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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