જો A = left[ {begin{array}{*{20}{c}}
1&1 | vedclass

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Question

$A=\left[\begin{array}{ll}{1} & {1} \ {1} & {1}\end{array}ight]$

$A^{2}=\left[\begin{array}{ll}{1} & {1} \ {1} & {1}\end{array}

Vedclass · Accepted Answer

$2$

Vedclass · Answer

$A=\left[\begin{array}{ll}{1} & {1} \ {1} & {1}\end{array}ight]$

$A^{2}=\left[\begin{array}{ll}{1} & {1} \ {1} & {1}\end{array}ight]\left[\begin{array}{ll}{1} & {1} \ {1} & {1}\end{array}ight]$

$A^{2}=\left[\begin{array}{ll}{2} & {2} \ {2} & {2}\end{array}ight]=2 \mathrm{A} \Rightarrow \mathrm{A}^{3}=2 \mathrm{A}^{2}$

$=2^{2} \mathrm{A}$

Similarly $\mathrm{A}^{4}=2^{3} \mathrm{A}$ and so on 

So $\mathrm{A}^{\mathrm{n}}=2^{\mathrm{n}-1} \mathrm{A}$

$\Rightarrow A^{n}=2^{n-1}\left[\begin{array}{ll}{1} & {1} \ {1} & {1}\end{array}ight]$

${{m{A}}^{m{n}}} - {m{I}} = \left[ {\begin{array}{*{20}{c}}
{{2^{{m{n}} - 1}} - 1}&{{2^{{m{n}} - 1}}}\
{{2^{{m{n}} - 1}}}&{{2^{{m{n}} - 1}} - 1}
\end{array}} ight]$

$\left| {{{m{A}}^{m{n}}} - {m{I}}} ight| = {\left( {{2^{{m{n}} - 1}} - 1} ight)^2} - {\left( {{2^{{m{n}} - 1}}} ight)^2}$

$=1-2^{\mathrm{n}} $

$ \Rightarrow \lambda =2$

જો $A = \left[ {\begin{array}{*{20}{c}}
1&1\\
1&1
\end{array}} \right]$ અને $\det ({A^n} - I) = 1 - {\lambda ^n}\,,\,n \in N$ તો $\lambda $ મેળવો.

Similar Questions

સમીકરણ $\left| {\,\begin{array}{*{20}{c}}{1 + x}&1&1\\1&{1 + x}&1\\1&1&{1 + x}\end{array}\,} \right| = 0$ ના બીજ મેળવો.

$\left| {\,\begin{array}{*{20}{c}}1&a&{{a^2} - bc}\\1&b&{{b^2} - ac}\\1&c&{{c^2} - ab}\end{array}\,} \right| = $

સમીકરણ સંહતિને $2{x_1} - 2{x_2} + {x_3} = \lambda {x_1}\;,\;2{x_1} - 3{x_2} + 2{x_3} = \lambda {x_2}\;\;,\;\; - {x_1} + 2{x_2} = \lambda {x_3}$ યોગ્ય ઉકેલ હોય તેવા બધાજ $\lambda $ ઓનો ગણ . . . . . . છે.

જો $A=\left[\begin{array}{lll}1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9\end{array}\right]$ હોય, તો $|A|$ શોધો.