3 and 4 .Determinants and Matrices
easy

यदि $A = \left[ {\begin{array}{*{20}{c}}1&0\\1&1\end{array}} \right]$ और $I = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]$, तो प्रत्येक $n \ge 1$ के लिये मान रखता है

A

${A^n} = nA + (n - 1)I$

B

${A^n} = {2^{n - 1}}A + (n - 1)I$

C

${A^n} = nA - (n - 1)I$

D

${A^n} = {2^{n - 1}}A - (n - 1)I$

(AIEEE-2005)

Solution

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

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

$\therefore $ ${A^n} = \left[ {\begin{array}{*{20}{c}}1&0\\n&1\end{array}} \right]$ $;$ $nA = \left[ {\begin{array}{*{20}{c}}n&0\\n&n\end{array}} \right],(n – 1)I = \left[ {\begin{array}{*{20}{c}}{n – 1}&0\\0&{n – 1}\end{array}} \right]$

$nA – (n – 1)I = \left[ {\begin{array}{*{20}{c}}1&0\\n&1\end{array}} \right] = {A^n}$.

Standard 12
Mathematics

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