JEE Main 2022 — Determinants Question with Solution

Let the matrix be $A=\left[\begin{array}{lll}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right]$

Now $A\left[\begin{array}{l}0\\ 0\\ 1\end{array}\right]=\left[\begin{array}{l}{c}_1\\ c_2\\ c_3\end{array}\right]=\left[\begin{array}{l}1\\ 1\\ 2\end{array}\right]$

$\Rightarrow c_1=1,c_2=1,c_3=2$

$A\left[\begin{array}{l}1\\ 0\\ 1\end{array}\right]=\left[\begin{array}{l}{c}_1+a_1\\ c_2+a_2\\ c_3+a_3\end{array}\right]=\left[\begin{array}{c}-1\\ 0\\ 1\end{array}\right]$

$\Rightarrow a_1=-2,a_2=-1,a_3=-1$

$A\left[\begin{array}{l}1\\ 1\\ 0\end{array}\right]=\left[\begin{array}{l}{a}_1+b_1\\ a_2+b_2\\ a_3+b_3\end{array}\right]=\left[\begin{array}{l}1\\ 1\\ 0\end{array}\right]$

$\Rightarrow b_1=3,b_2=2,b_3=1$

So matrix $A=\left[\begin{array}{lll}-2& 3& 1\\ -1& 2& 1\\ -1& 1& 2\end{array}\right]$

$\Rightarrow A-2I=\left[\begin{array}{lll}-4& 3& 1\\ -1& 0& 1\\ -1& 1& 0\end{array}\right]$

i.e. $\left|A-2 I\right|=0$

Now $X=\left[\begin{array}{l}{x}_1\\ x_2\\ x_3\end{array}\right]$

So, $\left[\begin{array}{lll}-4& 3& 1\\ -1& 0& 1\\ -1& 1& 0\end{array}\right]\left[\begin{array}{l}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{l}4\\ 1\\ 1\end{array}\right]$

$-4x_1+3x_2+x_3=4 ...\left(1\right)$

$-x_1+x_3=1 ...\left(2\right)$

$-x_1+x_2=1 ...\left(3\right)$

Solving the above system of equations we get infinite solutions.