Q. If$A = \left| \begin{array}{r} 1\ \ 0\ \ 0 \\ x\ \ 1\ \ 0 \\ x\ \ x\ \ 1 \end{array} \right|$and$\ I = \left| \begin{array}{r} 1\ \ 0\ \ 0 \\ 0\ \ 1\ \ 0 \\ 0\ \ 0\ \ 1 \end{array} \right|,\$then $A^{3} - 4A^{2} + 3A + I\$is equal to

A) $3I$
B) $I$
C) $- I$
D) $- 2I$
Explanation

(b)
Given, $A = \left| \begin{array}{r}
1\ \ 0\ \ 0 \\
x\ \ 1\ \ 0 \\
x\ \ x\ \ 1
\end{array} \right| \Rightarrow A = 1$
$\therefore A^{3} - 4A^{2} + 3A + I = (1)^{3} - 4(1)^{2} + 3(1) + I = I$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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