- Home
- Standard 12
- Mathematics
3 and 4 .Determinants and Matrices
easy
જો $X = \left[ {\begin{array}{*{20}{c}}3&{ - 4}\\1&{ - 1}\end{array}} \right]$, તો ${X^n}$ = . . .
A
$\left[ {\begin{array}{*{20}{c}}{3n}&{ - 4n}\\n&{ - n}\end{array}} \right]$
B
$\left[ {\begin{array}{*{20}{c}}{2 + n}&{5 - n}\\n&{ - n}\end{array}} \right]$
C
$\left[ {\begin{array}{*{20}{c}}{{3^n}}&{{{( - 4)}^n}}\\{{1^n}}&{{{( - 1)}^n}}\end{array}} \right]$
D
એકપણ નહી.
Solution
(d) $X = \left[ {\begin{array}{*{20}{c}}3&{ – 4}\\1&{ – 1}\end{array}} \right] \Rightarrow {X^2} = \left[ {\begin{array}{*{20}{c}}5&{ – 8}\\2&{ – 3}\end{array}} \right]$.
Clearly for $n = 2$, the matrices in $(a), (b), (c)$ do not tally with $\left[ {\begin{array}{*{20}{c}}5&{ – 8}\\2&{ – 3}\end{array}} \right]$.
Standard 12
Mathematics
Similar Questions
hard