3 and 4 .Determinants and Matrices
hard

आव्यूह $A = \left[ {\begin{array}{*{20}{c}}1&1&0\\1&2&1\\2&1&0\end{array}} \right]$ के लिये सत्य कथन होगा

A

${A^3} + 3{A^2} - I = O$

B

${A^3} - 3{A^2} - I = O$

C

${A^3} + 2{A^2} - I = O$

D

${A^3} - {A^2} + I = O$

Solution

(b) ${A^2} = AA = \left[ {\,\begin{array}{*{20}{c}}1&1&0\\1&2&1\\2&1&0\end{array}\,} \right]\,\left[ {\,\begin{array}{*{20}{c}}1&1&0\\1&2&1\\2&1&0\end{array}\,} \right]$= $\left[ {\,\begin{array}{*{20}{c}}2&3&1\\5&6&2\\3&4&1\end{array}\,} \right]$

==>${A^3} = {A^2}A = \left[ {\,\begin{array}{*{20}{c}}2&3&1\\5&6&2\\3&4&1\end{array}\,} \right]\,\,\left[ {\,\begin{array}{*{20}{c}}1&1&0\\1&2&1\\2&1&0\end{array}\,} \right]$=$\left[ {\,\begin{array}{*{20}{c}}7&9&3\\{15}&{19}&6\\9&{12}&4\end{array}\,} \right]$

अत:,${A^3} – 3{A^2} = \left[ {\,\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}\,} \right]\, = I$

==> ${A^3} – 3{A^2} – I = 0$.

Standard 12
Mathematics

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