k\end{array}\right]$is 2 and $\mathrm{k}$ is a real number, then $\mathrm{k}$ is a root of the following quadratic equation
$x^2+3 x+2=0$
$x^2+x-2=0$
$x^2+x-6=0$
$x^2-x-6=0$
Solution
$A=\left[\begin{array}{cccc}1 & 2 & 1 & -1 \\ -1 & 2 & 3 & 5 \\ 0 & 1 & k & k\end{array}\right]$
Given, rank of $\mathrm{A}=2$
$\Rightarrow$ there must be 2 rows | columns which are linearly dependent.
using Echelon transformation,
$\mathrm{R}_2 \rightarrow \mathrm{R}_2+\mathrm{R}_1$
$A=\left[\begin{array}{cccc}1 & 2 & 1 & -1 \\ 0 & 4 & 4 & 4 \\ 0 & 1 & k & k\end{array}\right]$
Clearly at $k=1, \mathrm{R}_2$ and $\mathrm{R}_3$ will be identical.
Which will make rank $=2$
Taking $k=1$
out of given options $k=1$ only.
satisfies $x^2+x-2=0$