Gujarati
3 and 4 .Determinants and Matrices
medium

Matrix $A$ is such that ${A^2} = 2A - I$, where $I$ is the identity matrix. Then for $n \ge 2,\,{A^n} = $`

A

$nA - (n - 1)I$

B

$nA - I$

C

${2^{n - 1}}A - (n - 1)I$

D

${2^{n - 1}}A - I$

Solution

(a) As we have ${A^2} = 2A – I \Rightarrow {A^2}.A = (2A – I)\,A$
$ \Rightarrow $${A^3} = 2{A^2} – IA = 2(2A – I) – A \Rightarrow {A^3} = 3A – 2I$

Similarly,${A^4} = 4A – 3I,\,{A^5} = 5A – 4I..$
Hence ${y^b} = {e^m},\,{x^c}{y^d} = {e^n}$.

Standard 12
Mathematics

Similar Questions

Start a Free Trial Now

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