(b) Given, $A = \left| \begin{array}{r} 1\ \ 0\ \ 0 \\ x\ \ 1\ \ 0 \\ x\ \ x\ \ 1 \end{array} \right| \Rightarrow A = 1$ $\therefore A^{3} - 4A^{2} + 3A + I = (1)^{3} - 4(1)^{2} + 3(1) + I = I$
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.