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

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