Question

If A=\(\begin{bmatrix}3&1\\-1&2\end{bmatrix}\),show that A²-5A+7I=0.Hence find A^-1.

Answer 1

Given A=\(\begin{bmatrix}3&1\\-1&2\end{bmatrix}\) 

A²=A.A =\(\begin{bmatrix}3&1\\-1&2\end{bmatrix}\)\(\begin{bmatrix}3&1\\-1&2\end{bmatrix}\) =\(\begin{bmatrix}9-1&3+2\\-3-2&-1+4\end{bmatrix}\)=\(\begin{bmatrix}8&5\\-5&3\end{bmatrix}\) 

therefore LHS=A²-5A+7I 

\(\Rightarrow\begin{bmatrix}8&5\\-5&3\end{bmatrix}\)-5\(\begin{bmatrix}3&1\\-1&2\end{bmatrix}\)+7\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\) 

=\(\begin{bmatrix}8&5\\-5&3\end{bmatrix}\)-\(\begin{bmatrix}15&5\\-5&10\end{bmatrix}\)+\(\begin{bmatrix}7&0\\0&7\end{bmatrix}\) 

=\(\begin{bmatrix}-7&0\\0&-7\end{bmatrix}\)+\(\begin{bmatrix}7&0\\0&7\end{bmatrix}\) =0

=RHS 

So A²-5A+7I=0
therefore A. A-5A=-7I
\(\Rightarrow\) A. A(A^-1)-5A(A^-1)=-7I(A^-1)     [post multiplying by A^-1 as IAI≠0]
\(\Rightarrow\) A(AA^-1)-5I=-7I(A^-1) 
\(\Rightarrow\) AI-5I=-7A^-1
\(\Rightarrow\) A^-1=-\(\frac{1}{7}\)(A-5I)
\(\Rightarrow\) A^-1=\(\frac{1}{7}\)(5I-A)

=\(\frac{41}{7}\) \(\bigg(\)\(\begin{bmatrix}5&0\\0&5\end{bmatrix}\)-\(\begin{bmatrix}3&1\\-1&2\end{bmatrix}\)\(\bigg)\)=\(\frac{1}{7}\begin{bmatrix}2&-1\\1&3\end{bmatrix}\)

\(\therefore\) A^-1=\(\frac{1}{7}\begin{bmatrix}2&-1\\1&3\end{bmatrix}\)