3 and 4 .Determinants and Matrices
normal

$A = \left[ {\begin{array}{*{20}{c}}1&0\\1&1\end{array}} \right]$ અને  $I = \left[ {\begin{array}{*{20}{c}}
1&0\\0&1\end{array}} \right]$ , તો આપેલ પૈકી ક્યૂ વિધાન $n \geq 2, n \in N$ માટે સત્ય છે . 

A

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

B

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

C

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

D

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

Solution

$A^2 = \left[ {\begin{array}{*{20}{c}}   1&0 \\    1&1  \end{array}} \right]\left[ {\begin{array}{*{20}{c}}   1&0 \\    1&1  \end{array}} \right] = \left[ {\begin{array}{*{20}{c}}   1&0 \\    2&1  \end{array}} \right]$ it satisfies only option $(D)$

Standard 12
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.