Question

Write A in terms of a and I we
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Cayley - Hamilton theorem if A = (1 0 0 5 ) 2×2 matrix ​

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{A=\left(\begin{array}{cc}1&0\\0&5\end{array}\right)}$

$\underline{\textbf{To find:}}$

$\mathsf{A^{-1}\;in\;terms\;of\;A\;and\;I}$

$\underline{\textbf{Solution:}}$

$\underline{\textbf{Cayley-Hamilton theorem:}}$

$\boxed{\textbf{Every square matrix satisfies its characteristic equation}}$

$\textsf{Charecteristic polynomial of A is}$

$\mathsf{|A-m\,I|=0}$

$\mathsf{\left|\left(\begin{array}{cc}1&0\\0&5\end{array}\right)-m\left(\begin{array}{cc}1&0\\0&1\end{array}\right)\right|=0}$

$\mathsf{\left|\left(\begin{array}{cc}1&0\\0&5\end{array}\right)-\left(\begin{array}{cc}m&0\\0&m\end{array}\right)\right|=0}$

$\mathsf{\left|\begin{array}{cc}1-m&0\\0&5-m\end{array}\right|=0}$

$\mathsf{(1-m)(5-m)=0}$

$\mathsf{(m-1)(m-5)=0}$

$\boxed{\mathsf{m^2-6\,m+5=0}}$

$\textsf{By Cayley-Hamilton theroem, we have}$

$\mathsf{A^2-6\,A+5\,I=0}$

$\mathsf{Multiply\;bothsides\;by\;A^{-1}}$

$\mathsf{A{-1}(A^2-6\,A+5\,I)=A^{-1}0}$

$\mathsf{A^{-1}A^2-6\,A^{-1}A+5\,A^{-1}I}=0}$

$\mathsf{A-6\,I+5\,A^{-1}}=0}$

$\mathsf{5\,A^{-1}}=6\,I-A}$

$\boxed{\mathsf{A^{-1}}=\dfrac{1}{5}(6\,I-A)}}$

$\underline{\textbf{Find more:}}$

Verify Cayley-Hamilton theorem for the matrix

1 0 2

A=. 0 2 1

2 0 3

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