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

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

જો $a, b, c > 0$ અને $\Delta = \left| \begin{gathered}
a + b\,\,b\,\,c \hfill \\
b\, + \,c\,\,c\,\,\,a \hfill \\
c + a\,\,a\,\,b \hfill \\
\end{gathered} \right| ,$ હોય તો આપલે પૈકી ક્યૂ વિધાન અસત્ય થાય.

$\left| {\begin{array}{*{20}{c}}
0&{x - y}&{x - z} \\
{y - x}&0&{y - z} \\
{z - x}&{z - y}&0
\end{array}} \right|$ મેળવો.

ધારો કે સમીકરણ સંહતિ $x+y+k z=2$ ; $2 x+3 y-z=1$ ; $3 x+4 y+2 z=k$ ને અસંખ્ય ઉકેલો છે. $( k +1) x +(2 k -1) y =7$ ; $(2 k +1) x +( k +5) y =10$ ને:

$k \in R$ ની કઈ કિમંત માટે આપેલ સમીકરણ સંહતિ $3 x-y+4 z=3$ ; $x+2 y-3 x=-2$ ; $6 x+5 y+k z=-3$ ને અનંત ઉકેલ ધરાવે છે.

જો $\left| {\,\begin{array}{*{20}{c}}a&b&c\\b&c&a\\c&a&b\end{array}\,} \right| = k(a + b + c)({a^2} + {b^2} + {c^2}$ $ - bc - ca - ab)$, તો $k =$

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.